Rs Aggarwal 2021 2022 for Class 10 Maths Chapter 4 - Quadratic Equations

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Rs Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 4 Quadratic Equations are provided here with simple step-by-step explanations. These solutions for Quadratic Equations are extremely popular among Class 10 students for Maths Quadratic Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2021 2022 Book of Class 10 Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2021 2022 Solutions. All Rs Aggarwal 2021 2022 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 182:

Answer:

$i) (x^{2} - x + 3) is a quadratic polynomial . ∴ x^{2} - x + 3 = 0 is a quadratic equation . ii) C learly, (2 x^{2} + \frac{5}{2} x - \sqrt{3}) is a quadratic polynomial . ∴ 2 x^{2} + \frac{5}{2} x - \sqrt{3} = 0 is a quadratic equation . iii) C learly, (\sqrt{2} x^{2} + 7 x + 5 \sqrt{2}) is a quadratic polynomial . ∴ \sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0 is a quadratic equation . iv) C learly, (\frac{1}{3} x^{2} + \frac{1}{5} x - 2) is a quadratic polynomial . ∴ \frac{1}{3} x^{2} + \frac{1}{5} x - 2 = 0 is a quadratic equation . v) (x^{2} - 3 x - \sqrt{x} + 4) contains a term with \sqrt{x}, i .e, x^{\frac{1}{2}}, where \frac{1}{2} is not a integer . Therefore, it is not a quadratic polynomial . ∴ x^{2} - 3 x - \sqrt{x} + 4 = 0 is not a quadratic equation . vi) x - \frac{6}{x} = 3 \Rightarrow x^{2} - 6 = 3 x \Rightarrow x^{2} - 3 x - 6 = 0 \begin{array}{l} (x^{2} - 3 x - 6) is a quadratic polynomial; therefore, the given \\ equation is quadratic . \end{array} vii) x + \frac{2}{x} = x^{2} \Rightarrow x^{2} + 2 = x^{3} \Rightarrow x^{3} - x^{2} - 2 = 0 (x^{3} - x^{2} - 2) is not a quadratic polynomial . ∴ x^{3} - x^{2} - 2 = 0 is not a quadratic equation . viii) x^{2} - \frac{1}{x^{2}} = 5 \Rightarrow x^{4} - 1 = 5 x^{2} \Rightarrow x^{4} - 5 x^{2} - 1 = 0 (x^{4} - 5 x^{2} - 1) is a polynomial with degree 4 . ∴ x^{4} - 5 x^{2} - 1 = 0 is not a quadratic equation .$
(ix) ${(x + 2)}^{3} = x^{3} - 8$
$\Rightarrow x^{3} + 6 x^{2} + 12 x + 8 = x^{3} - 8 \Rightarrow 6 x^{2} + 12 x + 16 = 0$
This is of the form ax²+ bx + c = 0.
Hence, the given equation is a quadratic equation.
(x) $(2 x + 3) (3 x + 2) = 6 (x - 1) (x - 2)$
$\Rightarrow 6 x^{2} + 4 x + 9 x + 6 = 6 (x^{2} - 3 x + 2) \Rightarrow 6 x^{2} + 13 x + 6 = 6 x^{2} - 18 x + 12 \Rightarrow 31 x - 6 = 0$
This is not of the form ax²+ bx + c = 0.
Hence, the given equation is not a quadratic equation.
(xi) ${(x + \frac{1}{x})}^{2} = 2 (x + \frac{1}{x}) + 3$
$\Rightarrow {(\frac{x^{2} + 1}{x})}^{2} = 2 (\frac{x^{2} + 1}{x}) + 3 \Rightarrow {(x^{2} + 1)}^{2} = 2 x (x^{2} + 1) + 3 x^{2} \Rightarrow x^{4} + 2 x^{2} + 1 = 2 x^{3} + 2 x + 3 x^{2} \Rightarrow x^{4} - 2 x^{3} - x^{2} - 2 x + 1 = 0$
This is not of the form ax²+ bx + c = 0.
Hence, the given equation is not a quadratic equation.

Page No 182:

Answer:

${The given equation is (3x}^{2} + 2 x - 1 = 0) . (i) x = (- 1) L .H .S . = x^{2} + 2 x - 1 = 3 \times {(- 1)}^{2} + 2 \times (- 1) - 1 = 3 - 2 - 1 = 0 = R .H .S . Thus, (- 1) is a root of {(3x}^{2} + 2 x - 1 = 0) . (ii) On substituting x = \frac{1}{3} in the given equation, we get : L .H .S . = 3 x^{2} + 2 x - 1 = 3 \times {(\frac{1}{3})}^{2} + 2 \times \frac{1}{3} - 1 = 3^{1} \times \frac{1}{9_{3}} + \frac{2}{3} - 1 = \frac{1 + 2 - 3}{3} = \frac{0}{3} = 0 = R .H .S . Thus, (\frac{1}{3}) is a root of {(3x}^{2} + 2 x - 1 = 0) . (iii) On substituting x = (- \frac{1}{2}) in the given equation, we get : L .H .S . = 3 x^{2} + 2 x - 1 = 3 \times {(- \frac{1}{2})}^{2} + 2^{1} \times (- \frac{1}{2_{1}}) - 1 = 3 \times \frac{1}{4} - 1 - 1 = \frac{3}{4} - 2 = \frac{3 - 8}{4} = \frac{- 5}{4} \neq 0 Thus, L .H .S . = R .H .S . Hence, (- \frac{1}{2}) is a not solution of (3 x^{2} + 2 x - 1 = 0) .$

Page No 182:

Answer:

(i)
$It is given that (x = 1) is a root of (x^{2} + k x + 3 = 0) . Therefore, (x = 1) must satisfy the equation . \Rightarrow {(1)}^{2} + k \times 1 + 3 = 0 \Rightarrow k + 4 = 0 \Rightarrow k = - 4 Hence, the required value of k is - 4 .$
So, the equation becomes $x^{2} - 4 x + 3 = 0$
On factorising we get;
$x^{2} - x - 3 x + 3 = 0 x (x - 1) - 3 (x - 1) = 0 (x - 1) (x - 3) = 0 \Rightarrow x - 1 = 0 or x - 3 = 0 \Rightarrow x = 1 or x = 3$
Hence, the other root is 3.

(ii)
$It is given that \frac{3}{4} is a root of a x^{2} + b x - 6 = 0; therefore, we have: a \times {(\frac{3}{4})}^{2} + b \times \frac{3}{4} - 6 = 0 \Rightarrow \frac{9 a}{16} + \frac{3 b}{4} = 6 \Rightarrow \frac{9 a + 12 b}{16} = 6 \Rightarrow 9 a + 12 b - 96 = 0 \Rightarrow 3 a + 4 b = 32 ... (i) Again, (- 2) is a root of a x^{2} + b x - 6 = 0; therefore, we have: a \times {(- 2)}^{2} + b \times (- 2) - 6 = 0 \Rightarrow 4 a - 2 b = 6 \Rightarrow 2 a - b = 3 ... (ii) On multiplying (ii) by 4 and adding the result with (i), w e get: \Rightarrow 3 a + 4 b + 8 a - 4 b = 32 + 12 \Rightarrow 11 a = 44 \Rightarrow a = 4 Putting the value of a in (ii), we get : 2 \times 4 - b = 3 \Rightarrow 8 - b = 3 \Rightarrow b = 5 Hence, the required values of a and b are 4 and 5, respectively.$

Page No 182:

Answer:

LHS;
Consider the quadratic equation;
$a d^{2} (\frac{a x}{b} + \frac{2 c}{d}) x + b c^{2} = 0$
Put $x = - \frac{b c}{a d}$ in the given equation.
$a d^{2} [\frac{a (- \frac{b c}{a d})}{b} + \frac{2 c}{d}] (- \frac{b c}{a d}) + b c^{2} = \frac{a d^{2}}{b d} [a d (- \frac{b c}{a d}) + 2 b c] (- \frac{b c}{a d}) + b c^{2} = \frac{a d^{2}}{b d} [- b c + 2 b c] (- \frac{b c}{a d}) + b c^{2} = \frac{a d^{2}}{b d} (b c) (- \frac{b c}{a d}) + b c^{2} = - \frac{a b^{2} c^{2} d^{2}}{a b d^{2}} + b c^{2} = - b c^{2} + b c^{2} = 0 = RHS$
Hence, $x = - \frac{b c}{a d}$ is a solution to the given quadratic equation.

Page No 182:

Answer:

(2x − 3)(3x + 1) = 0

⇒ 2x − 3 = 0 or 3x + 1 = 0

⇒ 2x = 3 or 3x = −1

⇒ x = $\frac{3}{2}$ or x = $- \frac{1}{3}$

Hence, the roots of the given equation are $\frac{3}{2}$ and $- \frac{1}{3}$ .

Page No 182:

Answer:

4x² + 5x = 0

⇒ x(4x + 5) = 0

⇒ x = 0 or 4x + 5 = 0

⇒ x = 0 or x = $- \frac{5}{4}$

Hence, the roots of the given equation are 0 and $- \frac{5}{4}$ .

Page No 182:

Answer:

$Given: 3 x^{2} - 243 = 0 \Rightarrow 3 (x^{2} - 81) = 0 \Rightarrow {(x)}^{2} - {(9)}^{2} = 0 \Rightarrow (x + 9) (x - 9) = 0 \Rightarrow x + 9 = 0 or x - 9 = 0 \Rightarrow x = - 9 or x = 9 Hence, - 9 and 9 are the roots of the equation 3 x^{2} - 243 = 0.$

Page No 182:

Answer:

We write, $x = 4 x - 3 x$ as $2 x^{2} \times (- 6) = - 12 x^{2} = 4 x \times (- 3 x)$

$∴ 2 x^{2} + x - 6 = 0 \Rightarrow 2 x^{2} + 4 x - 3 x - 6 = 0 \Rightarrow 2 x (x + 2) - 3 (x + 2) = 0 \Rightarrow (x + 2) (2 x - 3) = 0$

$\Rightarrow x + 2 = 0 or 2 x - 3 = 0 \Rightarrow x = - 2 or x = \frac{3}{2}$
Hence, the roots of the given equation are $- 2$ and $\frac{3}{2}$ .

Page No 182:

Answer:

We write, $6 x = x + 5 x$ as $x^{2} \times 5 = 5 x^{2} = x \times 5 x$

$∴ x^{2} + 6 x + 5 = 0 \Rightarrow x^{2} + x + 5 x + 5 = 0 \Rightarrow x (x + 1) + 5 (x + 1) = 0 \Rightarrow (x + 1) (x + 5) = 0$

$\Rightarrow x + 1 = 0 or x + 5 = 0 \Rightarrow x = - 1 or x = - 5$

Hence, the roots of the given equation are −1 and −5.

Page No 182:

Answer:

We write, $- 3 x = 3 x - 6 x$ as $9 x^{2} \times (- 2) = - 18 x^{2} = 3 x \times (- 6 x)$

$∴ 9 x^{2} - 3 x - 2 = 0 \Rightarrow 9 x^{2} + 3 x - 6 x - 2 = 0 \Rightarrow 3 x (3 x + 1) - 2 (3 x + 1) = 0 \Rightarrow (3 x + 1) (3 x - 2) = 0$
$\Rightarrow 3 x + 1 = 0 or 3 x - 2 = 0 \Rightarrow x = - \frac{1}{3} or x = \frac{2}{3}$
Hence, the roots of the given equation are $- \frac{1}{3}$ and $\frac{2}{3}$ .

Page No 182:

Answer:

$Given : x^{2} + 12 x + 35 = 0 \Rightarrow x^{2} + 7 x + 5 x + 35 = 0 \Rightarrow x (x + 7) + 5 (x + 7) = 0 \Rightarrow (x + 5) (x + 7) = 0 \Rightarrow x + 5 = 0 or x + 7 = 0 \Rightarrow x = - 5 or x = - 7 Hence, - 5 and - 7 are the roots of the equation x^{2} + 12 x + 35 = 0.$

Page No 182:

Answer:

$Given : x^{2} = 18 x - 77 \Rightarrow x^{2} - 18 x + 77 = 0 \Rightarrow x^{2} - (11 x + 7 x) + 77 = 0 \Rightarrow x^{2} - 11 x - 7 x + 77 = 0 \Rightarrow x (x - 11) - 7 (x - 11) = 0 \Rightarrow (x - 7) (x - 11) = 0 \Rightarrow x - 7 = 0 or x - 11 = 0 \Rightarrow x = 7 or x = 11 Hence, 7 and 11 are the roots of the equation x^{2} = 18 x - 77.$

Page No 183:

Answer:

$Given : 6 x^{2} + 11 x + 3 = 0 \Rightarrow 6 x^{2} + 9 x + 2 x + 3 = 0 \Rightarrow 3 x (2 x + 3) + 1 (2 x + 3) = 0 \Rightarrow (3 x + 1) (2 x + 3) = 0 \Rightarrow 3 x + 1 = 0 or 2 x + 3 = 0 \Rightarrow x = \frac{- 1}{3} or x = \frac{- 3}{2} Hence, \frac{- 1}{3} and \frac{- 3}{2} are the roots of the equation 6 x^{2} + 11 x + 3 = 0.$

Page No 183:

Answer:

$Given: 6 x^{2} + x - 12 = 0 \Rightarrow 6 x^{2} + 9 x - 8 x - 12 = 0 \Rightarrow 3 x (2 x + 3) - 4 (2 x + 3) = 0 \Rightarrow (3 x - 4) (2 x + 3) = 0 \Rightarrow 3 x - 4 = 0 or 2 x + 3 = 0 \Rightarrow x = \frac{4}{3} or x = \frac{- 3}{2} Hence, \frac{4}{3} and \frac{- 3}{2} are the roots of the equation 6 x^{2} + x - 12 = 0.$

Page No 183:

Answer:

We write, $- 2 x = - 3 x + x$ as $3 x^{2} \times (- 1) = - 3 x^{2} = (- 3 x) \times x$

$∴ 3 x^{2} - 2 x - 1 = 0 \Rightarrow 3 x^{2} - 3 x + x - 1 = 0 \Rightarrow 3 x (x - 1) + 1 (x - 1) = 0 \Rightarrow (x - 1) (3 x + 1) = 0$
$\Rightarrow x - 1 = 0 or 3 x + 1 = 0 \Rightarrow x = 1 or x = - \frac{1}{3}$
Hence, the roots of the given equation are $1$ and $- \frac{1}{3}$ .

Page No 183:

Answer:

$Given: 4 x^{2} - 9 x = 100 \Rightarrow 4 x^{2} - 9 x - 100 = 0 \Rightarrow 4 x^{2} - (25 x - 16 x) - 100 = 0 \Rightarrow 4 x^{2} - 25 x + 16 x - 100 = 0 \Rightarrow x (4 x - 25) + 4 (4 x - 25) = 0 \Rightarrow (4 x - 25) (x + 4) = 0 \Rightarrow 4 x - 25 = 0 or x + 4 = 0 \Rightarrow x = \frac{25}{4} or x = - 4 Hence, the roots of the equation are \frac{25}{4} and - 4 .$

Page No 183:

Answer:

$Given: 15 x^{2} - 28 = x \Rightarrow 15 x^{2} - x - 28 = 0 \Rightarrow 15 x^{2} - (21 x - 20 x) - 28 = 0 \Rightarrow 15 x^{2} - 21 x + 20 x - 28 = 0 \Rightarrow 3 x (5 x - 7) + 4 (5 x - 7) = 0 \Rightarrow (3 x + 4) (5 x - 7) = 0 \Rightarrow 3 x + 4 = 0 or 5 x - 7 = 0 \Rightarrow x = \frac{- 4}{3} or x = \frac{7}{5} H ence, the roots of the equation are \frac{- 4}{3} and \frac{7}{5} .$

Page No 183:

Answer:

$Given: 4 - 11 x = 3 x^{2} \Rightarrow 3 x^{2} + 11 x - 4 = 0 \Rightarrow 3 x^{2} + 12 x - x - 4 = 0 \Rightarrow 3 x (x + 4) - 1 (x + 4) = 0 \Rightarrow (x + 4) (3 x - 1) = 0 \Rightarrow x + 4 = 0 or 3 x - 1 = 0 \Rightarrow x = - 4 or x = \frac{1}{3} H ence, the roots of the equation are - 4 and \frac{1}{3} .$

Page No 183:

Answer:

$Given: 48 x^{2} - 13 x - 1 = 0 \Rightarrow 48 x^{2} - (16 x - 3 x) - 1 = 0 \Rightarrow 48 x^{2} - 16 x + 3 x - 1 = 0 \Rightarrow 16 x (3 x - 1) + 1 (3 x - 1) = 0 \Rightarrow (16 x + 1) (3 x - 1) = 0 \Rightarrow 16 x + 1 = 0 or 3 x - 1 = 0 \Rightarrow x = \frac{- 1}{16} or x = \frac{1}{3} Hence, the roots of the equation are \frac{- 1}{16} and \frac{1}{3} .$

Page No 183:

Answer:

We write, $2 \sqrt{2} x = 3 \sqrt{2} x - \sqrt{2} x$ as $x^{2} \times (- 6) = - 6 x^{2} = 3 \sqrt{2} x \times (- \sqrt{2} x)$

$∴ x^{2} + 2 \sqrt{2} x - 6 = 0 \Rightarrow x^{2} + 3 \sqrt{2} x - \sqrt{2} x - 6 = 0 \Rightarrow x (x + 3 \sqrt{2}) - \sqrt{2} (x + 3 \sqrt{2}) = 0 \Rightarrow (x + 3 \sqrt{2}) (x - \sqrt{2}) = 0$
$\Rightarrow x + 3 \sqrt{2} = 0 or x - \sqrt{2} = 0 \Rightarrow x = - 3 \sqrt{2} or x = \sqrt{2}$
Hence, the roots of the given equation are $- 3 \sqrt{2}$ and $\sqrt{2}$ .

Page No 183:

Answer:

Consider $\sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0$
Factorising by splitting the middle term;
$\sqrt{3} x^{2} + 12 x - 2 x - 8 \sqrt{3} = 0 \Rightarrow \sqrt{3} x (x + 4 \sqrt{3}) - 2 (x + 4 \sqrt{3}) = 0 \Rightarrow (\sqrt{3} x - 2) (x + 4 \sqrt{3}) = 0 \Rightarrow \sqrt{3} x - 2 = 0 or x + 4 \sqrt{3} = 0 \Rightarrow x = \frac{2}{\sqrt{3}} or x = - 4 \sqrt{3}$
Hence, the roots of the given equation are $\frac{2}{\sqrt{3}}$ and $- 4 \sqrt{3}$ .

Page No 183:

Answer:

$Given : \sqrt{3} x^{2} + 11 x + 6 \sqrt{3} = 0 \Rightarrow \sqrt{3} x^{2} + 9 x + 2 x + 6 \sqrt{3} = 0 \Rightarrow \sqrt{3} x (x + 3 \sqrt{3}) + 2 (x + 3 \sqrt{3}) = 0 \Rightarrow (x + 3 \sqrt{3}) (\sqrt{3} x + 2) = 0 \Rightarrow x + 3 \sqrt{3} = 0 or \sqrt{3} x + 2 = 0 \Rightarrow x = - 3 \sqrt{3} or x = \frac{- 2}{\sqrt{3}} = \frac{- 2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{- 2 \sqrt{3}}{3} Hence, the roots of the equation are - 3 \sqrt{3} and \frac{- 2 \sqrt{3}}{3} .$

Page No 183:

Answer:

$Given: 3 \sqrt{7} x^{2} + 4 x - \sqrt{7} = 0 \Rightarrow 3 \sqrt{7} x^{2} + 7 x - 3 x - \sqrt{7} = 0 \Rightarrow \sqrt{7} x (3 x + \sqrt{7}) - 1 (3 x + \sqrt{7}) = 0 \Rightarrow (3 x + \sqrt{7}) (\sqrt{7} x - 1) = 0 \Rightarrow 3 x + \sqrt{7} = 0 or \sqrt{7} x - 1 = 0 \Rightarrow x = \frac{- \sqrt{7}}{3} or x = \frac{1}{\sqrt{7}} = \frac{1 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \frac{\sqrt{7}}{7} H ence, the roots of the equation are \frac{- \sqrt{7}}{3} and \frac{\sqrt{7}}{7} .$

Page No 183:

Answer:

We write, $- 6 x = 7 x - 13 x$ as $\sqrt{7} x^{2} \times (- 13 \sqrt{7}) = - 91 x^{2} = 7 x \times (- 13 x)$

$∴ \sqrt{7} x^{2} - 6 x - 13 \sqrt{7} = 0 \Rightarrow \sqrt{7} x^{2} + 7 x - 13 x - 13 \sqrt{7} = 0 \Rightarrow \sqrt{7} x (x + \sqrt{7}) - 13 (x + \sqrt{7}) = 0 \Rightarrow (x + \sqrt{7}) (\sqrt{7} x - 13) = 0$
$\Rightarrow x + \sqrt{7} = 0 or \sqrt{7} x - 13 = 0 \Rightarrow x = - \sqrt{7} or x = \frac{13}{\sqrt{7}} = \frac{13 \sqrt{7}}{7}$
Hence, the roots of the given equation are $- \sqrt{7}$ and $\frac{13 \sqrt{7}}{7}$ .

Page No 183:

Answer:

$Given: 4 \sqrt{6} x^{2} - 13 x - 2 \sqrt{6} = 0 \Rightarrow 4 \sqrt{6} x^{2} - 16 x + 3 x - 2 \sqrt{6} = 0 \Rightarrow 4 \sqrt{2} x (\sqrt{3} x - 2 \sqrt{2}) + \sqrt{3} (\sqrt{3} x - 2 \sqrt{2}) = 0 \Rightarrow (4 \sqrt{2} x + \sqrt{3}) (\sqrt{3} x - 2 \sqrt{2}) = 0 \Rightarrow 4 \sqrt{2} x + \sqrt{3} = 0 or \sqrt{3} x - 2 \sqrt{2} = 0 \Rightarrow x = \frac{- \sqrt{3}}{4 \sqrt{2}} = \frac{- \sqrt{3} \times \sqrt{2}}{4 \sqrt{2} \times \sqrt{2}} = \frac{- \sqrt{6}}{8} or x = \frac{2 \sqrt{2}}{\sqrt{3}} = \frac{2 \sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{2 \sqrt{6}}{3} H ence, the roots of the equation are \frac{- \sqrt{6}}{8} and \frac{2 \sqrt{6}}{3} .$

Page No 183:

Answer:

We write, $- 2 \sqrt{6} x = - \sqrt{6} x - \sqrt{6} x$ as $3 x^{2} \times 2 = 6 x^{2} = (- \sqrt{6} x) \times (- \sqrt{6} x)$

$∴ 3 x^{2} - 2 \sqrt{6} x + 2 = 0 \Rightarrow 3 x^{2} - \sqrt{6} x - \sqrt{6} x + 2 = 0 \Rightarrow \sqrt{3} x (\sqrt{3} x - \sqrt{2}) - \sqrt{2} (\sqrt{3} x - \sqrt{2}) = 0 \Rightarrow (\sqrt{3} x - \sqrt{2}) (\sqrt{3} x - \sqrt{2}) = 0$
$\Rightarrow {(\sqrt{3} x - \sqrt{2})}^{2} = 0 \Rightarrow \sqrt{3} x - \sqrt{2} = 0 \Rightarrow x = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3}$
Hence, $\frac{\sqrt{6}}{3}$ is the repreated root of the given equation.

Page No 183:

Answer:

We write, $- 2 \sqrt{2} x = - 3 \sqrt{2} x + \sqrt{2} x$ as $\sqrt{3} x^{2} \times (- 2 \sqrt{3}) = - 6 x^{2} = (- 3 \sqrt{2} x) \times (\sqrt{2} x)$

$∴ \sqrt{3} x^{2} - 2 \sqrt{2} x - 2 \sqrt{3} = 0 \Rightarrow \sqrt{3} x^{2} - 3 \sqrt{2} x + \sqrt{2} x - 2 \sqrt{3} = 0 \Rightarrow \sqrt{3} x (x - \sqrt{6}) + \sqrt{2} (x - \sqrt{6}) = 0 \Rightarrow (x - \sqrt{6}) (\sqrt{3} x + \sqrt{2}) = 0$
$\Rightarrow x - \sqrt{6} = 0 or \sqrt{3} x + \sqrt{2} = 0 \Rightarrow x = \sqrt{6} or x = - \frac{\sqrt{2}}{\sqrt{3}} = - \frac{\sqrt{6}}{3}$
Hence, the roots of the given equation are $\sqrt{6}$ and $- \frac{\sqrt{6}}{3}$ .

Page No 183:

Answer:

We write, $- 3 \sqrt{5} x = - 2 \sqrt{5} x - \sqrt{5} x$ as $x^{2} \times 10 = 10 x^{2} = (- 2 \sqrt{5} x) \times (- \sqrt{5} x)$

$∴ x^{2} - 3 \sqrt{5} x + 10 = 0 \Rightarrow x^{2} - 2 \sqrt{5} x - \sqrt{5} x + 10 = 0 \Rightarrow x (x - 2 \sqrt{5}) - \sqrt{5} (x - 2 \sqrt{5}) = 0 \Rightarrow (x - 2 \sqrt{5}) (x - \sqrt{5}) = 0$
$\Rightarrow x - \sqrt{5} = 0 or x - 2 \sqrt{5} = 0 \Rightarrow x = \sqrt{5} or x = 2 \sqrt{5}$
Hence, the roots of the given equation are $\sqrt{5}$ and $2 \sqrt{5}$ .

Page No 183:

Answer:

$x^{2} - (\sqrt{3} + 1) x + \sqrt{3} = 0 \Rightarrow x^{2} - \sqrt{3} x - x + \sqrt{3} = 0 \Rightarrow x (x - \sqrt{3}) - 1 (x - \sqrt{3}) = 0 \Rightarrow (x - \sqrt{3}) (x - 1) = 0$
$\Rightarrow x - \sqrt{3} = 0 or x - 1 = 0 \Rightarrow x = \sqrt{3} or x = 1$
Hence, 1 and $\sqrt{3}$ are the roots of the given equation.

Page No 183:

Answer:

We write, $3 \sqrt{3} x = 5 \sqrt{3} x - 2 \sqrt{3} x$ as $x^{2} \times (- 30) = - 30 x^{2} = 5 \sqrt{3} x \times (- 2 \sqrt{3} x)$

$∴ x^{2} + 3 \sqrt{3} x - 30 = 0 \Rightarrow x^{2} + 5 \sqrt{3} x - 2 \sqrt{3} x - 30 = 0 \Rightarrow x (x + 5 \sqrt{3}) - 2 \sqrt{3} (x + 5 \sqrt{3}) = 0 \Rightarrow (x + 5 \sqrt{3}) (x - 2 \sqrt{3}) = 0$
$\Rightarrow x + 5 \sqrt{3} = 0 or x - 2 \sqrt{3} = 0 \Rightarrow x = - 5 \sqrt{3} or x = 2 \sqrt{3}$
Hence, the roots of the given equation are $- 5 \sqrt{3}$ and $2 \sqrt{3}$ .

Page No 183:

Answer:

We write, $7 x = 5 x + 2 x$ as $\sqrt{2} x^{2} \times 5 \sqrt{2} = 10 x^{2} = 5 x \times 2 x$

$∴ \sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0 \Rightarrow \sqrt{2} x^{2} + 5 x + 2 x + 5 \sqrt{2} = 0 \Rightarrow x (\sqrt{2} x + 5) + \sqrt{2} (\sqrt{2} x + 5) = 0 \Rightarrow (\sqrt{2} x + 5) (x + \sqrt{2}) = 0$
$\Rightarrow x + \sqrt{2} = 0 or \sqrt{2} x + 5 = 0 \Rightarrow x = - \sqrt{2} or x = - \frac{5}{\sqrt{2}} = - \frac{5 \sqrt{2}}{2}$
Hence, the roots of the given equation are $- \sqrt{2}$ and $- \frac{5 \sqrt{2}}{2}$ .

Page No 183:

Answer:

We write, 13x = 5x + 8x as $5 x^{2} \times 8 = 40 x^{2} = 5 x \times 8 x$
$∴ 5 x^{2} + 13 x + 8 = 0 \Rightarrow 5 x^{2} + 5 x + 8 x + 8 = 0 \Rightarrow 5 x (x + 1) + 8 (x + 1) = 0 \Rightarrow (x + 1) (5 x + 8) = 0$
$\Rightarrow x + 1 = 0 or 5 x + 8 = 0 \Rightarrow x = - 1 or x = - \frac{8}{5}$
Hence, $- 1$ and $- \frac{8}{5}$ are the roots of the given equation.

Page No 183:

Answer:

$x^{2} - (1 + \sqrt{2}) x + \sqrt{2} = 0 x^{2} - x - \sqrt{2} x + \sqrt{2} = 0 x (x - 1) - \sqrt{2} (x - 1) = 0 (x - \sqrt{2}) (x - 1) = 0 \Rightarrow x - \sqrt{2} = 0 and x - 1 = 0 \Rightarrow x = \sqrt{2} and x = 1$

Page No 183:

Answer:

$Given : 9 x^{2} + 6 x + 1 = 0 \Rightarrow 9 x^{2} + 3 x + 3 x + 1 = 0 \Rightarrow 3 x (3 x + 1) + 1 (3 x + 1) = 0 \Rightarrow (3 x + 1) (3 x + 1) = 0 \Rightarrow 3 x + 1 = 0 or 3 x + 1 = 0 \Rightarrow x = \frac{- 1}{3} or x = \frac{- 1}{3} Hence, \frac{- 1}{3} is the root of the equation 9 x^{2} + 6 x + 1 = 0 .$

Page No 183:

Answer:

We write, $- 20 x = - 10 x - 10 x$ as $100 x^{2} \times 1 = 100 x^{2} = (- 10 x) \times (- 10 x)$
$∴ 100 x^{2} - 20 x + 1 = 0 \Rightarrow 100 x^{2} - 10 x - 10 x + 1 = 0 \Rightarrow 10 x (10 x - 1) - 1 (10 x - 1) = 0 \Rightarrow (10 x - 1) (10 x - 1) = 0$
$\Rightarrow {(10 x - 1)}^{2} = 0 \Rightarrow 10 x - 1 = 0 \Rightarrow x = \frac{1}{10}$
Hence, $\frac{1}{10}$ is the repreated root of the given equation.

Page No 183:

Answer:

We write, $- x = - \frac{x}{2} - \frac{x}{2}$ as $2 x^{2} \times \frac{1}{8} = \frac{x^{2}}{4} = (- \frac{x}{2}) \times (- \frac{x}{2})$
$∴ 2 x^{2} - x + \frac{1}{8} = 0 \Rightarrow 2 x^{2} - \frac{x}{2} - \frac{x}{2} + \frac{1}{8} = 0 \Rightarrow 2 x (x - \frac{1}{4}) - \frac{1}{2} (x - \frac{1}{4}) = 0 \Rightarrow (x - \frac{1}{4}) (2 x - \frac{1}{2}) = 0$
$\Rightarrow x - \frac{1}{4} = 0 or 2 x - \frac{1}{2} = 0 \Rightarrow x = \frac{1}{4} or x = \frac{1}{4}$
Hence, $\frac{1}{4}$ is the repeated root of the given equation.

Page No 183:

Answer:

$Given: 10 x - \frac{1}{x} = 3 \Rightarrow 10 x^{2} - 1 = 3 x [Multiplying both sides by x] \Rightarrow 10 x^{2} - 3 x - 1 = 0 \Rightarrow 10 x^{2} - (5 x - 2 x) - 1 = 0 \Rightarrow 10 x^{2} - 5 x + 2 x - 1 = 0 \Rightarrow 5 x (2 x - 1) + 1 (2 x - 1) = 0 \Rightarrow (2 x - 1) (5 x + 1) = 0 \Rightarrow 2 x - 1 = 0 or 5 x + 1 = 0 \Rightarrow x = \frac{1}{2} or x = \frac{- 1}{5} H ence, the roots of the equation are \frac{1}{2} and \frac{- 1}{5} .$

Page No 183:

Answer:

$Given: \frac{2}{x^{2}} - \frac{5}{x} + 2 = 0 \Rightarrow 2 - 5 x + 2 x^{2} = 0 [Multiplying both side by x^{2}] \Rightarrow 2 x^{2} - 5 x + 2 = 0 \Rightarrow 2 x^{2} - (4 x + x) + 2 = 0 \Rightarrow 2 x^{2} - 4 x - x + 2 = 0 \Rightarrow 2 x (x - 2) - 1 (x - 2) = 0 \Rightarrow (2 x - 1) (x - 2) = 0 \Rightarrow 2 x - 1 = 0 or x - 2 = 0 \Rightarrow x = \frac{1}{2} or x = 2 H ence, the roots of the equation are \frac{1}{2} and 2 .$

Page No 183:

Answer:

We write, $a x = 2 a x - a x$ as $2 x^{2} \times (- a^{2}) = - 2 a^{2} x^{2} = 2 a x \times (- a x)$
$∴ 2 x^{2} + a x - a^{2} = 0 \Rightarrow 2 x^{2} + 2 a x - a x - a^{2} = 0 \Rightarrow 2 x (x + a) - a (x + a) = 0 \Rightarrow (x + a) (2 x - a) = 0$
$\Rightarrow x + a = 0 or 2 x - a = 0 \Rightarrow x = - a or x = \frac{a}{2}$
Hence, $- a$ and $\frac{a}{2}$ are the roots of the given equation.

Page No 183:

Answer:

We write, $4 b x = 2 (a + b) x - 2 (a - b) x$ as $4 x^{2} \times [- (a^{2} - b^{2})] = - 4 (a^{2} - b^{2}) x^{2} = 2 (a + b) x \times [- 2 (a - b) x]$
$∴ 4 x^{2} + 4 b x - (a^{2} - b^{2}) = 0 \Rightarrow 4 x^{2} + 2 (a + b) x - 2 (a - b) x - (a - b) (a + b) = 0 \Rightarrow 2 x [2 x + (a + b)] - (a - b) [2 x + (a + b)] = 0 \Rightarrow [2 x + (a + b)] [2 x - (a - b)] = 0$
$\Rightarrow 2 x + (a + b) = 0 or 2 x - (a - b) = 0 \Rightarrow x = - \frac{a + b}{2} or x = \frac{a - b}{2}$
Hence, $- \frac{a + b}{2}$ and $\frac{a - b}{2}$ are the roots of the given equation.

Page No 183:

Answer:

We write, $- 4 a^{2} x = - 2 (a^{2} + b^{2}) x - 2 (a^{2} - b^{2}) x$ as $4 x^{2} \times (a^{4} - b^{4}) = 4 (a^{4} - b^{4}) x^{2} = [- 2 (a^{2} + b^{2})] x \times [- 2 (a^{2} - b^{2})] x$
$∴ 4 x^{2} - 4 a^{2} x + (a^{4} - b^{4}) = 0 \Rightarrow 4 x^{2} - 2 (a^{2} + b^{2}) x - 2 (a^{2} - b^{2}) x + (a^{2} - b^{2}) (a^{2} + b^{2}) = 0 \Rightarrow 2 x [2 x - (a^{2} + b^{2})] - (a^{2} - b^{2}) [2 x - (a^{2} + b^{2})] = 0 \Rightarrow [2 x - (a^{2} + b^{2})] [2 x - (a^{2} - b^{2})] = 0$
$\Rightarrow 2 x - (a^{2} + b^{2}) = 0 or 2 x - (a^{2} - b^{2}) = 0 \Rightarrow x = \frac{a^{2} + b^{2}}{2} or x = \frac{a^{2} - b^{2}}{2}$
Hence, $\frac{a^{2} + b^{2}}{2}$ and $\frac{a^{2} - b^{2}}{2}$ are the roots of the given equation.

Page No 183:

Answer:

We write, $5 x = (a + 3) x - (a - 2) x$ as $x^{2} \times [- (a^{2} + a - 6)] = - (a^{2} + a - 6) x^{2} = (a + 3) x \times [- (a - 2) x]$
$∴ x^{2} + 5 x - (a^{2} + a - 6) = 0 \Rightarrow x^{2} + (a + 3) x - (a - 2) x - (a + 3) (a - 2) = 0 \Rightarrow x [x + (a + 3)] - (a - 2) [x + (a + 3)] = 0 \Rightarrow [x + (a + 3)] [x - (a - 2)] = 0$
$\Rightarrow x + (a + 3) = 0 or x - (a - 2) = 0 \Rightarrow x = - (a + 3) or x = a - 2$
Hence, $- (a + 3)$ and $(a - 2)$ are the roots of the given equation.

Page No 183:

Answer:

We write, $- 2 a x = (2 b - a) x - (2 b + a) x$ as $x^{2} \times [- (4 b^{2} - a^{2})] = - (4 b^{2} - a^{2}) x^{2} = (2 b - a) x \times [- (2 b + a) x]$
$∴ x^{2} - 2 a x - (4 b^{2} - a^{2}) = 0 \Rightarrow x^{2} + (2 b - a) x - (2 b + a) x - (2 b - a) (2 b + a) = 0 \Rightarrow x [x + (2 b - a)] - (2 b + a) [x + (2 b - a)] = 0 \Rightarrow [x + (2 b - a)] [x - (2 b + a)] = 0$
$\Rightarrow x + (2 b - a) = 0 or x - (2 b + a) = 0 \Rightarrow x = - (2 b - a) or x = 2 b + a \Rightarrow x = a - 2 b or x = a + 2 b$
Hence, $a - 2 b$ and $a + 2 b$ are the roots of the given equation.

Page No 183:

Answer:

We write, $- (2 b - 1) x = - (b - 5) x - (b + 4) x$ as $x^{2} \times (b^{2} - b - 20) = (b^{2} - b - 20) x^{2} = [- (b - 5) x] \times [- (b + 4) x]$
$∴ x^{2} - (2 b - 1) x + (b^{2} - b - 20) = 0 \Rightarrow x^{2} - (b - 5) x - (b + 4) x + (b - 5) (b + 4) = 0 \Rightarrow x [x - (b - 5)] - (b + 4) [x - (b - 5)] = 0 \Rightarrow [x - (b - 5)] [x - (b + 4)] = 0$
$\Rightarrow x - (b - 5) = 0 or x - (b + 4) = 0 \Rightarrow x = b - 5 or x = b + 4$
Hence, $b - 5$ and $b + 4$ are the roots of the given equation.

Page No 183:

Answer:

We write, $6 x = (a + 4) x - (a - 2) x$ as $x^{2} \times [- (a^{2} + 2 a - 8)] = - (a^{2} + 2 a - 8) x^{2} = (a + 4) x \times [- (a - 2) x]$
$∴ x^{2} + 6 x - (a^{2} + 2 a - 8) = 0 \Rightarrow x^{2} + (a + 4) x - (a - 2) x - (a + 4) (a - 2) = 0 \Rightarrow x [x + (a + 4)] - (a - 2) [x + (a + 4)] = 0 \Rightarrow [x + (a + 4)] [x - (a - 2)] = 0$
$\Rightarrow x + (a + 4) = 0 or x - (a - 2) = 0 \Rightarrow x = - (a + 4) or x = a - 2$
Hence, $- (a + 4)$ and $(a - 2)$ are the roots of the given equation.

Page No 183:

Answer:

$Given: a b x^{2} + (b^{2} - a c) x - b c = 0 \Rightarrow a b x^{2} + b^{2} x - a c x - b c = 0 \Rightarrow b x (a x + b) - c (a x + b) = 0 \Rightarrow (b x - c) (a x + b) = 0 \Rightarrow b x - c = 0 or a x + b = 0 \Rightarrow x = \frac{c}{b} or x = \frac{- b}{a} H ence, the roots of the equation are \frac{c}{b} and \frac{- b}{a} .$

Page No 183:

Answer:

We write, $- 4 a x = - (b + 2 a) x + (b - 2 a) x$ as $x^{2} \times (- b^{2} + 4 a^{2}) = (- b^{2} + 4 a^{2}) x^{2} = - (b + 2 a) x \times (b - 2 a) x$
$∴ x^{2} - 4 a x - b^{2} + 4 a^{2} = 0 \Rightarrow x^{2} - (b + 2 a) x + (b - 2 a) x - (b - 2 a) (b + 2 a) = 0 \Rightarrow x [x - (b + 2 a)] + (b - 2 a) [x - (b + 2 a)] = 0 \Rightarrow [x - (b + 2 a)] [x + (b - 2 a)] = 0$
$\Rightarrow x - (b + 2 a) = 0 or x + (b - 2 a) = 0 \Rightarrow x = 2 a + b or x = - (b - 2 a) \Rightarrow x = 2 a + b or x = 2 a - b$
Hence, $(2 a + b)$ and $(2 a - b)$ are the roots of the given equation.

Page No 183:

Answer:

$Given : 4 x^{2} - 2 (a^{2} + b^{2}) x + a^{2} b^{2} = 0 \Rightarrow 4 x^{2} - 2 a^{2} x - 2 b^{2} x + a^{2} b^{2} = 0 \Rightarrow 2 x (2 x - a^{2}) - b^{2} (2 x - a^{2}) = 0 \Rightarrow (2 x - b^{2}) (2 x - a^{2}) = 0 \Rightarrow 2 x - b^{2} = 0 or 2 x - a^{2} = 0 \Rightarrow x = \frac{b^{2}}{2} or x = \frac{a^{2}}{2} H ence, the roots of the equation are \frac{b^{2}}{2} and \frac{a^{2}}{2} .$

Page No 183:

Answer:

$Given : 12 a b x^{2} - (9 a^{2} - 8 b^{2}) x - 6 a b = 0 \Rightarrow 12 a b x^{2} - 9 a^{2} x + 8 b^{2} x - 6 a b = 0 \Rightarrow 3 a x (4 b x - 3 a) + 2 b (4 b x - 3 a) = 0 \Rightarrow (3 a x + 2 b) (4 b x - 3 a) = 0 \Rightarrow 3 a x + 2 b = 0 or 4 b x - 3 a = 0 \Rightarrow x = \frac{- 2 b}{3 a} or x = \frac{3 a}{4 b} H ence, the roots of the equation are \frac{- 2 b}{3 a} and \frac{3 a}{4 b} .$

Page No 183:

Answer:

$\begin{array}{l} Given : \\ a^{2} b^{2} x^{2} + b^{2} x - a^{2} x - 1 = 0 \Rightarrow b^{2} x (a^{2} x + 1) - 1 (a^{2} x + 1) = 0 \Rightarrow (b^{2} x - 1) (a^{2} x + 1) = 0 \Rightarrow (b^{2} x - 1) = 0 or (a^{2} x + 1) = 0 \Rightarrow x = \frac{1}{b^{2}} or x = \frac{- 1}{a^{2}} H ence, \frac{1}{b^{2}} and \frac{- 1}{a^{2}} are the roots of the given equation. \end{array}$

Page No 183:

Answer:

We write, $- 9 (a + b) x = - 3 (2 a + b) x - 3 (a + 2 b) x$ as $9 x^{2} \times (2 a^{2} + 5 a b + 2 b^{2}) = 9 (2 a^{2} + 5 a b + 2 b^{2}) x^{2} = [- 3 (2 a + b) x] \times [- 3 (a + 2 b) x]$
$∴ 9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0 \Rightarrow 9 x^{2} - 3 (2 a + b) x - 3 (a + 2 b) x + (2 a + b) (a + 2 b) = 0 \Rightarrow 3 x [3 x - (2 a + b)] - (a + 2 b) [3 x - (2 a + b)] = 0 \Rightarrow [3 x - (2 a + b)] [3 x - (a + 2 b)] = 0$
$\Rightarrow 3 x - (2 a + b) = 0 or 3 x - (a + 2 b) = 0 \Rightarrow x = \frac{2 a + b}{3} or x = \frac{a + 2 b}{3}$
Hence, $\frac{2 a + b}{3}$ and $\frac{a + 2 b}{3}$ are the roots of the given equation.

Page No 183:

Answer:

$\frac{16}{x} - 1 = \frac{15}{x + 1}, x \neq 0, - 1 \Rightarrow \frac{16}{x} - \frac{15}{x + 1} = 1 \Rightarrow \frac{16 x + 16 - 15 x}{x (x + 1)} = 1 \Rightarrow \frac{x + 16}{x^{2} + x} = 1$
$\Rightarrow x^{2} + x = x + 16 (Cross multiplication) \Rightarrow x^{2} - 16 = 0 \Rightarrow (x + 4) (x - 4) = 0 \Rightarrow x + 4 = 0 or x - 4 = 0$
$\Rightarrow x = - 4 or x = 4$
Hence, −4 and 4 are the roots of the given equation.

Page No 183:

Answer:

$\frac{4}{x} - 3 = \frac{5}{2 x + 3}, x \neq 0, - \frac{3}{2} \Rightarrow \frac{4}{x} - \frac{5}{2 x + 3} = 3 \Rightarrow \frac{8 x + 12 - 5 x}{x (2 x + 3)} = 3 \Rightarrow \frac{3 x + 12}{2 x^{2} + 3 x} = 3$
$\Rightarrow \frac{x + 4}{2 x^{2} + 3 x} = 1 \Rightarrow 2 x^{2} + 3 x = x + 4 (Cross multiplication) \Rightarrow 2 x^{2} + 2 x - 4 = 0 \Rightarrow x^{2} + x - 2 = 0$
$\Rightarrow x^{2} + 2 x - x - 2 = 0 \Rightarrow x (x + 2) - 1 (x + 2) = 0 \Rightarrow (x + 2) (x - 1) = 0 \Rightarrow x + 2 = 0 or x - 1 = 0$
$\Rightarrow x = - 2 or x = 1$
Hence, −2 and 1 are the roots of the given equation.

Page No 183:

Answer:

$\frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3 x - 1}, x \neq - 1, \frac{1}{3} \Rightarrow \frac{3}{x + 1} - \frac{2}{3 x - 1} = \frac{1}{2} \Rightarrow \frac{9 x - 3 - 2 x - 2}{(x + 1) (3 x - 1)} = \frac{1}{2} \Rightarrow \frac{7 x - 5}{3 x^{2} + 2 x - 1} = \frac{1}{2}$
$\Rightarrow 3 x^{2} + 2 x - 1 = 14 x - 10 (Cross multiplication) \Rightarrow 3 x^{2} - 12 x + 9 = 0 \Rightarrow x^{2} - 4 x + 3 = 0 \Rightarrow x^{2} - 3 x - x + 3 = 0$
$\Rightarrow x (x - 3) - 1 (x - 3) = 0 \Rightarrow (x - 3) (x - 1) = 0 \Rightarrow x - 3 = 0 or x - 1 = 0 \Rightarrow x = 3 or x = 1$
Hence, 1 and 3 are the roots of the given equation.

Page No 184:

Answer:

(i)
$\frac{1}{x - 1} - \frac{1}{x + 5} = \frac{6}{7}, x \neq 1, - 5 \Rightarrow \frac{x + 5 - x + 1}{(x - 1) (x + 5)} = \frac{6}{7} \Rightarrow \frac{6}{x^{2} + 4 x - 5} = \frac{6}{7} \Rightarrow x^{2} + 4 x - 5 = 7$
$\Rightarrow x^{2} + 4 x - 12 = 0 \Rightarrow x^{2} + 6 x - 2 x - 12 = 0 \Rightarrow x (x + 6) - 2 (x + 6) = 0 \Rightarrow (x + 6) (x - 2) = 0$
$\Rightarrow x + 6 = 0 or x - 2 = 0 \Rightarrow x = - 6 or x = 2$
Hence, −6 and 2 are the roots of the given equation.

(ii)
$\frac{1}{2 x - 3} + \frac{1}{x - 5} = 1 \frac{1}{9} \Rightarrow \frac{1}{2 x - 3} + \frac{1}{x - 5} = \frac{10}{9} \Rightarrow \frac{(x - 5) + (2 x - 3)}{(2 x - 3) (x - 5)} = \frac{10}{9} \Rightarrow \frac{3 x - 8}{2 x^{2} - 3 x - 10 x + 15} = \frac{10}{9} \Rightarrow \frac{3 x - 8}{2 x^{2} - 13 x + 15} = \frac{10}{9} \Rightarrow 27 x - 72 = 20 x^{2} - 130 x + 150 \Rightarrow 20 x^{2} - 157 x + 222 = 0 \Rightarrow 20 x^{2} - 120 x - 37 x + 222 = 0 \Rightarrow 20 x (x - 6) - 37 (x - 6) = 0 \Rightarrow (20 x - 37) (x - 6) = 0 \Rightarrow 20 x - 37 = 0 or x - 6 = 0 \Rightarrow x = \frac{37}{20} or x = 6$

Page No 184:

Answer:

$\frac{1}{2 a + b + 2 x} = \frac{1}{2 a} + \frac{1}{b} + \frac{1}{2 x} \Rightarrow \frac{1}{2 a + b + 2 x} - \frac{1}{2 x} = \frac{1}{2 a} + \frac{1}{b} \Rightarrow \frac{2 x - 2 a - b - 2 x}{2 x (2 a + b + 2 x)} = \frac{2 a + b}{2 a b} \Rightarrow \frac{- (2 a + b)}{4 x^{2} + 4 a x + 2 b x} = \frac{2 a + b}{2 a b}$
$\Rightarrow 4 x^{2} + 4 a x + 2 b x = - 2 a b \Rightarrow 4 x^{2} + 4 a x + 2 b x + 2 a b = 0 \Rightarrow 4 x (x + a) + 2 b (x + a) = 0 \Rightarrow (x + a) (4 x + 2 b) = 0$
$\Rightarrow x + a = 0 or 4 x + 2 b = 0 \Rightarrow x = - a or x = - \frac{b}{2}$
Hence, $- a$ and $- \frac{b}{2}$ are the roots of the given equation.

Page No 184:

Answer:

$Given : \frac{x + 3}{x - 2} - \frac{1 - x}{x} = 4 \frac{1}{4} \Rightarrow \frac{(x + 3)}{(x - 2)} - \frac{(1 - x)}{x} = \frac{17}{4} \Rightarrow \frac{x (x + 3) - (1 - x) (x - 2)}{(x - 2) x} = \frac{17}{4} \Rightarrow \frac{x^{2} + 3 x - (x - 2 - x^{2} + 2 x)}{x^{2} - 2 x} = \frac{17}{4} \Rightarrow \frac{x^{2} + 3 x + x^{2} - 3 x + 2}{x^{2} - 2 x} = \frac{17}{4} \Rightarrow \frac{2 x^{2} + 2}{x^{2} - 2 x} = \frac{17}{4} \Rightarrow 8 x^{2} + 8 = 17 x^{2} - 34 x [On cross multiplying] \Rightarrow - 9 x^{2} + 34 x + 8 = 0 \Rightarrow 9 x^{2} - 34 x - 8 = 0 \Rightarrow 9 x^{2} - 36 x + 2 x - 8 = 0 \Rightarrow 9 x (x - 4) + 2 (x - 4) = 0 \Rightarrow (x - 4) (9 x + 2) = 0 \Rightarrow x - 4 = 0 or 9 x + 2 = 0 \Rightarrow x = 4 or x = \frac{- 2}{9} H ence, the roots of the equation are 4 and \frac{- 2}{9} .$

Page No 184:

Answer:

$\frac{3 x - 4}{7} + \frac{7}{3 x - 4} = \frac{5}{2}, x \neq \frac{4}{3} \Rightarrow \frac{{(3 x - 4)}^{2} + 49}{7 (3 x - 4)} = \frac{5}{2} \Rightarrow \frac{9 x^{2} - 24 x + 16 + 49}{21 x - 28} = \frac{5}{2} \Rightarrow \frac{9 x^{2} - 24 x + 65}{21 x - 28} = \frac{5}{2}$
$\Rightarrow 18 x^{2} - 48 x + 130 = 105 x - 140 \Rightarrow 18 x^{2} - 153 x + 270 = 0 \Rightarrow 2 x^{2} - 17 x + 30 = 0 \Rightarrow 2 x^{2} - 12 x - 5 x + 30 = 0$
$\Rightarrow 2 x (x - 6) - 5 (x - 6) = 0 \Rightarrow (x - 6) (2 x - 5) = 0 \Rightarrow x - 6 = 0 or 2 x - 5 = 0 \Rightarrow x = 6 or x = \frac{5}{2}$
Hence, 6 and $\frac{5}{2}$ are the roots of the given equation.

Page No 184:

Answer:

(i)
$\frac{x}{x - 1} + \frac{x - 1}{x} = 4 \frac{1}{4}, x \neq 0, 1 \Rightarrow \frac{x^{2} + {(x - 1)}^{2}}{x (x - 1)} = \frac{17}{4} \Rightarrow \frac{x^{2} + x^{2} - 2 x + 1}{x^{2} - x} = \frac{17}{4} \Rightarrow \frac{2 x^{2} - 2 x + 1}{x^{2} - x} = \frac{17}{4}$
$\Rightarrow 8 x^{2} - 8 x + 4 = 17 x^{2} - 17 x \Rightarrow 9 x^{2} - 9 x - 4 = 0 \Rightarrow 9 x^{2} - 12 x + 3 x - 4 = 0 \Rightarrow 3 x (3 x - 4) + 1 (3 x - 4) = 0$
$\Rightarrow (3 x - 4) (3 x + 1) = 0 \Rightarrow 3 x - 4 = 0 or 3 x + 1 = 0 \Rightarrow x = \frac{4}{3} or x = - \frac{1}{3}$
Hence, $\frac{4}{3}$ and $- \frac{1}{3}$ are the roots of the given equation.

(ii)
$\frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = 2 \Rightarrow \frac{(x - 1)^{2} + (2 x + 1)^{2}}{(2 x + 1) (x - 1)} = 2 \Rightarrow (x^{2} + 1 - 2 x) + (4 x^{2} + 1 + 4 x) = 2 (2 x + 1) (x - 1) \Rightarrow 5 x^{2} + 2 x + 2 = 2 (2 x^{2} - x - 1) \Rightarrow 5 x^{2} + 2 x + 2 = 4 x^{2} - 2 x - 2 \Rightarrow x^{2} + 4 x + 4 = 0 \Rightarrow x^{2} + 2 x + 2 x + 4 = 0 \Rightarrow x (x + 2) + 2 (x + 2) = 0 \Rightarrow (x + 2) (x + 2) = 0 \Rightarrow (x + 2) = 0 or (x + 2) = 0 \Rightarrow x = - 2 or x = - 2 \Rightarrow x = - 2$

Page No 184:

Answer:

$\frac{x}{x + 1} + \frac{x + 1}{x} = 2 \frac{4}{15}, x \neq 0, - 1 \Rightarrow \frac{x^{2} + {(x + 1)}^{2}}{x (x + 1)} = \frac{34}{15} \Rightarrow \frac{x^{2} + x^{2} + 2 x + 1}{x^{2} + x} = \frac{34}{15} \Rightarrow \frac{2 x^{2} + 2 x + 1}{x^{2} + x} = \frac{34}{15}$
$\Rightarrow 30 x^{2} + 30 x + 15 = 34 x^{2} + 34 x \Rightarrow 4 x^{2} + 4 x - 15 = 0 \Rightarrow 4 x^{2} + 10 x - 6 x - 15 = 0 \Rightarrow 2 x (2 x + 5) - 3 (2 x + 5) = 0$
$\Rightarrow (2 x + 5) (2 x - 3) = 0 \Rightarrow 2 x + 5 = 0 or 2 x - 3 = 0 \Rightarrow x = - \frac{5}{2} or x = \frac{3}{2}$
Hence, $- \frac{5}{2}$ and $\frac{3}{2}$ are the roots of the given equation.

Page No 184:

Answer:

$\frac{x - 4}{x - 5} + \frac{x - 6}{x - 7} = 3 \frac{1}{3}, x \neq 5, 7 \Rightarrow \frac{(x - 4) (x - 7) + (x - 5) (x - 6)}{(x - 5) (x - 7)} = \frac{10}{3} \Rightarrow \frac{x^{2} - 11 x + 28 + x^{2} - 11 x + 30}{x^{2} - 12 x + 35} = \frac{10}{3} \Rightarrow \frac{2 x^{2} - 22 x + 58}{x^{2} - 12 x + 35} = \frac{10}{3}$
$\Rightarrow \frac{x^{2} - 11 x + 29}{x^{2} - 12 x + 35} = \frac{5}{3} \Rightarrow 3 x^{2} - 33 x + 87 = 5 x^{2} - 60 x + 175 \Rightarrow 2 x^{2} - 27 x + 88 = 0$
$\Rightarrow 2 x^{2} - 16 x - 11 x + 88 = 0 \Rightarrow 2 x (x - 8) - 11 (x - 8) = 0 \Rightarrow (x - 8) (2 x - 11) = 0$
$\Rightarrow x - 8 = 0 or 2 x - 11 = 0 \Rightarrow x = 8 or x = \frac{11}{2}$
Hence, 8 and $\frac{11}{2}$ are the roots of the given equation.

Page No 184:

Answer:

$\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3 \frac{1}{3}, x \neq 2, 4 \Rightarrow \frac{(x - 1) (x - 4) + (x - 2) (x - 3)}{(x - 2) (x - 4)} = \frac{10}{3} \Rightarrow \frac{x^{2} - 5 x + 4 + x^{2} - 5 x + 6}{x^{2} - 6 x + 8} = \frac{10}{3} \Rightarrow \frac{2 x^{2} - 10 x + 10}{x^{2} - 6 x + 8} = \frac{10}{3}$
$\Rightarrow \frac{x^{2} - 5 x + 5}{x^{2} - 6 x + 8} = \frac{5}{3} \Rightarrow 3 x^{2} - 15 x + 15 = 5 x^{2} - 30 x + 40 \Rightarrow 2 x^{2} - 15 x + 25 = 0 \Rightarrow 2 x^{2} - 10 x - 5 x + 25 = 0$
$\Rightarrow 2 x (x - 5) - 5 (x - 5) = 0 \Rightarrow (x - 5) (2 x - 5) = 0 \Rightarrow x - 5 = 0 or 2 x - 5 = 0 \Rightarrow x = 5 or x = \frac{5}{2}$
Hence, 5 and $\frac{5}{2}$ are the roots of the given equation.

Page No 184:

Answer:

$Given : \frac{1}{(x - 2)} + \frac{2}{(x - 1)} = \frac{6}{x} \Rightarrow \frac{(x - 1) + 2 (x - 2)}{(x - 1) (x - 2)} = \frac{6}{x} \Rightarrow \frac{3 x - 5}{x^{2} - 3 x + 2} = \frac{6}{x} \Rightarrow 3 x^{2} - 5 x = 6 x^{2} - 18 x + 12 [O n cross multiplying] \Rightarrow 3 x^{2} - 13 x + 12 = 0 \Rightarrow 3 x^{2} - (9 + 4) x + 12 = 0 \Rightarrow 3 x^{2} - 9 x - 4 x + 12 = 0 \Rightarrow 3 x (x - 3) - 4 (x - 3) = 0 \Rightarrow (3 x - 4) (x - 3) = 0 \Rightarrow 3 x - 4 = 0 or x - 3 = 0 \Rightarrow x = \frac{4}{3} or x = 3 H ence, the roots of the equation are \frac{4}{3} and 3.$

Page No 184:

Answer:

(i)

$\frac{1}{x + 1} + \frac{2}{x + 2} = \frac{5}{x + 4}, x \neq - 1, - 2, - 4 \Rightarrow \frac{x + 2 + 2 x + 2}{(x + 1) (x + 2)} = \frac{5}{x + 4} \Rightarrow \frac{3 x + 4}{x^{2} + 3 x + 2} = \frac{5}{x + 4} \Rightarrow (3 x + 4) (x + 4) = 5 (x^{2} + 3 x + 2)$
$\Rightarrow 3 x^{2} + 16 x + 16 = 5 x^{2} + 15 x + 10 \Rightarrow 2 x^{2} - x - 6 = 0 \Rightarrow 2 x^{2} - 4 x + 3 x - 6 = 0 \Rightarrow 2 x (x - 2) + 3 (x - 2) = 0$
$\Rightarrow (x - 2) (2 x + 3) = 0 \Rightarrow x - 2 = 0 or 2 x + 3 = 0 \Rightarrow x = 2 or x = - \frac{3}{2}$
Hence, 2 and $- \frac{3}{2}$ are the roots of the given equation.

(ii)

$\frac{1}{x + 1} + \frac{3}{5 x + 1} = \frac{5}{x + 4} \frac{(5 x + 1) + (3 x + 3)}{(x + 1) (5 x + 1)} = \frac{5}{x + 4} (x + 4) [(5 x + 1) + (3 x + 3)] = 5 (x + 1) (5 x + 1) (x + 4) [8 x + 4] = 25 x^{2} + 30 x + 5 8 x^{2} + 4 x + 32 x + 16 = 25 x^{2} + 30 x + 5 8 x^{2} + 36 x + 16 = 25 x^{2} + 30 x + 5 17 x^{2} - 6 x - 11 = 0 17 x^{2} - 17 x + 11 x - 11 = 0 17 x (x - 1) + 11 (x - 1) = 0 (17 x + 11) (x - 1) = 0 (17 x + 11) = 0 or (x - 1) = 0 x = - \frac{11}{17} or x = 1$

Page No 184:

Answer:

$3 (\frac{3 x - 1}{2 x + 3}) - 2 (\frac{2 x + 3}{3 x - 1}) = 5, x \neq \frac{1}{3}, - \frac{3}{2} \Rightarrow \frac{3 {(3 x - 1)}^{2} - 2 {(2 x + 3)}^{2}}{(2 x + 3) (3 x - 1)} = 5 \Rightarrow \frac{3 (9 x^{2} - 6 x + 1) - 2 (4 x^{2} + 12 x + 9)}{6 x^{2} + 7 x - 3} = 5 \Rightarrow \frac{27 x^{2} - 18 x + 3 - 8 x^{2} - 24 x - 18}{6 x^{2} + 7 x - 3} = 5$
$\Rightarrow \frac{19 x^{2} - 42 x - 15}{6 x^{2} + 7 x - 3} = 5 \Rightarrow 19 x^{2} - 42 x - 15 = 30 x^{2} + 35 x - 15 \Rightarrow 11 x^{2} + 77 x = 0 \Rightarrow 11 x (x + 7) = 0$
$\Rightarrow x = 0 or x + 7 = 0 \Rightarrow x = 0 or x = - 7$
Hence, 0 and −7 are the roots of the given equation.

Page No 184:

Answer:

$3 (\frac{7 x + 1}{5 x - 3}) - 4 (\frac{5 x - 3}{7 x + 1}) = 11, x \neq \frac{3}{5}, - \frac{1}{7} \Rightarrow \frac{3 {(7 x + 1)}^{2} - 4 {(5 x - 3)}^{2}}{(5 x - 3) (7 x + 1)} = 11 \Rightarrow \frac{3 (49 x^{2} + 14 x + 1) - 4 (25 x^{2} - 30 x + 9)}{35 x^{2} - 16 x - 3} = 11 \Rightarrow \frac{147 x^{2} + 42 x + 3 - 100 x^{2} + 120 x - 36}{35 x^{2} - 16 x - 3} = 11$
$\Rightarrow \frac{47 x^{2} + 162 x - 33}{35 x^{2} - 16 x - 3} = 11 \Rightarrow 47 x^{2} + 162 x - 33 = 385 x^{2} - 176 x - 33 \Rightarrow 338 x^{2} - 338 x = 0 \Rightarrow 338 x (x - 1) = 0$
$\Rightarrow x = 0 or x - 1 = 0 \Rightarrow x = 0 or x = 1$
Hence, 0 and 1 are the roots of the given equation.

Page No 184:

Answer:

$Given : (\frac{4 x - 3}{2 x + 1}) - 10 (\frac{2 x + 1}{4 x - 3}) = 3 P utting \frac{4 x - 3}{2 x + 1} = y, we get: y - \frac{10}{y} = 3 \Rightarrow \frac{y^{2} - 10}{y} = 3 \Rightarrow y^{2} - 10 = 3 y [On cross multiplying] \Rightarrow y^{2} - 3 y - 10 = 0 \Rightarrow y^{2} - (5 - 2) y - 10 = 0 \Rightarrow y^{2} - 5 y + 2 y - 10 = 0 \Rightarrow y (y - 5) + 2 (y - 5) = 0 \Rightarrow (y - 5) (y + 2) = 0 \Rightarrow y - 5 = 0 or y + 2 = 0 \Rightarrow y = 5 or y = - 2 Case I If y = 5, we get : \frac{4 x - 3}{2 x + 1} = 5 \Rightarrow 4 x - 3 = 5 (2 x + 1) [On cross multiplying] \Rightarrow 4 x - 3 = 10 x + 5 \Rightarrow - 6 x = 8 \Rightarrow - 6 x = 8 \Rightarrow x = - \frac{8^{4}}{6^{3}} \Rightarrow x = - \frac{4}{3} Case II If y = - 2, we get : \frac{4 x - 3}{2 x + 1} = - 2 \begin{array}{l} \Rightarrow 4 x - 3 = - 2 (2 x + 1) \\ \Rightarrow 4 x - 3 = - 4 x - 2 \Rightarrow 8 x = 1 \Rightarrow x = \frac{1}{8} Hence, the roots of the equation are - \frac{4}{3} and \frac{1}{8} . \end{array}$

Page No 184:

Answer:

$Given : {(\frac{x}{x + 1})}^{2} - 5 (\frac{x}{x + 1}) + 6 = 0 P utting \frac{x}{x + 1} = y, we get : y^{2} - 5 y + 6 = 0 \Rightarrow y^{2} - 5 y + 6 = 0 \Rightarrow y^{2} - (3 + 2) y + 6 = 0 \Rightarrow y^{2} - 3 y - 2 y + 6 = 0 \Rightarrow y (y - 3) - 2 (y - 3) = 0 \Rightarrow (y - 3) (y - 2) = 0 \Rightarrow y - 3 = 0 or y - 2 = 0 \Rightarrow y = 3 or y = 2 Case I If y = 3, w e g e t : \frac{x}{x + 1} = 3 \Rightarrow x = 3 (x + 1) [On cross multiplying] \Rightarrow x = 3 x + 3 \Rightarrow x = \frac{- 3}{2} Case II If y = 2, we get : \frac{x}{x + 1} = 2 \Rightarrow x = 2 (x + 1) \Rightarrow x = 2 x + 2 \Rightarrow - x = 2 \Rightarrow x = - 2 Hence, the roots of the equation are \frac{- 3}{2} and - 2 .$

Page No 184:

Answer:

$\begin{array}{l} \frac{a}{(x - b)} + \frac{b}{(x - a)} = 2 \\ \begin{array}{l} \Rightarrow [\frac{a}{(x - b)} - 1] + [\frac{b}{(x - a)} - 1] = 0 \\ \Rightarrow \frac{a - (x - b)}{x - b} + \frac{b - (x - a)}{x - a} = 0 \end{array} \end{array} \Rightarrow \frac{a - x + b}{x - b} + \frac{a - x + b}{x - a} = 0 \Rightarrow (a - x + b) [\frac{1}{(x - b)} + \frac{1}{(x - a)}] = 0 \Rightarrow (a - x + b) [\frac{(x - a) + (x - b)}{(x - b) (x - a)}] = 0 \Rightarrow (a - x + b) [\frac{2 x - (a + b)}{(x - b) (x - a)}] = 0 \Rightarrow (a - x + b) [2 x - (a + b)] = 0 \Rightarrow a - x + b = 0 or 2 x - (a + b) = 0 \Rightarrow x = a + b or x = \frac{a + b}{2} Hence, the roots of the equation are (a + b) and (\frac{a + b}{2}) .$

Page No 184:

Answer:

$\frac{a}{(a x - 1)} + \frac{b}{(b x - 1)} = (a + b) \Rightarrow [\frac{a}{(a x - 1)} - b] + [\frac{b}{(b x - 1)} - a] = 0 \Rightarrow \frac{a - b (a x - 1)}{a x - 1} + \frac{b - a (b x - 1)}{b x - 1} = 0 \Rightarrow \frac{a - a b x + b}{a x - 1} + \frac{a - a b x + b}{b x - 1} = 0 \Rightarrow (a - a b x + b) [\frac{1}{(a x - 1)} + \frac{1}{(b x - 1)}] = 0 \Rightarrow (a - a b x + b) [\frac{(b x - 1) + (a x - 1)}{(a x - 1) (b x - 1)}] = 0 \Rightarrow (a - a b x + b) [\frac{(a + b) x - 2}{(a x - 1) (b x - 1)}] = 0 \Rightarrow (a - a b x + b) [(a + b) x - 2] = 0 \Rightarrow a - a b x + b = 0 or (a + b) x - 2 = 0 \Rightarrow x = \frac{(a + b)}{a b} or x = \frac{2}{(a + b)} Hence, the roots of the equation are \frac{(a + b)}{a b} and \frac{2}{(a + b)} .$

Page No 184:

Answer:

$3^{(x + 2)} + 3^{- x} = 10 3^{x} . 9 + \frac{1}{3^{x}} = 10 Let 3^{x} be equal to y . ∴ 9 y + \frac{1}{y} = 10 \Rightarrow 9 y^{2} + 1 = 10 y \Rightarrow 9 y^{2} - 10 y + 1 = 0 \Rightarrow (y - 1) (9 y - 1) = 0 \Rightarrow y - 1 = 0 or 9 y - 1 = 0 \Rightarrow y = 1 or y = \frac{1}{9} \Rightarrow 3^{x} = 1 {or 3}^{x} = \frac{1}{9} \Rightarrow 3^{x} = 3^{0} {or 3}^{x} = {3^{-}}^{2} \Rightarrow x = 0 or x = - 2 Hence, 0 and - 2 are the roots of the given equation .$

Page No 184:

Answer:

$Given : 4^{(x + 1)} + 4^{(1 - x)} = 10 \Rightarrow 4^{x} .4 + 4^{1} . \frac{1}{4^{x}} = 10 {Let 4}^{x} be y . \begin{array}{l} ∴ 4 y + \frac{4}{y} = 10 \\ \Rightarrow 4 y^{2} - 10 y + 4 = 0 \end{array} \Rightarrow 4 y^{2} - 8 y - 2 y + 4 = 0 \Rightarrow 4 y (y - 2) - 2 (y - 2) = 0 \begin{array}{l} \Rightarrow y = 2 or y = \frac{2}{4} = \frac{1}{2} \\ \Rightarrow 4^{x} = 2 or \frac{1}{2} \end{array} \Rightarrow 4^{x} = 2^{2 x} = 2^{1} or 2^{2 x} = 2^{- 1} \Rightarrow x = \frac{1}{2} or x = \frac{- 1}{2} Hence, \frac{1}{2} and \frac{- 1}{2} are roots of the given equation .$

Page No 184:

Answer:

$Given : 2^{2 x} - {3.2}^{(x + 2)} + 32 = 0 \Rightarrow {(2^{x})}^{2} - {3.2}^{x} {.2}^{2} + 32 = 0 {Let 2}^{x} be y . ∴ y^{2} - 12 y + 32 = 0 \Rightarrow y^{2} - 8 y - 4 y + 32 = 0 \Rightarrow y (y - 8) - 4 (y - 8) = 0 \Rightarrow (y - 8) = 0 or (y - 4) = 0 \Rightarrow y = 8 or y = 4 ∴ 2^{x} = 8 or 2^{x} = 4 \Rightarrow 2^{x} = 2^{3} or 2^{x} = 2^{2} \Rightarrow x = 2 or 3 Hence, 2 and 3 are the roots of the given equation .$

Page No 193:

Answer:

(i) 2 x^{2} - 7 x + 6 = 0 Here, a = 2, b = - 7, c = 6 Discriminant D is diven by : D = b^{2} - 4 a c = {(- 7)}^{2} - 4 \times 2 \times 6 = 49 - 48 = 1

(ii) 3 x^{2} - 2 x + 8 = 0 Here, a = 3, b = - 2, c = 8 Discriminant D is given by : D = b^{2} - 4 a c = {(- 2)}^{2} - 4 \times 3 \times 8 = 4 - 96 = - 92

(iii) 2 x^{2} - 5 \sqrt{2 x} + 4 = 0 Here, a = 2, b = - 5 \sqrt{2}, c = 4 Discriminant D is given by : D = b^{2} - 4 a c = {(- 5 \sqrt{2})}^{2} - 4 \times 2 \times 4 = (25 \times 2) - 32 = 50 - 32 = 18

(iv) \sqrt{3} x^{2} + 2 \sqrt{2} x - 2 \sqrt{3} = 0 Here, a = \sqrt{3}, b = 2 \sqrt{2}, c = - 2 \sqrt{3} Discriminant D is given by : D = b^{2} - 4 a c = {(2 \sqrt{2})}^{2} - 4 \times \sqrt{3} \times (- 2 \sqrt{3}) = (4 \times 2) + (8 \times 3) = 8 + 24 = 32

(v)

(x - 1) (2 x - 1) = 0

\Rightarrow 2 x^{2} - 3 x + 1 = 0

Comparing it with

a x^{2} + b x + c = 0

, we get

a = 2, b = −3 and c = 1

∴ Discriminant, D =

b^{2} - 4 a c = {(- 3)}^{2} - 4 \times 2 \times 1 = 9 - 8 = 1

(vi) 1 - x = 2 x^{2} \Rightarrow 2 x^{2} + x - 1 = 0 Here, a = 2, b = 1, c = - 1 Discriminant D is given by : D = b^{2} - 4 a c {= 1}^{2} - 4 \times 2 (- 1) = 1 + 8 = 9

Page No 193:

Answer:

$Given : x^{2} - 4 x - 1 = 0 On c omparing it with a x^{2} + b x + c = 0, we get : a = 1, b = - 4 and c = - 1 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 4)}^{2} - 4 \times 1 \times (- 1) = 16 + 4 = 20 = 20 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 4) + \sqrt{20}}{2 \times 1} = \frac{4 + 2 \sqrt{5}}{2} = \frac{2 (2 + \sqrt{5})}{2} = (2 + \sqrt{5}) β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 4) - \sqrt{20}}{2} = \frac{4 - 2 \sqrt{5}}{2} = \frac{2 (2 - \sqrt{5})}{2} = (2 - \sqrt{5}) Thus, the roots of the equation are (2 + \sqrt{5}) and (2 - \sqrt{5}) .$

Page No 193:

Answer:

$Given : x^{2} - 6 x + 4 = 0 On c omparing it with a x^{2} + b x + c = 0, we get: a = 1, b = - 6 and c = 4 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 6)}^{2} - 4 \times 1 \times 4 = 36 - 16 = 20 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 6) + \sqrt{20}}{2 \times 1} = \frac{6 + 2 \sqrt{5}}{2} = \frac{2 (3 + \sqrt{5})}{2} = (3 + \sqrt{5}) \begin{array}{l} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 6) - \sqrt{20}}{2 \times 1} = \frac{6 - 2 \sqrt{5}}{2} = \frac{2 (3 - \sqrt{5})}{2} = (3 - \sqrt{5}) \\ Thus, the roots of the equation are (3 + 2 \sqrt{5}) and (3 - 2 \sqrt{5}) . \end{array}$

Page No 193:

Answer:

The given equation is $2 x^{2} + x - 4 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 2, b = 1 and c = −4

∴ Discriminant, D = $b^{2} - 4 a c = {(1)}^{2} - 4 \times 2 \times (- 4) = 1 + 32 = 33 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{33}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 1 + \sqrt{33}}{2 \times 2} = \frac{- 1 + \sqrt{33}}{4} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 1 - \sqrt{33}}{2 \times 2} = \frac{- 1 - \sqrt{33}}{4}$

Hence, $\frac{- 1 + \sqrt{33}}{4}$ and $\frac{- 1 - \sqrt{33}}{4}$ are the roots of the given equation.

Page No 193:

Answer:

$Given: 25 x^{2} + 30 x + 7 = 0 On comparing it with a x^{2} + b x + c = 0, we get: a = 25, b = 30 and c = 7 Discriminant D is given by : D = (b^{2} - 4 a c) {= 30}^{2} - 4 \times 25 \times 7 = 900 - 700 = 200 = 200 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 30 + \sqrt{200}}{2 \times 25} = \frac{- 30 + 10 \sqrt{2}}{50} = \frac{10 (- 3 + \sqrt{2})}{50} = \frac{(- 3 + \sqrt{2})}{5} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 30 - \sqrt{200}}{2 \times 25} = \frac{- 30 - 10 \sqrt{2}}{50} = \frac{10 (- 3 - \sqrt{2})}{50} = \frac{(- 3 - \sqrt{2})}{5} Thus, the roots of the equation are \frac{(- 3 + \sqrt{2})}{5} and \frac{(- 3 - \sqrt{2})}{5} .$

Page No 193:

Answer:

$Given : 16 x^{2} = 24 x + 1 ⇒ 16 x^{2} - 24 x - 1 = 0 On c omparing it with a x^{2} + b x + c = 0 a = 16, b = - 24 and c = - 1 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 24)}^{2} - 4 \times 16 \times (- 1) = 576 + (64) = 640 > 0 Hence, the roots of the equation are real, Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 24) + \sqrt{640}}{2 \times 16} = \frac{24 + 8 \sqrt{10}}{32} = \frac{8 (3 + \sqrt{10})}{32} = \frac{(3 + \sqrt{10})}{4} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 24) - \sqrt{640}}{2 \times 16} = \frac{24 - 8 \sqrt{10}}{32} = \frac{8 (3 - \sqrt{10})}{32} = \frac{(3 - \sqrt{10})}{4} Thus, the roots of the equation are \frac{(3 + \sqrt{10})}{4} and \frac{(3 - \sqrt{10})}{4} .$

Page No 193:

Answer:

$Given: 15 x^{2} - 28 = x ⇒ 15 x^{2} - x - 28 = 0 On c omparing it with a x^{2} + b x + c = 0, we get: a = 15, b = - 1 and c = - 28 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 1)}^{2} - 4 \times 15 \times (- 28) = 1 - (- 1680) = 1 + 1680 = 1681 = 1681 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 1) + \sqrt{1681}}{2 \times 15} = \frac{1 + 41}{30} = \frac{42}{30} = \frac{7}{5} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 1) - \sqrt{1681}}{2 \times 15} = \frac{1 - 41}{30} = \frac{- 40}{30} = \frac{- 4}{3} Thus, the roots of the equation are \frac{7}{5} and \frac{- 4}{3} .$

Page No 193:

Answer:

The given equation is $2 x^{2} - 2 \sqrt{2} x + 1 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 2, b = $- 2 \sqrt{2}$ and c = 1

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{2})}^{2} - 4 \times 2 \times 1 = 8 - 8 = 0$

So, the given equation has real roots.

Now, $\sqrt{D} = 0$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) + \sqrt{0}}{2 \times 2} = \frac{2 \sqrt{2}}{4} = \frac{\sqrt{2}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) - \sqrt{0}}{2 \times 2} = \frac{2 \sqrt{2}}{4} = \frac{\sqrt{2}}{2}$

Hence, $\frac{\sqrt{2}}{2}$ is the repeated root of the given equation.

Page No 193:

Answer:

The given equation is $\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = $\sqrt{2}$ , b = 7 and c = $5 \sqrt{2}$

∴ Discriminant, D = $b^{2} - 4 a c = {(7)}^{2} - 4 \times \sqrt{2} \times 5 \sqrt{2} = 49 - 40 = 9 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{9} = 3$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 7 + 3}{2 \times \sqrt{2}} = \frac{- 4}{2 \sqrt{2}} = - \sqrt{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 7 - 3}{2 \times \sqrt{2}} = \frac{- 10}{2 \sqrt{2}} = - \frac{5 \sqrt{2}}{2}$

Hence, $- \sqrt{2}$ and $- \frac{5 \sqrt{2}}{2}$ are the roots of the given equation.

Page No 193:

Answer:

$Given : \sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0 On c omparing it with a x^{2} + b x + c = 0, we get: a = \sqrt{3}, b = 10 and c = - 8 \sqrt{3} Discriminant D is given by : D = (b^{2} - 4 a c) = {(10)}^{2} - 4 \times \sqrt{3} \times (- 8 \sqrt{3}) = 100 + 96 = 196 > 0 Hence, the roots of the equation are real. Roots α and β are given by: α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 10 + \sqrt{196}}{2 \sqrt{3}} = \frac{- 10 + 14}{2 \sqrt{3}} = \frac{4}{2 \sqrt{3}} = \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2 \sqrt{3}}{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 10 - \sqrt{196}}{2 \sqrt{3}} = \frac{- 10 - 14}{2 \sqrt{3}} = \frac{- 24}{2 \sqrt{3}} = \frac{- 12}{\sqrt{3}} = \frac{- 12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{- 12 \sqrt{3}}{3} = - 4 \sqrt{3} Thus, the roots of the equation are \frac{2 \sqrt{3}}{3} and - 4 \sqrt{3} .$

Page No 193:

Answer:

The given equation is $\sqrt{3} x^{2} - 2 \sqrt{2} x - 2 \sqrt{3} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = $\sqrt{3}$ , b = $- 2 \sqrt{2}$ and c = $- 2 \sqrt{3}$

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{2})}^{2} - 4 \times \sqrt{3} \times (- 2 \sqrt{3}) = 8 + 24 = 32 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{32} = 4 \sqrt{2}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) + 4 \sqrt{2}}{2 \times \sqrt{3}} = \frac{6 \sqrt{2}}{2 \sqrt{3}} = \sqrt{6} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) - 4 \sqrt{2}}{2 \times \sqrt{3}} = \frac{- 2 \sqrt{2}}{2 \sqrt{3}} = - \frac{\sqrt{6}}{3}$

Hence, $\sqrt{6}$ and $- \frac{\sqrt{6}}{3}$ are the roots of the given equation.

Page No 193:

Answer:

The given equation is $2 x^{2} + 6 \sqrt{3} x - 60 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 2, b = $6 \sqrt{3}$ and c = $- 60$

∴ Discriminant, D = $b^{2} - 4 a c = {(6 \sqrt{3})}^{2} - 4 \times 2 \times (- 60) = 108 + 480 = 588 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{588} = 14 \sqrt{3}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 6 \sqrt{3} + 14 \sqrt{3}}{2 \times 2} = \frac{8 \sqrt{3}}{4} = 2 \sqrt{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 6 \sqrt{3} - 14 \sqrt{3}}{2 \times 2} = \frac{- 20 \sqrt{3}}{4} = - 5 \sqrt{3}$

Hence, $2 \sqrt{3}$ and $- 5 \sqrt{3}$ are the roots of the given equation.

Page No 193:

Answer:

The given equation is $4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = $4 \sqrt{3}$ , b = 5 and c = $- 2 \sqrt{3}$

∴ Discriminant, D = $b^{2} - 4 a c = 5^{2} - 4 \times 4 \sqrt{3} \times (- 2 \sqrt{3}) = 25 + 96 = 121 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{121} = 11$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 5 + 11}{2 \times 4 \sqrt{3}} = \frac{6}{8 \sqrt{3}} = \frac{\sqrt{3}}{4} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 5 - 11}{2 \times 4 \sqrt{3}} = \frac{- 16}{8 \sqrt{3}} = - \frac{2 \sqrt{3}}{3}$

Hence, $\frac{\sqrt{3}}{4}$ and $- \frac{2 \sqrt{3}}{3}$ are the roots of the given equation.

Page No 193:

Answer:

The given equation is $3 x^{2} - 2 \sqrt{6} x + 2 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 3, b = $- 2 \sqrt{6}$ and c = 2

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{6})}^{2} - 4 \times 3 \times 2 = 24 - 24 = 0$

So, the given equation has real roots.

Now, $\sqrt{D} = 0$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{6}) + 0}{2 \times 3} = \frac{2 \sqrt{6}}{6} = \frac{\sqrt{6}}{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{6}) - 0}{2 \times 3} = \frac{2 \sqrt{6}}{6} = \frac{\sqrt{6}}{3}$

Hence, $\frac{\sqrt{6}}{3}$ is the repeated root of the given equation.

Page No 193:

Answer:

The given equation is $2 \sqrt{3} x^{2} - 5 x + \sqrt{3} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = $2 \sqrt{3}$ , b = $- 5$ and c = $\sqrt{3}$

∴ Discriminant, D = $b^{2} - 4 a c = {(- 5)}^{2} - 4 \times 2 \sqrt{3} \times \sqrt{3} = 25 - 24 = 1 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{1} = 1$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 5) + 1}{2 \times 2 \sqrt{3}} = \frac{6}{4 \sqrt{3}} = \frac{\sqrt{3}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 5) - 1}{2 \times 2 \sqrt{3}} = \frac{4}{4 \sqrt{3}} = \frac{\sqrt{3}}{3}$

Hence, $\frac{\sqrt{3}}{2}$ and $\frac{\sqrt{3}}{3}$ are the roots of the given equation.

Page No 193:

Answer:

The given equation is $x^{2} + x + 2 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 1, b = 1 and c = 2

∴ Discriminant, D = $b^{2} - 4 a c = 1^{2} - 4 \times 1 \times 2 = 1 - 8 = - 7 < 0$

Hence, the given equation has no real roots (or real roots does not exist).

Page No 193:

Answer:

The given equation is $2 x^{2} + a x - a^{2} = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 2, B = a and C = $- a^{2}$

∴ Discriminant, D = $B^{2} - 4 A C = a^{2} - 4 \times 2 \times - a^{2} = a^{2} + 8 a^{2} = 9 a^{2} \geq 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{9 a^{2}} = 3 a$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- a + 3 a}{2 \times 2} = \frac{2 a}{4} = \frac{a}{2} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- a - 3 a}{2 \times 2} = \frac{- 4 a}{4} = - a$

Hence, $\frac{a}{2}$ and $- a$ are the roots of the given equation.

Page No 193:

Answer:

The given equation is $x^{2} - (\sqrt{3} + 1) x + \sqrt{3} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 1, b = $- (\sqrt{3} + 1)$ and c = $\sqrt{3}$

∴ Discriminant, D = $b^{2} - 4 a c = {[- (\sqrt{3} + 1)]}^{2} - 4 \times 1 \times \sqrt{3} = 3 + 1 + 2 \sqrt{3} - 4 \sqrt{3} = 3 - 2 \sqrt{3} + 1 = {(\sqrt{3} - 1)}^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{{(\sqrt{3} - 1)}^{2}} = \sqrt{3} - 1$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- [- (\sqrt{3} + 1)] + (\sqrt{3} - 1)}{2 \times 1} = \frac{\sqrt{3} + 1 + \sqrt{3} - 1}{2} = \frac{2 \sqrt{3}}{2} = \sqrt{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- [- (\sqrt{3} + 1)] - (\sqrt{3} - 1)}{2 \times 1} = \frac{\sqrt{3} + 1 - \sqrt{3} + 1}{2} = \frac{2}{2} = 1$

Hence, $\sqrt{3}$ and 1 are the roots of the given equation.

Page No 193:

Answer:

The given equation is $2 x^{2} + 5 \sqrt{3} x + 6 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 2, b = $5 \sqrt{3}$ and c = 6

∴ Discriminant, D = $b^{2} - 4 a c = {(5 \sqrt{3})}^{2} - 4 \times 2 \times 6 = 75 - 48 = 27 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{27} = 3 \sqrt{3}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 5 \sqrt{3} + 3 \sqrt{3}}{2 \times 2} = \frac{- 2 \sqrt{3}}{4} = - \frac{\sqrt{3}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 5 \sqrt{3} - 3 \sqrt{3}}{2 \times 2} = \frac{- 8 \sqrt{3}}{4} = - 2 \sqrt{3}$

Hence, $- \frac{\sqrt{3}}{2}$ and $- 2 \sqrt{3}$ are the roots of the given equation.

Page No 193:

Answer:

The given equation is $3 x^{2} - 2 x + 2 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 3, b = −2 and c = 2

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2)}^{2} - 4 \times 3 \times 2 = 4 - 24 = - 20 < 0$

Hence, the given equation has no real roots (or real roots does not exist).

Page No 193:

Answer:

The given equation is

$x + \frac{1}{x} = 3, x \neq 0 \Rightarrow \frac{x^{2} + 1}{x} = 3 \Rightarrow x^{2} + 1 = 3 x \Rightarrow x^{2} - 3 x + 1 = 0$

This equation is of the form $a x^{2} + b x + c = 0$ , where a = 1, b = −3 and c = 1.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 3)}^{2} - 4 \times 1 \times 1 = 9 - 4 = 5 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{5}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 3) + \sqrt{5}}{2 \times 1} = \frac{3 + \sqrt{5}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 3) - \sqrt{5}}{2 \times 1} = \frac{3 - \sqrt{5}}{2}$

Hence, $\frac{3 + \sqrt{5}}{2}$ and $\frac{3 - \sqrt{5}}{2}$ are the roots of the given equation.

Page No 193:

Answer:

The given equation is

$\frac{1}{x} - \frac{1}{x - 2} = 3, x \neq 0, 2 \Rightarrow \frac{x - 2 - x}{x (x - 2)} = 3 \Rightarrow \frac{- 2}{x^{2} - 2 x} = 3 \Rightarrow - 2 = 3 x^{2} - 6 x$

$\Rightarrow 3 x^{2} - 6 x + 2 = 0$

This equation is of the form $a x^{2} + b x + c = 0$ , where a = 3, b = −6 and c = 2.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 6)}^{2} - 4 \times 3 \times 2 = 36 - 24 = 12 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{12} = 2 \sqrt{3}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 6) + 2 \sqrt{3}}{2 \times 3} = \frac{6 + 2 \sqrt{3}}{6} = \frac{3 + \sqrt{3}}{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 6) - 2 \sqrt{3}}{2 \times 3} = \frac{6 - 2 \sqrt{3}}{6} = \frac{3 - \sqrt{3}}{3}$

Hence, $\frac{3 + \sqrt{3}}{3}$ and $\frac{3 - \sqrt{3}}{3}$ are the roots of the given equation.

Page No 193:

Answer:

The given equation is

$x - \frac{1}{x} = 3, x \neq 0 \Rightarrow \frac{x^{2} - 1}{x} = 3 \Rightarrow x^{2} - 1 = 3 x \Rightarrow x^{2} - 3 x - 1 = 0$

This equation is of the form $a x^{2} + b x + c = 0$ , where a = 1, b = −3 and c = −1.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 3)}^{2} - 4 \times 1 \times (- 1) = 9 + 4 = 13 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{13}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 3) + \sqrt{13}}{2 \times 1} = \frac{3 + \sqrt{13}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 3) - \sqrt{13}}{2 \times 1} = \frac{3 - \sqrt{13}}{2}$

Hence, $\frac{3 + \sqrt{13}}{2}$ and $\frac{3 - \sqrt{13}}{2}$ are the roots of the given equation.

Page No 193:

Answer:

The given equation is

$\frac{m}{n} x^{2} + \frac{n}{m} = 1 - 2 x \Rightarrow \frac{m^{2} x^{2} + n^{2}}{m n} = 1 - 2 x \Rightarrow m^{2} x^{2} + n^{2} = m n - 2 m n x \Rightarrow m^{2} x^{2} + 2 m n x + n^{2} - m n = 0$

This equation is of the form $a x^{2} + b x + c = 0$ , where a = $m^{2}$ , b = 2mn and c = $n^{2} - m n$ .

∴ Discriminant, D = $b^{2} - 4 a c = {(2 m n)}^{2} - 4 \times m^{2} \times (n^{2} - m n) = 4 m^{2} n^{2} - 4 m^{2} n^{2} + 4 m^{3} n = 4 m^{3} n > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{4 m^{3} n} = 2 m \sqrt{m n}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 2 m n + 2 m \sqrt{m n}}{2 \times m^{2}} = \frac{2 m (- n + \sqrt{m n})}{2 m^{2}} = \frac{- n + \sqrt{m n}}{m} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 2 m n - 2 m \sqrt{m n}}{2 \times m^{2}} = \frac{- 2 m (n + \sqrt{m n})}{2 m^{2}} = \frac{- n - \sqrt{m n}}{m}$

Hence, $\frac{- n + \sqrt{m n}}{m}$ and $\frac{- n - \sqrt{m n}}{m}$ are the roots of the given equation.

Page No 193:

Answer:

The given equation is $36 x^{2} - 12 a x + (a^{2} - b^{2}) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 36, B = $- 12 a$ and C = $a^{2} - b^{2}$

∴ Discriminant, D = $B^{2} - 4 A C = {(- 12 a)}^{2} - 4 \times 36 \times (a^{2} - b^{2}) = 144 a^{2} - 144 a^{2} + 144 b^{2} = 144 b^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{144 b^{2}} = 12 b$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- (- 12 a) + 12 b}{2 \times 36} = \frac{12 (a + b)}{72} = \frac{a + b}{6} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- (- 12 a) - 12 b}{2 \times 36} = \frac{12 (a - b)}{72} = \frac{a - b}{6}$

Hence, $\frac{a + b}{6}$ and $\frac{a - b}{6}$ are the roots of the given equation.

Page No 193:

Answer:

$Given: x^{2} - 2 a x + (a^{2} - b^{2}) = 0 On c omparing it with A x^{2} + B x + C = 0, we get: A = 1, B = - 2 a and C = (a^{2} - b^{2}) Discriminant D is given by : D = B^{2} - 4 A C = {(- 2 a)}^{2} - 4 \times 1 \times (a^{2} - b^{2}) = 4 a^{2} - 4 a^{2} + 4 b^{2} = 4 b^{2} > 0 Hence, the roots of the equation are real. Roots α and β are given by: α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 a) + \sqrt{4 b^{2}}}{2 \times 1} = \frac{2 a + 2 b}{2} = \frac{2 (a + b)}{2} = (a + b) β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 a) - \sqrt{4 b^{2}}}{2 \times 1} = \frac{2 a - 2 b}{2} = \frac{2 (a - b)}{2} = (a - b) Hence, the roots of the equation are (a + b) and (a - b) .$

Page No 193:

Answer:

The given equation is $x^{2} - 2 a x - (4 b^{2} - a^{2}) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = $- 2 a$ and C = $- (4 b^{2} - a^{2})$

∴ Discriminant, D = $B^{2} - 4 A C = {(- 2 a)}^{2} - 4 \times 1 \times [- (4 b^{2} - a^{2})] = 4 a^{2} + 16 b^{2} - 4 a^{2} = 16 b^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{16 b^{2}} = 4 b$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- (- 2 a) + 4 b}{2 \times 1} = \frac{2 (a + 2 b)}{2} = a + 2 b β = \frac{- B - \sqrt{D}}{2 A} = \frac{- (- 2 a) - 4 b}{2 \times 1} = \frac{2 (a - 2 b)}{2} = a - 2 b$

Hence, $a + 2 b$ and $a - 2 b$ are the roots of the given equation.

Page No 194:

Answer:

The given equation is $x^{2} + 6 x - (a^{2} + 2 a - 8) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = 6 and C = $- (a^{2} + 2 a - 8)$

∴ Discriminant, D = $B^{2} - 4 A C = 6^{2} - 4 \times 1 \times [- (a^{2} + 2 a - 8)] = 36 + 4 a^{2} + 8 a - 32 = 4 a^{2} + 8 a + 4 = 4 (a^{2} + 2 a + 1) = 4 {(a + 1)}^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{4 {(a + 1)}^{2}} = 2 (a + 1)$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- 6 + 2 (a + 1)}{2 \times 1} = \frac{2 a - 4}{2} = a - 2 β = \frac{- B - \sqrt{D}}{2 A} = \frac{- 6 - 2 (a + 1)}{2 \times 1} = \frac{- 2 a - 8}{2} = - a - 4 = - (a + 4)$

Hence, $(a - 2)$ and $- (a + 4)$ are the roots of the given equation.

Page No 194:

Answer:

The given equation is $x^{2} + 5 x - (a^{2} + a - 6) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = 5 and C = $- (a^{2} + a - 6)$

∴ Discriminant, D = $B^{2} - 4 A C = 5^{2} - 4 \times 1 \times [- (a^{2} + a - 6)] = 25 + 4 a^{2} + 4 a - 24 = 4 a^{2} + 4 a + 1 = {(2 a + 1)}^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{{(2 a + 1)}^{2}} = 2 a + 1$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- 5 + 2 a + 1}{2 \times 1} = \frac{2 a - 4}{2} = a - 2 β = \frac{- B - \sqrt{D}}{2 A} = \frac{- 5 - (2 a + 1)}{2 \times 1} = \frac{- 2 a - 6}{2} = - a - 3 = - (a + 3)$

Hence, $(a - 2)$ and $- (a + 3)$ are the roots of the given equation.

Page No 194:

Answer:

The given equation is $x^{2} - 4 a x - b^{2} + 4 a^{2} = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = −4a and C = $- b^{2} + 4 a^{2}$

∴ Discriminant, D = $B^{2} - 4 A C = {(- 4 a)}^{2} - 4 \times 1 \times (- b^{2} + 4 a^{2}) = 16 a^{2} + 4 b^{2} - 16 a^{2} = 4 b^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{4 b^{2}} = 2 b$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- (- 4 a) + 2 b}{2 \times 1} = \frac{4 a + 2 b}{2} = 2 a + b β = \frac{- B - \sqrt{D}}{2 A} = \frac{- (- 4 a) - 2 b}{2 \times 1} = \frac{4 a - 2 b}{2} = 2 a - b$

Hence, $(2 a + b)$ and $(2 a - b)$ are the roots of the given equation.

Page No 194:

Answer:

The given equation is $4 x^{2} - 4 a^{2} x + (a^{4} - b^{4}) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 4, B = −4a² and C = $a^{4} - b^{4}$

∴ Discriminant, D = $B^{2} - 4 A C = {(- 4 a^{2})}^{2} - 4 \times 4 \times (a^{4} - b^{4}) = 16 a^{4} - 16 a^{4} + 16 b^{4} = 16 b^{4} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{16 b^{4}} = 4 b^{2}$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- (- 4 a^{2}) + 4 b^{2}}{2 \times 4} = \frac{4 (a^{2} + b^{2})}{8} = \frac{a^{2} + b^{2}}{2} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- (- 4 a^{2}) - 4 b^{2}}{2 \times 4} = \frac{4 (a^{2} - b^{2})}{8} = \frac{a^{2} - b^{2}}{2}$

Hence, $\frac{1}{2} (a^{2} + b^{2})$ and $\frac{1}{2} (a^{2} - b^{2})$ are the roots of the given equation.

Page No 194:

Answer:

The given equation is $4 x^{2} + 4 b x - (a^{2} - b^{2}) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 4, B = 4b and C = $- (a^{2} - b^{2})$

∴ Discriminant, D = $B^{2} - 4 A C = {(4 b)}^{2} - 4 \times 4 \times [- (a^{2} - b^{2})] = 16 b^{2} + 16 a^{2} - 16 b^{2} = 16 a^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{16 a^{2}} = 4 a$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- 4 b + 4 a}{2 \times 4} = \frac{4 (a - b)}{8} = \frac{a - b}{2} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- 4 b - 4 a}{2 \times 4} = \frac{- 4 (a + b)}{8} = - \frac{a + b}{2}$

Hence, $\frac{1}{2} (a - b)$ and $- \frac{1}{2} (a + b)$ are the roots of the given equation.

Page No 194:

Answer:

The given equation is $x^{2} - (2 b - 1) x + (b^{2} - b - 20) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = $- (2 b - 1)$ and C = $b^{2} - b - 20$

∴ Discriminant, D = $B^{2} - 4 A C = {[- (2 b - 1)]}^{2} - 4 \times 1 \times (b^{2} - b - 20) = 4 b^{2} - 4 b + 1 - 4 b^{2} + 4 b + 80 = 81 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{81} = 9$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- [- (2 b - 1)] + 9}{2 \times 1} = \frac{2 b + 8}{2} = b + 4 β = \frac{- B - \sqrt{D}}{2 A} = \frac{- [- (2 b - 1)] - 9}{2 \times 1} = \frac{2 b - 10}{2} = b - 5$

Hence, $(b + 4)$ and $(b - 5)$ are the roots of the given equation.

Page No 194:

Answer:

$Given: {3a}^{2} x^{2} + 8 a b x + 4 b^{2} = 0 On c omparing it with A x^{2} + B x + C = 0, we get: A = 3 a^{2}, B = 8 a b and C = 4 b^{2} Discriminant D is given by : D = (B^{2} - 4 A C) = {(8 a b)}^{2} - 4 \times 3 a^{2} \times 4 b^{2} {= 16a}^{2} b^{2} > 0 Hence, the roots of the equation are real. Roots α and β are given by: α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 8 a b + \sqrt{16 a^{2} b^{2}}}{2 \times 3 a^{2}} = \frac{- 8 a b + 4 a b}{6 a^{2}} = \frac{- 4 a b}{6 a^{2}} = \frac{- 2 b}{3 a} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 8 a b - \sqrt{16 a^{2} b^{2}}}{2 \times 3 a^{2}} = \frac{- 8 a b - 4 a b}{6 a^{2}} = \frac{- 12 a b}{6 a^{2}} = \frac{- 2 b}{a} Thus, the roots of the equation are \frac{- 2 b}{3 a} and \frac{- 2 b}{a} .$

Page No 194:

Answer:

The given equation is $a^{2} b^{2} x^{2} - (4 b^{4} - 3 a^{4}) x - 12 a^{2} b^{2} = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = $a^{2} b^{2}$ , B = $- (4 b^{4} - 3 a^{4})$ and C = $- 12 a^{2} b^{2}$

∴ Discriminant, D = $B^{2} - 4 A C = {[- (4 b^{4} - 3 a^{4})]}^{2} - 4 \times a^{2} b^{2} \times (- 12 a^{2} b^{2}) = 16 b^{8} - 24 a^{4} b^{4} + 9 a^{8} + 48 a^{4} b^{4} = 16 b^{8} + 24 a^{4} b^{4} + 9 a^{8} = {(4 b^{4} + 3 a^{4})}^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{{(4 b^{4} + 3 a^{4})}^{2}} = 4 b^{4} + 3 a^{4}$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- [- (4 b^{4} - 3 a^{4})] + (4 b^{4} + 3 a^{4})}{2 \times a^{2} b^{2}} = \frac{8 b^{4}}{2 a^{2} b^{2}} = \frac{4 b^{2}}{a^{2}} β = \frac{- B - \sqrt{D}}{2 A} = = \frac{- [- (4 b^{4} - 3 a^{4})] - (4 b^{4} + 3 a^{4})}{2 \times a^{2} b^{2}} = \frac{- 6 a^{4}}{2 a^{2} b^{2}} = - \frac{3 a^{2}}{b^{2}}$

Hence, $\frac{4 b^{2}}{a^{2}}$ and $- \frac{3 a^{2}}{b^{2}}$ are the roots of the given equation.

Page No 194:

Answer:

$Given: 12 a b x^{2} - (9 a^{2} - 8 b^{2}) x - 6 a b = 0 On c omparing it with A x^{2} + B x + C = 0, we get: A = 12 a b, B = - (9 a^{2} - 8 b^{2}) and C = - 6 a b Discriminant D is given by : D = B^{2} - 4 A C = {[- (9 a^{2} - 8 b^{2})]}^{2} - 4 \times 12 a b \times (- 6 a b) = 81 a^{4} - 144 a^{2} b^{2} + 64 b^{4} + 288 a^{2} b^{2} = 81 a^{^{4}} + 144 a^{2} b^{2} + 64 b^{4} = {(9 a^{2} + 8 b^{2})}^{2} > 0 Hence, the roots of the equation are equal. Roots α and β are given by: α = \frac{- B + \sqrt{D}}{2 A} = \frac{- [- (9 a^{2} - 8 b^{2})] + \sqrt{{(9 a^{2} + 8 b^{2})}^{2}}}{2 \times 12 a b} = \frac{9 a^{2} - 8 b^{2} + 9 a^{2} + 8 b^{2}}{24 a b} = \frac{18 a^{2}}{24 a b} = \frac{3 a}{4 b} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- [- (9 a^{2} - 8 b^{2})] - \sqrt{{(9 a^{2} + 8 b^{2})}^{2}}}{2 \times 12 a b} = \frac{9 a^{2} - 8 b^{2} - 9 a^{2} - 8 b^{2}}{24 a b} = \frac{- 16 b^{2}}{24 a b} = \frac{- 2 b}{3 a} Thus, the roots of the equation are \frac{3 a}{4 b} and \frac{- 2 b}{3 a} .$

Page No 201:

Answer:

(i) The given equation is $2 x^{2} - 8 x + 5 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 2, b = −8 and c = 5.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 8)}^{2} - 4 \times 2 \times 5 = 64 - 40 = 24 > 0$

Hence, the given equation has real and unequal roots.

(ii) The given equation is $3 x^{2} - 2 \sqrt{6} x + 2 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 3, b = $- 2 \sqrt{6}$ and c = 2.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{6})}^{2} - 4 \times 3 \times 2 = 24 - 24 = 0$

Hence, the given equation has real and equal roots.

(iii) The given equation is $5 x^{2} - 4 x + 1 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 5, b = −4 and c = 1.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 4)}^{2} - 4 \times 5 \times 1 = 16 - 20 = - 4 < 0$

Hence, the given equation has no real roots.

(iv) The given equation is

$5 x (x - 2) + 6 = 0 \Rightarrow 5 x^{2} - 10 x + 6 = 0$

This is of the form $a x^{2} + b x + c = 0$ , where a = 5, b = −10 and c = 6.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 10)}^{2} - 4 \times 5 \times 6 = 100 - 120 = - 20 < 0$

Hence, the given equation has no real roots.

(v) The given equation is $12 x^{2} - 4 \sqrt{15} x + 5 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 12, b = $- 4 \sqrt{15}$ and c = 5.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 4 \sqrt{15})}^{2} - 4 \times 12 \times 5 = 240 - 240 = 0$

Hence, the given equation has real and equal roots.

(vi) The given equation is $x^{2} - x + 2 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 1, b = −1 and c = 2.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 1)}^{2} - 4 \times 1 \times 2 = 1 - 8 = - 7 < 0$

Hence, the given equation has no real roots.

Page No 201:

Answer:

The given equation is $2 (a^{2} + b^{2}) x^{2} + 2 (a + b) x + 1 = 0$ .
$∴ D = {[2 (a + b)]}^{2} - 4 \times 2 (a^{2} + b^{2}) \times 1 = 4 (a^{2} + 2 a b + b^{2}) - 8 (a^{2} + b^{2}) = 4 a^{2} + 8 a b + 4 b^{2} - 8 a^{2} - 8 b^{2} = - 4 a^{2} + 8 a b - 4 b^{2} = - 4 (a^{2} - 2 a b + b^{2}) = - 4 {(a - b)}^{2} < 0$
Hence, the given equation has no real roots.

Page No 202:

Answer:

$Given: x^{2} + p x - q^{2} = 0 Here, a = 1, b = p and c = - q^{2} Discriminant D is given by : D = (b^{2} - 4 a c) = p^{2} - 4 \times 1 \times (- q^{2}) = (p^{2} + 4 q^{2}) > 0 D > 0 for all real values of p and q . Thus, the roots of the equation are real .$

Page No 202:

Answer:

$Given: 3 x^{2} + 2 k x + 27 = 0 Here, a = 3, b = 2 k and c = 27 It is given that the roots of the equation are real and equal; therefore, we have: D = 0 \Rightarrow {(2 k)}^{2} - 4 \times 3 \times 27 = 0 \Rightarrow 4 k^{2} - 324 = 0 \Rightarrow 4 k^{2} = 324 \Rightarrow k^{2} = 81 \Rightarrow k = \pm 9 ∴ k = 9 or k = - 9$

Page No 202:

Answer:

The given equation is

$k x (x - 2 \sqrt{5}) + 10 = 0 \Rightarrow k x^{2} - 2 \sqrt{5} k x + 10 = 0$

This is of the form $a x^{2} + b x + c = 0$ , where a = k, b = $- 2 \sqrt{5} k$ and c = 10.

$∴ D = b^{2} - 4 a c = {(- 2 \sqrt{5} k)}^{2} - 4 \times k \times 10 = 20 k^{2} - 40 k$

The given equation will have real and equal roots if D = 0.

$∴ 20 k^{2} - 40 k = 0 \Rightarrow 20 k (k - 2) = 0 \Rightarrow k = 0 or k - 2 = 0 \Rightarrow k = 0 or k = 2$

But, for k = 0, we get 10 = 0, which is not true.

Hence, 2 is the required value of k.

Page No 202:

Answer:

The given equation is $4 x^{2} + p x + 3 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 4, b = p and c = 3.

$∴ D = b^{2} - 4 a c = p^{2} - 4 \times 4 \times 3 = p^{2} - 48$

The given equation will have real and equal roots if D = 0.

$∴ p^{2} - 48 = 0 \Rightarrow p^{2} = 48 \Rightarrow p = \pm \sqrt{48} = \pm 4 \sqrt{3}$

Hence, $4 \sqrt{3}$ and $- 4 \sqrt{3}$ are the required values of p.

Page No 202:

Answer:

The given equation is $9 x^{2} - 3 k x + k = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 9, b = −3k and c = k.

$∴ D = b^{2} - 4 a c = {(- 3 k)}^{2} - 4 \times 9 \times k = 9 k^{2} - 36 k$

The given equation will have real and equal roots if D = 0.

$∴ 9 k^{2} - 36 k = 0 \Rightarrow 9 k (k - 4) = 0 \Rightarrow k = 0 or k - 4 = 0 \Rightarrow k = 0 or k = 4$

But, k ≠ 0 (Given)

Hence, the required value of k is 4.

Page No 202:

Answer:

(i)
The given equation is $(3 k + 1) x^{2} + 2 (k + 1) x + 1 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 3k +1, b = 2(k + 1) and c = 1.

$∴ D = b^{2} - 4 a c = {[2 (k + 1)]}^{2} - 4 \times (3 k + 1) \times 1 = 4 (k^{2} + 2 k + 1) - 4 (3 k + 1) = 4 k^{2} + 8 k + 4 - 12 k - 4$
$= 4 k^{2} - 4 k$

The given equation will have real and equal roots if D = 0.

$∴ 4 k^{2} - 4 k = 0 \Rightarrow 4 k (k - 1) = 0 \Rightarrow k = 0 or k - 1 = 0 \Rightarrow k = 0 or k = 1$

Hence, 0 and 1 are the required values of k.

(ii)
$x^{2} + k (2 x + k - 1) + 2 = 0 x^{2} + 2 k x + k^{2} - k + 2 = 0 For real and equal roots D = 0 D = b^{2} - 4 a c = 0 D = {(2 k)}^{2} - 4 (1) (k^{2} - k + 2) D = 4 k^{2} - 4 k^{2} + 4 k - 8 = 0 \Rightarrow 4 k - 8 = 0 \Rightarrow k = 2$

Page No 202:

Answer:

The given equation is $(2 p + 1) x^{2} - (7 p + 2) x + (7 p - 3) = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 2p +1, b = −(7p + 2) and c = 7p − 3.

$∴ D = b^{2} - 4 a c = {[- (7 p + 2)]}^{2} - 4 \times (2 p + 1) \times (7 p - 3) = (49 p^{2} + 28 p + 4) - 4 (14 p^{2} + p - 3) = 49 p^{2} + 28 p + 4 - 56 p^{2} - 4 p + 12$
$= - 7 p^{2} + 24 p + 16$

The given equation will have real and equal roots if D = 0.

$∴ - 7 p^{2} + 24 p + 16 = 0 \Rightarrow 7 p^{2} - 24 p - 16 = 0 \Rightarrow 7 p^{2} - 28 p + 4 p - 16 = 0 \Rightarrow 7 p (p - 4) + 4 (p - 4) = 0$
$\Rightarrow (p - 4) (7 p + 4) = 0 \Rightarrow p - 4 = 0 or 7 p + 4 = 0 \Rightarrow p = 4 or p = - \frac{4}{7}$

Hence, 4 and $- \frac{4}{7}$ are the required values of p.

Page No 202:

Answer:

The given equation is $(p + 1) x^{2} - 6 (p + 1) x + 3 (p + 9) = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = p +1, b = −6(p + 1) and c = 3(p + 9).

$∴ D = b^{2} - 4 a c = {[- 6 (p + 1)]}^{2} - 4 \times (p + 1) \times 3 (p + 9) = 12 (p + 1) [3 (p + 1) - (p + 9)] = 12 (p + 1) (2 p - 6)$

The given equation will have real and equal roots if D = 0.

$∴ 12 (p + 1) (2 p - 6) = 0 \Rightarrow p + 1 = 0 or 2 p - 6 = 0 \Rightarrow p = - 1 or p = 3$

But, $p \neq - 1$ (Given)

Thus, the value of p is 3.

Putting p = 3, the given equation becomes $4 x^{2} - 24 x + 36 = 0$ .

$4 x^{2} - 24 x + 36 = 0 \Rightarrow 4 (x^{2} - 6 x + 9) = 0 \Rightarrow {(x - 3)}^{2} = 0 \Rightarrow x - 3 = 0$
$\Rightarrow x = 3$

Hence, 3 is the repeated root of this equation.

Page No 202:

Answer:

It is given that −5 is a root of the quadratic equation $2 x^{2} + p x - 15 = 0$ .

$∴ 2 {(- 5)}^{2} + p \times (- 5) - 15 = 0 \Rightarrow - 5 p + 35 = 0 \Rightarrow p = 7$

The roots of the equation $p x^{2} + p x + k = 0$ = 0 are equal.

$∴ D = 0 \Rightarrow p^{2} - 4 p k = 0 \Rightarrow {(7)}^{2} - 4 \times 7 \times k = 0 \Rightarrow 49 - 28 k = 0$
$\Rightarrow k = \frac{49}{28} = \frac{7}{4}$

Thus, the value of k is $\frac{7}{4}$ .

Page No 202:

Answer:

It is given that 3 is a root of the quadratic equation $x^{2} - x + k = 0$ .

$∴ {(3)}^{2} - 3 + k = 0 \Rightarrow k + 6 = 0 \Rightarrow k = - 6$

The roots of the equation $x^{2} + 2 k x + (k^{2} + 2 k + p) = 0$ are equal.

$∴ D = 0 \Rightarrow {(2 k)}^{2} - 4 \times 1 \times (k^{2} + 2 k + p) = 0 \Rightarrow 4 k^{2} - 4 k^{2} - 8 k - 4 p = 0 \Rightarrow - 8 k - 4 p = 0$
$\Rightarrow p = \frac{8 k}{- 4} = - 2 k \Rightarrow p = - 2 \times (- 6) = 12$

Hence, the value of p is 12.

Page No 202:

Answer:

It is given that −4 is a root of the quadratic equation $x^{2} + 2 x + 4 p = 0$ .

$∴ {(- 4)}^{2} + 2 \times (- 4) + 4 p = 0 \Rightarrow 16 - 8 + 4 p = 0 \Rightarrow 4 p + 8 = 0 \Rightarrow p = - 2$

The equation $x^{2} + p x (1 + 3 k) + 7 (3 + 2 k) = 0$ has equal roots.

$∴ D = 0 \Rightarrow {[p (1 + 3 k)]}^{2} - 4 \times 1 \times 7 (3 + 2 k) = 0 \Rightarrow {[- 2 (1 + 3 k)]}^{2} - 28 (3 + 2 k) = 0 \Rightarrow 4 (1 + 6 k + 9 k^{2}) - 28 (3 + 2 k) = 0$
$\Rightarrow 4 (1 + 6 k + 9 k^{2} - 21 - 14 k) = 0 \Rightarrow 9 k^{2} - 8 k - 20 = 0 \Rightarrow 9 k^{2} - 18 k + 10 k - 20 = 0 \Rightarrow 9 k (k - 2) + 10 (k - 2) = 0$
$\Rightarrow (k - 2) (9 k + 10) = 0 \Rightarrow k - 2 = 0 or 9 k + 10 = 0 \Rightarrow k = 2 or k = - \frac{10}{9}$

Hence, the required value of k is 2 or $- \frac{10}{9}$ .

Page No 202:

Answer:

$Given: (1 + m^{2}) x^{2} + 2 m c x + (c^{2} - a^{2}) = 0 Here, a = (1 + m^{2}), b = 2 m c and c = (c^{2} - a^{2}) It is given that the roots of the equation are equal; therefore, we have: D = 0 \Rightarrow (b^{2} - 4 a c) = 0 \Rightarrow {(2 m c)}^{2} - 4 \times (1 + m^{2}) \times (c^{2} - a^{2}) = 0 \Rightarrow 4 m^{2} c^{2} - 4 (c^{2} - a^{2} + m^{2} c^{2} - m^{2} a^{2}) = 0 \Rightarrow 4 m^{2} c^{2} - 4 c^{2} + 4 a^{2} - 4 m^{2} c^{2} + 4 m^{2} a^{2} = 0 \Rightarrow - 4 c^{2} + 4 a^{2} + 4 m^{2} a^{2} = 0 \Rightarrow a^{2} + m^{2} a^{2} = c^{2} \Rightarrow a^{2} (1 + m^{2}) = c^{2} \Rightarrow c^{2} = a^{2} (1 + m^{2}) Hence proved .$

Page No 202:

Answer:

$Given: (c^{2} - a b) x^{2} - 2 (a^{2} - b c) x + (b^{2} - a c) = 0 Here, a = (c^{2} - a b), b = - 2 (a^{2} - b c), c = (b^{2} - a c) It is given that the roots of the equation are real and equal; therefore, we have: D = 0 \Rightarrow (b^{2} - 4 a c) = 0 \Rightarrow {- 2 (a^{2} - b c)}^{2} - 4 \times (c^{2} - a b) \times (b^{2} - a c) = 0 \Rightarrow 4 (a^{4} - 2 a^{2} b c + b^{2} c^{2}) - 4 (b^{2} c^{2} - a c^{3} - a b^{3} + a^{2} b c) = 0 \Rightarrow a^{4} - 2 a^{2} b c + b^{2} c^{2} - b^{2} c^{2} + a c^{3} + a b^{3} - a^{2} b c = 0 \Rightarrow a^{4} - 3 a^{2} b c + a c^{3} + a b^{3} = 0 \Rightarrow a (a^{3} - 3 a b c + c^{3} + b^{3}) = 0 Now, a = 0 or a^{3} - 3 a b c + c^{3} + b^{3} = 0 a = 0 {or a}^{3} + b^{3} + c^{3} = 3 a b c$

Page No 202:

Answer:

$Given: 2 x^{2} + p x + 8 = 0 Here, a = 2, b = p and c = 8 Discriminant D is given by : D = (b^{2} - 4 a c) = p^{2} - 4 \times 2 \times 8 = (p^{2} - 64) If D \geq 0, t he roots of the equation will be real. \Rightarrow (p^{2} - 64) \geq 0 \Rightarrow (p + 8) (p - 8) \geq 0 \Rightarrow p \geq 8 and p \leq - 8 T hus, the roots of the equation are real for p \geq 8 and p \leq - 8 .$

Page No 202:

Answer:

$Given: (α - 12) x^{2} + 2 (α - 12) x + 2 = 0 Here, a = (α - 12), b = 2 (α - 12) and c = 2 It is given that the roots of the equation are equal; therefore, we have: D = 0 \Rightarrow (b^{2} - 4 a c) = 0 \Rightarrow {2 (α - 12)}^{2} - 4 \times (α - 12) \times 2 = 0 \Rightarrow 4 (α^{2} - 24 α + 144) - 8 (α - 12) = 0 \Rightarrow 4 α^{2} - 96 α + 576 - 8 α + 96 = 0 \Rightarrow 4 α^{2} - 104 α + 672 = 0 \Rightarrow α^{2} - 26 α + 168 = 0 \Rightarrow α^{2} - 14 α - 12 α + 168 = 0 \Rightarrow α (α - 14) - 12 (α - 14) = 0 \Rightarrow (α - 14) (α - 12) = 0 ∴ α = 14 or α = 12 If the value of α is 12, the given equation becomes non-quadratic. Therefore, the valueof α will be 14 for the equation to have equal roots.$

Page No 203:

Answer:

$Given: 9 x^{2} + 8 k x + 16 = 0 Here, a = 9, b = 8 k and c = 16 It is given that the roots of the equation are real and equal; therefore, we have: D = 0 \Rightarrow (b^{2} - 4 a c) = 0 \Rightarrow {(8 k)}^{2} - 4 \times 9 \times 16 = 0 \Rightarrow 64 k^{2} - 576 = 0 \Rightarrow 64 k^{2} = 576 \Rightarrow k^{2} = 9 \Rightarrow k = \pm 3 ∴ k = 3 or k = - 3$

Page No 203:

Answer:

(i) The given equation is $k x^{2} + 6 x + 1 = 0$ .

$∴ D = 6^{2} - 4 \times k \times 1 = 36 - 4 k$

The given equation has real and distinct roots if D > 0.

$∴ 36 - 4 k > 0 \Rightarrow 4 k < 36 \Rightarrow k < 9$

(ii) The given equation is $x^{2} - k x + 9 = 0$ .

$∴ D = {(- k)}^{2} - 4 \times 1 \times 9 = k^{2} - 36$

The given equation has real and distinct roots if D > 0.

$∴ k^{2} - 36 > 0 \Rightarrow (k - 6) (k + 6) > 0 \Rightarrow k < - 6 or k > 6$

(iii) The given equation is $9 x^{2} + 3 k x + 4 = 0$ .

$∴ D = {(3 k)}^{2} - 4 \times 9 \times 4 = 9 k^{2} - 144$

The given equation has real and distinct roots if D > 0.

$∴ 9 k^{2} - 144 > 0 \Rightarrow 9 (k^{2} - 16) > 0 \Rightarrow (k - 4) (k + 4) > 0 \Rightarrow k < - 4 or k > 4$

(iv) The given equation is $5 x^{2} - k x + 1 = 0$ .

$∴ D = {(- k)}^{2} - 4 \times 5 \times 1 = k^{2} - 20$

The given equation has real and distinct roots if D > 0.

$∴ k^{2} - 20 > 0 \Rightarrow k^{2} - {(2 \sqrt{5})}^{2} > 0 \Rightarrow (k - 2 \sqrt{5}) (k + 2 \sqrt{5}) > 0 \Rightarrow k < - 2 \sqrt{5} or k > 2 \sqrt{5}$

Page No 203:

Answer:

The given equation is $(a - b) x^{2} + 5 (a + b) x - 2 (a - b) = 0$ .

$∴ D = {[5 (a + b)]}^{2} - 4 \times (a - b) \times [- 2 (a - b)] = 25 {(a + b)}^{2} + 8 {(a - b)}^{2}$
Since a and b are real and a ≠ b, so ${(a - b)}^{2} > 0$ and ${(a + b)}^{2} > 0$ .

∴ $8 {(a - b)}^{2} > 0$ .....(1)               (Product of two positive numbers is always positive)

Also, $25 {(a + b)}^{2} > 0$              .....(2)              (Product of two positive numbers is always positive)

Adding (1) and (2), we get

$25 {(a + b)}^{2} + 8 {(a - b)}^{2} > 0$                  (Sum of two positive numbers is always positive)

$\Rightarrow D > 0$

Hence, the roots of the given equation are real and unequal.

Page No 203:

Answer:

Ans

Page No 203:

Answer:

Ans

Page No 203:

Answer:

It is given that the roots of the equation $a x^{2} + 2 b x + c = 0$ are real.
$∴ D_{1} = {(2 b)}^{2} - 4 \times a \times c \geq 0 \Rightarrow 4 (b^{2} - a c) \geq 0 \Rightarrow b^{2} - a c \geq 0 . . . . . (1)$

Also, the roots of the equation $b x^{2} - 2 \sqrt{a c} x + b = 0$ are real.
$∴ D_{2} = {(- 2 \sqrt{a c})}^{2} - 4 \times b \times b \geq 0 \Rightarrow 4 (a c - b^{2}) \geq 0 \Rightarrow - 4 (b^{2} - a c) \geq 0$
$\Rightarrow b^{2} - a c \leq 0 . . . . . (2)$

The roots of the given equations are simultaneously real if (1) and (2) holds true together. This is possible if
$b^{2} - a c = 0 \Rightarrow b^{2} = a c$

Page No 203:

Answer:

Since, $x = \frac{- 1}{2}$ is a root of the quadratic equation 3x² + 2kx + 3 = 0,
then, it must satisfies the equation.

$3 {(- \frac{1}{2})}^{2} + 2 k (- \frac{1}{2}) + 3 = 0 \Rightarrow 3 (\frac{1}{4}) - k + 3 = 0 \Rightarrow \frac{3}{4} - k + 3 = 0 \Rightarrow \frac{3 - 4 k + 12}{4} = 0 \Rightarrow 3 - 4 k + 12 = 0 \Rightarrow 4 k = 15 \Rightarrow k = \frac{15}{4} Hence, the value of k is \frac{15}{4} .$

Page No 203:

Answer:

Since, $x = \frac{- 1}{3}$ is a root of the quadratic equation 2x² + 2x + k = 0,
then, it must satisfies the equation.

$2 {(- \frac{1}{3})}^{2} + 2 (- \frac{1}{3}) + k = 0 \Rightarrow 2 (\frac{1}{9}) - \frac{2}{3} + k = 0 \Rightarrow \frac{2}{9} - \frac{2}{3} + k = 0 \Rightarrow \frac{2 - 6 + 9 k}{9} = 0 \Rightarrow - 4 + 9 k = 0 \Rightarrow 9 k = 4 \Rightarrow k = \frac{4}{9} Hence, the value of k is \frac{4}{9} .$

Page No 203:

Answer:

Given: x² – 8x + 18 = 0

$x^{2} - 8 x + 18 = 0 Adding and subtracting {(\frac{1}{2} \times 8)}^{2}, we get \Rightarrow x^{2} - 8 x + 18 + 4^{2} - 4^{2} = 0 \Rightarrow x^{2} - 8 x + 16 + 18 - 16 = 0 \Rightarrow {(x - 4)}^{2} + 2 = 0 \Rightarrow {(x - 4)}^{2} = - 2 \Rightarrow (x - 4) = \pm \sqrt{- 2} But, \sqrt{- 2} is not a real number .$

Hence, the quadratic equation x² – 8x + 18 = 0 has no real solution.

Page No 203:

Answer:

Let 4x² –12x – k = 0 be a quadratic equation.

It is given that, it has no real roots.

$\Rightarrow Discriminant < 0 \Rightarrow b^{2} - 4 a c < 0 \Rightarrow {(- 12)}^{2} - 4 (4) (- k) < 0 \Rightarrow 144 + 16 k < 0 \Rightarrow 16 k < - 144 \Rightarrow k < - 9$

Hence, the values of k must be less than –9.

Page No 203:

Answer:

Let one root be $α$ and the other root be $\frac{1}{α}$ .

The given equation is 3x²– 10x + k = 0.

Product of roots = $\frac{k}{3}$
$\Rightarrow α \times \frac{1}{α} = \frac{k}{3} \Rightarrow 1 = \frac{k}{3} \Rightarrow k = 3$

Hence, the value of k for which the roots of the equation 3x² – 10x + k = 0 are reciprocal of each other is 3.

Page No 203:

Answer:

Let one root be $α$ and the other root be $\frac{1}{α}$ .

The given equation is 5x² +13x + k = 0.

Product of roots = $\frac{k}{5}$
$\Rightarrow α \times \frac{1}{α} = \frac{k}{5} \Rightarrow 1 = \frac{k}{5} \Rightarrow k = 5$

Hence, the value of k is 5.

Page No 203:

Answer:

Let 3x² + kx + 3 = 0 be a quadratic equation.

It is given that, it has real and equal roots.

$\Rightarrow Discriminant = 0 \Rightarrow b^{2} - 4 a c = 0 \Rightarrow {(k)}^{2} - 4 (3) (3) = 0 \Rightarrow k^{2} = 36 \Rightarrow k = \pm 6$

Hence, the values of k are –6 and 6.

Page No 203:

Answer:

Let 9x² – 3ax + 1 = 0 be a quadratic equation.

It is given that, it has real and equal roots.

$\Rightarrow Discriminant = 0 \Rightarrow b^{2} - 4 a c = 0 \Rightarrow {(- 3 a)}^{2} - 4 (9) (1) = 0 \Rightarrow 9 a^{2} = 36 \Rightarrow a^{2} = 4 \Rightarrow a = \pm 2$

Hence, the values of a are –2 and 2.

Page No 203:

Answer:

Let x² + k(2x + k – 1) + 2 = 0 be a quadratic equation.

x² + k(2x + k – 1) + 2 = 0
x² + 2xk + k² – k + 2 = 0

It is given that, it has real and equal roots.

$\Rightarrow Discriminant = 0 \Rightarrow b^{2} - 4 a c = 0 \Rightarrow {(2 k)}^{2} - 4 (1) (k^{2} - k + 2) = 0 \Rightarrow 4 k^{2} - 4 k^{2} + 4 k - 8 = 0 \Rightarrow 4 k = 8 \Rightarrow k = 2$

Hence, the value of k is 2.

Page No 224:

Answer:

Let the required natural number be x.

According to the given condition,

$x + x^{2} = 156 \Rightarrow x^{2} + x - 156 = 0 \Rightarrow x^{2} + 13 x - 12 x - 156 = 0 \Rightarrow x (x + 13) - 12 (x + 13) = 0$
$\Rightarrow (x + 13) (x - 12) = 0 \Rightarrow x + 13 = 0 or x - 12 = 0 \Rightarrow x = - 13 or x = 12$

∴ x = 12 (x cannot be negative)

Hence, the required natural number is 12.

Page No 224:

Answer:

Let the required natural number be x.

According to the given condition,

x + \sqrt{x} = 132

Putting

\sqrt{x} = y

or x = y², we get

y^{2} + y = 132 \Rightarrow y^{2} + y - 132 = 0 \Rightarrow y^{2} + 12 y - 11 y - 132 = 0 \Rightarrow y (y + 12) - 11 (y + 12) = 0

\Rightarrow (y + 12) (y - 11) = 0 \Rightarrow y + 12 = 0 or y - 11 = 0 \Rightarrow y = - 12 or y = 11

∴ y = 11 (y cannot be negative)

Now,

\sqrt{x} = 11 \Rightarrow x = {(11)}^{2} = 121

Hence, the required natural number is 121.

Page No 224:

Answer:

Let the required numbers be x and (28 − x).

According to the given condition,

$x (28 - x) = 192 \Rightarrow 28 x - x^{2} = 192 \Rightarrow x^{2} - 28 x + 192 = 0 \Rightarrow x^{2} - 16 x - 12 x + 192 = 0$
$\Rightarrow x (x - 16) - 12 (x - 16) = 0 \Rightarrow (x - 12) (x - 16) = 0 \Rightarrow x - 12 = 0 or x - 16 = 0 \Rightarrow x = 12 or x = 16$

When x = 12,

28 − x = 28 − 12 = 16

When x = 16,

28 − x = 28 − 16 = 12

Hence, the required numbers are 12 and 16.

Page No 224:

Answer:

Let the required two consecutive positive integers be x and (x + 1).

According to the given condition,
$x^{2} + {(x + 1)}^{2} = 365 \Rightarrow x^{2} + x^{2} + 2 x + 1 = 365 \Rightarrow 2 x^{2} + 2 x - 364 = 0 \Rightarrow x^{2} + x - 182 = 0$
$\Rightarrow x^{2} + 14 x - 13 x - 182 = 0 \Rightarrow x (x + 14) - 13 (x + 14) = 0 \Rightarrow (x + 14) (x - 13) = 0 \Rightarrow x + 14 = 0 or x - 13 = 0$
$\Rightarrow x = - 14 or x = 13$
∴ x = 13 (x is a positive integer)

When x = 13,
x + 1 = 13 + 1 = 14

Hence, the required positive integers are 13 and 14.

Page No 224:

Answer:

Let the two consecutive positive odd numbers be x and (x + 2).

According to the given condition,
$x^{2} + {(x + 2)}^{2} = 514 \Rightarrow x^{2} + x^{2} + 4 x + 4 = 514 \Rightarrow 2 x^{2} + 4 x - 510 = 0 \Rightarrow x^{2} + 2 x - 255 = 0$
$\Rightarrow x^{2} + 17 x - 15 x - 255 = 0 \Rightarrow x (x + 17) - 15 (x + 17) = 0 \Rightarrow (x + 17) (x - 15) = 0 \Rightarrow x + 17 = 0 or x - 15 = 0$
$\Rightarrow x = - 17 or x = 15$
∴ x = 15 (x is a positive odd number)

When x = 15,
x + 2 = 15 + 2 = 17

Hence, the required numbers are 15 and 17.

Page No 224:

Answer:

Let the two consecutive positive even numbers be x and (x + 2).

According to the given condition,
$x^{2} + {(x + 2)}^{2} = 452 \Rightarrow x^{2} + x^{2} + 4 x + 4 = 452 \Rightarrow 2 x^{2} + 4 x - 448 = 0 \Rightarrow x^{2} + 2 x - 224 = 0$
$\Rightarrow x^{2} + 16 x - 14 x - 224 = 0 \Rightarrow x (x + 16) - 14 (x + 16) = 0 \Rightarrow (x + 16) (x - 14) = 0 \Rightarrow x + 16 = 0 or x - 14 = 0$
$\Rightarrow x = - 16 or x = 14$
∴ x = 14 (x is a positive even number)

When x = 14,
x + 2 = 14 + 2 = 16

Hence, the required numbers are 14 and 16.

Page No 224:

Answer:

Let the two consecutive positive integers be x and (x + 1).

According to the given condition,
$x (x + 1) = 306 \Rightarrow x^{2} + x - 306 = 0 \Rightarrow x^{2} + 18 x - 17 x - 306 = 0 \Rightarrow x (x + 18) - 17 (x + 18) = 0$
$\Rightarrow (x + 18) (x - 17) = 0 \Rightarrow x + 18 = 0 or x - 17 = 0 \Rightarrow x = - 18 or x = 17$
∴ x = 17 (x is a positive integer)

When x = 17,
x + 1 = 17 + 1 = 18

Hence, the required integers are 17 and 18.

Page No 224:

Answer:

$Let the required numbers be x and (x + 3) . According to the question: x (x + 3) = 504 \Rightarrow x^{2} + 3 x = 504 \Rightarrow x^{2} + 3 x - 504 = 0 \Rightarrow x^{2} + (24 - 21) x - 504 = 0 \Rightarrow x^{2} + 24 x - 21 x - 504 = 0 \Rightarrow x (x + 24) - 21 (x + 24) = 0 \Rightarrow (x + 24) (x - 21) = 0 \Rightarrow x + 24 = 0 or x - 21 = 0 \Rightarrow x = - 24 or x = 21 If x = - 24, the numbers are - 24 and {(- 24 + 3) = - 21}. If x = 21, the numbers are 21 and {(21 + 3) = 24}. Hence, the numbers are (- 24, - 21) and (21, 24) .$

Page No 224:

Answer:

Let the required consecutive multiples of 3 be 3x and 3(x + 1).

According to the given condition,
$3 x \times 3 (x + 1) = 648 \Rightarrow 9 (x^{2} + x) = 648 \Rightarrow x^{2} + x = 72 \Rightarrow x^{2} + x - 72 = 0$
$\Rightarrow x^{2} + 9 x - 8 x - 72 = 0 \Rightarrow x (x + 9) - 8 (x + 9) = 0 \Rightarrow (x + 9) (x - 8) = 0 \Rightarrow x + 9 = 0 or x - 8 = 0$
$\Rightarrow x = - 9 or x = 8$
∴ x = 8 (Neglecting the negative value)

When x = 8,
3x = 3 × 8 = 24
3(x + 1) = 3 × (8 + 1) = 3 × 9 = 27

Hence, the required multiples are 24 and 27.

Page No 224:

Answer:

Let the two consecutive positive odd integers be x and (x + 2).

According to the given condition,
$x (x + 2) = 483 \Rightarrow x^{2} + 2 x - 483 = 0 \Rightarrow x^{2} + 23 x - 21 x - 483 = 0 \Rightarrow x (x + 23) - 21 (x + 23) = 0$
$\Rightarrow (x + 23) (x - 21) = 0 \Rightarrow x + 23 = 0 or x - 21 = 0 \Rightarrow x = - 23 or x = 21$
∴ x = 21 (x is a positive odd integer)

When x = 21,
x + 2 = 21 + 2 = 23

Hence, the required integers are 21 and 23.

Page No 224:

Answer:

Let the two consecutive positive even integers be x and (x + 2).

According to the given condition,
$x (x + 2) = 288 \Rightarrow x^{2} + 2 x - 288 = 0 \Rightarrow x^{2} + 18 x - 16 x - 288 = 0 \Rightarrow x (x + 18) - 16 (x + 18) = 0$
$\Rightarrow (x + 18) (x - 16) = 0 \Rightarrow x + 18 = 0 or x - 16 = 0 \Rightarrow x = - 18 or x = 16$
∴ x = 16 (x is a positive even integer)

When x = 16,
x + 2 = 16 + 2 = 18

Hence, the required integers are 16 and 18.

Page No 224:

Answer:

Let the required natural numbers be x and (9 − x).

According to the given condition,
$\frac{1}{x} + \frac{1}{9 - x} = \frac{1}{2} \Rightarrow \frac{9 - x + x}{x (9 - x)} = \frac{1}{2} \Rightarrow \frac{9}{9 x - x^{2}} = \frac{1}{2} \Rightarrow 9 x - x^{2} = 18$
$\Rightarrow x^{2} - 9 x + 18 = 0 \Rightarrow x^{2} - 6 x - 3 x + 18 = 0 \Rightarrow x (x - 6) - 3 (x - 6) = 0 \Rightarrow (x - 3) (x - 6) = 0$
$\Rightarrow x - 3 = 0 or x - 6 = 0 \Rightarrow x = 3 or x = 6$

When x = 3,
9 − x = 9 − 3 = 6

When x = 6,
9 − x = 9 − 6 = 3

Hence, the required natural numbers are 3 and 6.

Page No 224:

Answer:

Let the required natural numbers be x and (15 − x).

According to the given condition,
$\frac{1}{x} + \frac{1}{15 - x} = \frac{3}{10} \Rightarrow \frac{15 - x + x}{x (15 - x)} = \frac{3}{10} \Rightarrow \frac{15}{15 x - x^{2}} = \frac{3}{10} \Rightarrow 15 x - x^{2} = 50$
$\Rightarrow x^{2} - 15 x + 50 = 0 \Rightarrow x^{2} - 10 x - 5 x + 50 = 0 \Rightarrow x (x - 10) - 5 (x - 10) = 0 \Rightarrow (x - 5) (x - 10) = 0$
$\Rightarrow x - 5 = 0 or x - 10 = 0 \Rightarrow x = 5 or x = 10$

When x = 5,
15 − x = 15 − 5 = 10

When x = 10,
15 − x = 15 − 10 = 5

Hence, the required natural numbers are 5 and 10.

Page No 224:

Answer:

Let the required natural numbers be x and (x + 3).

Now, x < x + 3
$∴ \frac{1}{x} > \frac{1}{x + 3}$

According to the given condition,
$\frac{1}{x} - \frac{1}{x + 3} = \frac{3}{28} \Rightarrow \frac{x + 3 - x}{x (x + 3)} = \frac{3}{28} \Rightarrow \frac{3}{x^{2} + 3 x} = \frac{3}{28} \Rightarrow x^{2} + 3 x = 28$
$\Rightarrow x^{2} + 3 x - 28 = 0 \Rightarrow x^{2} + 7 x - 4 x - 28 = 0 \Rightarrow x (x + 7) - 4 (x + 7) = 0 \Rightarrow (x + 7) (x - 4) = 0$
$\Rightarrow x + 7 = 0 or x - 4 = 0 \Rightarrow x = - 7 or x = 4$
∴ x = 4 (−7 is not a natural number)

When x = 4,
x + 3 = 4 + 3 = 7

Hence, the required natural numbers are 4 and 7.

Page No 224:

Answer:

Let the required natural numbers be x and (x + 5).

Now, x < x + 5
$∴ \frac{1}{x} > \frac{1}{x + 5}$

According to the given condition,
$\frac{1}{x} - \frac{1}{x + 5} = \frac{5}{14} \Rightarrow \frac{x + 5 - x}{x (x + 5)} = \frac{5}{14} \Rightarrow \frac{5}{x^{2} + 5 x} = \frac{5}{14} \Rightarrow x^{2} + 5 x = 14$
$\Rightarrow x^{2} + 5 x - 14 = 0 \Rightarrow x^{2} + 7 x - 2 x - 14 = 0 \Rightarrow x (x + 7) - 2 (x + 7) = 0 \Rightarrow (x + 7) (x - 2) = 0$
$\Rightarrow x + 7 = 0 or x - 2 = 0 \Rightarrow x = - 7 or x = 2$
∴ x = 2 (−7 is not a natural number)

When x = 2,
x + 5 = 2 + 5 = 7

Hence, the required natural numbers are 2 and 7.

Page No 225:

Answer:

Let the required consecutive multiples of 7 be 7x and 7(x + 1).

According to the given condition,
${(7 x)}^{2} + {[7 (x + 1)]}^{2} = 1225 \Rightarrow 49 x^{2} + 49 (x^{2} + 2 x + 1) = 1225 \Rightarrow 49 x^{2} + 49 x^{2} + 98 x + 49 = 1225 \Rightarrow 98 x^{2} + 98 x - 1176 = 0$
$\Rightarrow x^{2} + x - 12 = 0 \Rightarrow x^{2} + 4 x - 3 x - 12 = 0 \Rightarrow x (x + 4) - 3 (x + 4) = 0 \Rightarrow (x + 4) (x - 3) = 0$
$\Rightarrow x + 4 = 0 or x - 3 = 0 \Rightarrow x = - 4 or x = 3$
∴ x = 3 (Neglecting the negative value)

When x = 3,
7x = 7 × 3 = 21
7(x + 1) = 7(3 + 1) = 7 × 4 = 28

Hence, the required multiples are 21 and 28.

Page No 225:

Answer:

Let the natural number be x.

According to the given condition,
$x + \frac{1}{x} = \frac{65}{8} \Rightarrow \frac{x^{2} + 1}{x} = \frac{65}{8} \Rightarrow 8 x^{2} + 8 = 65 x \Rightarrow 8 x^{2} - 65 x + 8 = 0$
$\Rightarrow 8 x^{2} - 64 x - x + 8 = 0 \Rightarrow 8 x (x - 8) - 1 (x - 8) = 0 \Rightarrow (x - 8) (8 x - 1) = 0 \Rightarrow x - 8 = 0 or 8 x - 1 = 0$
$\Rightarrow x = 8 or x = \frac{1}{8}$
∴ x = 8 (x is a natural number)

Hence, the required number is 8.

Page No 225:

Answer:

Let the two parts be x and (57 − x).

According to the given condition,
$x (57 - x) = 680 \Rightarrow 57 x - x^{2} = 680 \Rightarrow x^{2} - 57 x + 680 = 0 \Rightarrow x^{2} - 40 x - 17 x + 680 = 0$
$\Rightarrow x (x - 40) - 17 (x - 40) = 0 \Rightarrow (x - 40) (x - 17) = 0 \Rightarrow x - 40 = 0 or x - 17 = 0 \Rightarrow x = 40 or x = 17$

When x = 40,
57 − x = 57 − 40 = 17

When x = 17,
57 − x = 57 − 17 = 40

Hence, the required parts are 17 and 40.

Page No 225:

Answer:

Let the two parts be x and (27 − x).

According to the given condition,
$\frac{1}{x} + \frac{1}{27 - x} = \frac{3}{20} \Rightarrow \frac{27 - x + x}{x (27 - x)} = \frac{3}{20} \Rightarrow \frac{27}{27 x - x^{2}} = \frac{3}{20} \Rightarrow 27 x - x^{2} = 180$
$\Rightarrow x^{2} - 27 x + 180 = 0 \Rightarrow x^{2} - 15 x - 12 x + 180 = 0 \Rightarrow x (x - 15) - 12 (x - 15) = 0 \Rightarrow (x - 12) (x - 15) = 0$
$\Rightarrow x - 12 = 0 or x - 15 = 0 \Rightarrow x = 12 or x = 15$

When x = 12,
27 − x = 27 − 12 = 15

When x = 15,
27 − x = 27 − 15 = 12

Hence, the required parts are 12 and 15.

Page No 225:

Answer:

$Let the larger and smaller parts be x and y, respectively . According to the question: x + y = 16 ... (i) 2 x^{2} = y^{2} + 164 ... (ii) From (i), we get: x = 16 - y ... (iii) From (ii) and (iii), we get: 2 {(16 - y)}^{2} = y^{2} + 164 \Rightarrow 2 (256 - 32 y + y^{2}) = y^{2} + 164 \Rightarrow 512 - 64 y + 2 y^{2} = y^{2} + 164 \Rightarrow y^{2} - 64 y + 348 = 0 \Rightarrow y^{2} - (58 + 6) y + 348 = 0 \Rightarrow y^{2} - 58 y - 6 y + 348 = 0 \Rightarrow y (y - 58) - 6 (y - 58) = 0 \Rightarrow (y - 58) (y - 6) = 0 \Rightarrow y - 58 = 0 or y - 6 = 0 \Rightarrow y = 6 (∵ y < 16) Putting the value of y in equation (iii), we get : x = 16 - 6 = 10 Hence, the two natural numbers are 6 and 10 .$

Page No 225:

Answer:

$Let the two natural numbers be x and y . According to the question: x^{2} + y^{2} = 25 (x + y) . . . (i) x^{2} + y^{2} = 50 (x - y) . . . (ii) From (i) and (ii), we get: 25 (x + y) = 50 (x - y) \Rightarrow x + y = 2 (x - y) \Rightarrow x + y = 2 x - 2 y \Rightarrow y + 2 y = 2 x - x \Rightarrow 3 y = x . . . (iii) From (ii) and (iii), we get : {(3 y)}^{2} + y^{2} = 50 (3 y - y) \Rightarrow 9 y^{2} + y^{2} = 100 y \Rightarrow 10 y^{2} = 100 y \Rightarrow y = 10 From (iii), we have : 3 \times 10 = x \Rightarrow 30 = x Hence, the two natural numbers are 30 and 10 .$

Page No 225:

Answer:

$Let the greater number be x and the smaller number be y . According to the question: x^{2} - y^{2} = 45 . . . (i) y^{2} = 4 x . . . (ii) From (i) and (ii), we get: x^{2} - 4 x = 45 \Rightarrow x^{2} - 4 x - 45 = 0 \Rightarrow x^{2} - (9 - 5) x - 45 = 0 \Rightarrow x^{2} - 9 x + 5 x - 45 = 0 \Rightarrow x (x - 9) + 5 (x - 9) = 0 \Rightarrow (x - 9) (x + 5) = 0 \Rightarrow x - 9 = 0 or x + 5 = 0 \Rightarrow x = 9 or x = - 5 \Rightarrow x = 9 (∵ x is a natural number) Putting the value of x in equation (ii), we get : y^{2} = 4 \times 9 \Rightarrow y^{2} = 36 \Rightarrow y = 6 Hence, the two numbers are 9 and 6 .$

Page No 225:

Answer:

Let the three consecutive positive integers be x, x + 1 and x + 2.

According to the given condition,
$x^{2} + (x + 1) (x + 2) = 46 \Rightarrow x^{2} + x^{2} + 3 x + 2 = 46 \Rightarrow 2 x^{2} + 3 x - 44 = 0 \Rightarrow 2 x^{2} + 11 x - 8 x - 44 = 0$
$\Rightarrow x (2 x + 11) - 4 (2 x + 11) = 0 \Rightarrow (2 x + 11) (x - 4) = 0 \Rightarrow 2 x + 11 = 0 or x - 4 = 0 \Rightarrow x = - \frac{11}{2} or x = 4$
∴ x = 4 (x is a positive integer)

When x = 4,
x + 1 = 4 + 1 = 5
x + 2 = 4 + 2 = 6

Hence, the required integers are 4, 5 and 6.

Page No 225:

Answer:

$Let the digits at units and tens places be x and y, respectively . ∴ O riginal number = 10 y + x According to the question: 10 y + x = 4 (x + y) \Rightarrow 10 y + x = 4 x + 4 y \Rightarrow 3 x - 6 y = 0 \Rightarrow 3 x = 6 y \Rightarrow x = 2 y .... (i) Also, 10 y + x = 2 x y \Rightarrow 10 y + 2 y = 2 .2 y . y [From (i)] \Rightarrow 12 y = 4 y^{2} \Rightarrow y = 3 From (i), we get: x = 2 \times 3 = 6 ∴ Original number = 10 \times 3 + 6 = 36$

Page No 225:

Answer:

$Let the digits at units and tens places be x and y, respectively . ∴ x y =14 ⇒ y = \frac{14}{x} ... (i) According to the question: (10 y + x) + 45 = 10 x + y \Rightarrow 9 y - 9 x = - 45 \Rightarrow y - x = - 5 ... (ii) From (i) and (ii), we get: \frac{14}{x} - x = - 5 \Rightarrow \frac{14 - x^{2}}{x} = - 5 \Rightarrow 14 - x^{2} = - 5 x \Rightarrow x^{2} - 5 x - 14 = 0 \Rightarrow x^{2} - (7 - 2) x - 14 = 0 \Rightarrow x^{2} - 7 x + 2 x - 14 = 0 \Rightarrow x (x - 7) + 2 (x - 7) = 0 \Rightarrow (x - 7) (x + 2) = 0 \Rightarrow x - 7 = 0 or x +2=0 \Rightarrow x = 7 or x = - 2 \Rightarrow x = 7 (∵ the digit cannot be negative) Putting x =7 in equation (i), we get: y =2 ∴ Required number=10 \times 2+7=27$

Page No 225:

Answer:

$Let the numerator be x . ∴ Denominator = x + 3 ∴ Original number = \frac{x}{x + 3} According to the question: \frac{x}{x + 3} + \frac{1}{(\frac{x}{x + 3})} = 2 \frac{9}{10} \Rightarrow \frac{x}{x + 3} + \frac{x + 3}{x} = \frac{29}{10} \Rightarrow \frac{x^{2} + {(x + 3)}^{2}}{x (x + 3)} = \frac{29}{10} \Rightarrow \frac{x^{2} + x^{2} + 6 x + 9}{x^{2} + 3 x} = \frac{29}{10} \Rightarrow \frac{2 x^{2} + 6 x + 9}{x^{2} + 3 x} = \frac{29}{10} \Rightarrow 29 x^{2} + 87 x = 20 x^{2} + 60 x + 90 \Rightarrow 9 x^{2} + 27 x - 90 = 0 \Rightarrow 9 (x^{2} + 3 x - 10) = 0 \Rightarrow x^{2} + 3 x - 10 = 0 \Rightarrow x^{2} + 5 x - 2 x - 10 = 0 \Rightarrow x (x + 5) - 2 (x + 5) = 0 \Rightarrow (x - 2) (x + 5) = 0 \Rightarrow x - 2 = 0 or x + 5 = 0 \Rightarrow x = 2 or x = - 5 (rejected) So, numerator = x = 2 denominator = x + 3 = 2 + 3 = 5 So, required fraction = \frac{2}{5}$

Page No 225:

Answer:

Let the denominator of the required fraction be x.

Numerator of the required fraction = x − 3

∴ Original fraction = $\frac{x - 3}{x}$
If 1 is added to the denominator, then the new fraction obtained is $\frac{x - 3}{x + 1}$ .

According to the given condition,
$\frac{x - 3}{x + 1} = \frac{x - 3}{x} - \frac{1}{15} \Rightarrow \frac{x - 3}{x} - \frac{x - 3}{x + 1} = \frac{1}{15} \Rightarrow \frac{(x - 3) (x + 1) - x (x - 3)}{x (x + 1)} = \frac{1}{15} \Rightarrow \frac{x^{2} - 2 x - 3 - x^{2} + 3 x}{x^{2} + x} = \frac{1}{15}$
$\Rightarrow \frac{x - 3}{x^{2} + x} = \frac{1}{15} \Rightarrow x^{2} + x = 15 x - 45 \Rightarrow x^{2} - 14 x + 45 = 0 \Rightarrow x^{2} - 9 x - 5 x + 45 = 0$
$\Rightarrow x (x - 9) - 5 (x - 9) = 0 \Rightarrow (x - 5) (x - 9) = 0 \Rightarrow x - 5 = 0 or x - 9 = 0 \Rightarrow x = 5 or x = 9$

When x = 5,
$\frac{x - 3}{x} = \frac{5 - 3}{5} = \frac{2}{5}$

When x = 9,
$\frac{x - 3}{x} = \frac{9 - 3}{9} = \frac{6}{9} = \frac{2}{3}$ (This fraction is neglected because this does not satisfies the given condition.)

Hence, the required fraction is $\frac{2}{5}$ .

Page No 225:

Answer:

Let the required number be x.

According to the given condition,
$x + \frac{1}{x} = 2 \frac{1}{30} \Rightarrow \frac{x^{2} + 1}{x} = \frac{61}{30} \Rightarrow 30 x^{2} + 30 = 61 x \Rightarrow 30 x^{2} - 61 x + 30 = 0$
$\Rightarrow 30 x^{2} - 36 x - 25 x + 30 = 0 \Rightarrow 6 x (5 x - 6) - 5 (5 x - 6) = 0 \Rightarrow (5 x - 6) (6 x - 5) = 0 \Rightarrow 5 x - 6 = 0 or 6 x - 5 = 0$
$\Rightarrow x = \frac{6}{5} or x = \frac{5}{6}$
Hence, the required number is $\frac{5}{6}$ or $\frac{6}{5}$ .

Page No 225:

Answer:

$Let there be x rows. Then, the number of students in each row will also be x . ∴ T otal number of students = (x^{2} + 24) According to the question: {(x + 1)}^{2} - 25 = x^{2} + 24 \Rightarrow x^{2} + 2 x + 1 - 25 - x^{2} - 24 = 0 \Rightarrow 2 x - 48 = 0 \Rightarrow 2 x = 48 \Rightarrow x = 24 ∴ {Total number of students = 24}^{2} + 24 = 576 + 24 = 600$

Page No 225:

Answer:

$\begin{array}{l} Let the total number of students be x . \end{array} According to the question : \frac{300}{x} - \frac{300}{x + 10} = 1 \Rightarrow \frac{300 (x + 10) - 300 x}{x (x + 10)} = 1 \Rightarrow \frac{300 x + 3000 - 300 x}{x^{2} + 10 x} = 1 \Rightarrow 3000 = x^{2} + 10 x \Rightarrow x^{2} + 10 x - 3000 = 0 \Rightarrow x^{2} + (60 - 50) x - 3000 = 0 \Rightarrow x^{2} + 60 x - 50 x - 3000 = 0 \Rightarrow x (x + 60) - 50 (x + 60) = 0 \Rightarrow (x + 60) (x - 50) = 0 \Rightarrow x = 50 or x = - 60 x cannot be negative; therefore, the total number of students is 50.$

Page No 226:

Answer:

$\begin{array}{l} Let the marks of Kamal in mathematics and english be x and y, respectively. \end{array} According to the question : x + y = 40 . . . (i) Also, (x + 3) (y - 4) = 360 \Rightarrow (x + 3) (40 - x - 4) = 360 [F rom (i)] \Rightarrow (x + 3) (36 - x) = 360 \Rightarrow 36 x - x^{2} + 108 - 3 x = 360 \Rightarrow 33 x - x^{2} - 252 = 0 \Rightarrow - x^{2} + 33 x - 252 = 0 \Rightarrow x^{2} - 33 x + 252 = 0 \Rightarrow x^{2} - (21 + 12) x + 252 = 0 \Rightarrow x^{2} - 21 x - 12 x + 252 = 0 \Rightarrow x (x - 21) - 12 (x - 21) = 0 \Rightarrow (x - 21) (x - 12) = 0 \Rightarrow x = 21 or x = 12 If x = 21, y = 40 - 21 = 19 Thus, Kamal scored 21 and 19 m arks in mathematics and english, respectively. If x = 12, y = 40 - 12 = 28 Thus, Kamal scored 12 and 28 marks in mathematics and english, respectively.$

Page No 226:

Answer:

Let x be the number of students who planned a picnic.

∴ Original cost of food for each member = ₹ $\frac{2000}{x}$

Five students failed to attend the picnic. So, (x − 5) students attended the picnic.

∴ New cost of food for each member = ₹ $\frac{2000}{(x - 5)}$

According to the given condition,
₹ $\frac{2000}{x - 5}$ − ₹ $\frac{2000}{x}$ = ₹20
$\Rightarrow \frac{2000 x - 2000 x + 10000}{x (x - 5)} = 20 \Rightarrow \frac{10000}{x^{2} - 5 x} = 20 \Rightarrow x^{2} - 5 x = 500 \Rightarrow x^{2} - 5 x - 500 = 0$
$\Rightarrow x^{2} - 25 x + 20 x - 500 = 0 \Rightarrow x (x - 25) + 20 (x - 25) = 0 \Rightarrow (x - 25) (x + 20) = 0 \Rightarrow x - 25 = 0 or x + 20 = 0$
$\Rightarrow x = 25 or x = - 20$
∴ x = 25 (Number of students cannot be negative)

Number of students who attended the picnic = x − 5 = 25 − 5 = 20

Amount paid by each student for the food = ₹ $\frac{2000}{(25 - 5)}$ = ₹ $\frac{2000}{20}$ = ₹100

Page No 226:

Answer:

Let the original price of the book be ₹x.

∴ Number of books bought at original price for ₹600 = $\frac{600}{x}$

If the price of a book is reduced by ₹5, then the new price of the book is ₹(x − 5).

∴ Number of books bought at reduced price for ₹600 = $\frac{600}{x - 5}$

According to the given condition,
$\frac{600}{x - 5} - \frac{600}{x} = 4 \Rightarrow \frac{600 x - 600 x + 3000}{x (x - 5)} = 4 \Rightarrow \frac{3000}{x^{2} - 5 x} = 4 \Rightarrow x^{2} - 5 x = 750$
$\Rightarrow x^{2} - 5 x - 750 = 0 \Rightarrow x^{2} - 30 x + 25 x - 750 = 0 \Rightarrow x (x - 30) + 25 (x - 30) = 0 \Rightarrow (x - 30) (x + 25) = 0$
$\Rightarrow x - 30 = 0 or x + 25 = 0 \Rightarrow x = 30 or x = - 25$
∴ x = 30 (Price cannot be negative)

Hence, the original price of the book is ₹30.

Page No 226:

Answer:

Let the original duration of the tour be x days.

∴ Original daily expenses = ₹ $\frac{10, 800}{x}$

If he extends his tour by 4 days, then his new daily expenses = ₹ $\frac{10, 800}{x + 4}$

According to the given condition,

₹ $\frac{10, 800}{x}$ − ₹ $\frac{10, 800}{x + 4}$ = ₹90

$\Rightarrow \frac{10800 x + 43200 - 10800 x}{x (x + 4)} = 90 \Rightarrow \frac{43200}{x^{2} + 4 x} = 90 \Rightarrow x^{2} + 4 x = 480 \Rightarrow x^{2} + 4 x - 480 = 0$
$\Rightarrow x^{2} + 24 x - 20 x - 480 = 0 \Rightarrow x (x + 24) - 20 (x + 24) = 0 \Rightarrow (x + 24) (x - 20) = 0 \Rightarrow x + 24 = 0 or x - 20 = 0$
$\Rightarrow x = - 24 or x = 20$
∴ x = 20 (Number of days cannot be negative)

Hence, the original duration of the tour is 20 days.

Page No 226:

Answer:

Let the marks obtained by P in mathematics and science be x and (28 − x), respectively.

According to the given condition,
$(x + 3) (28 - x - 4) = 180 \Rightarrow (x + 3) (24 - x) = 180 \Rightarrow - x^{2} + 21 x + 72 = 180 \Rightarrow x^{2} - 21 x + 108 = 0$
$\Rightarrow x^{2} - 12 x - 9 x + 108 = 0 \Rightarrow x (x - 12) - 9 (x - 12) = 0 \Rightarrow (x - 12) (x - 9) = 0 \Rightarrow x - 12 = 0 or x - 9 = 0$
$\Rightarrow x = 12 or x = 9$

When x = 12,
28 − x = 28 − 12 = 16

When x = 9,
28 − x = 28 − 9 = 19

Hence, he obtained 12 marks in mathematics and 16 marks in science or 9 marks in mathematics and 19 marks in science.

Page No 226:

Answer:

Let number of pens bought = x
Let the price per pen = Rs y
It is given that the price of x pens is Rs 180
$\Rightarrow x y = 180 . . . (i) \Rightarrow x = \frac{180}{y}$
Also, given that if 3 more pens are purchased for the same amount, the price per pen gets reduced by Rs 3
$\Rightarrow (y - 3) (x + 3) = 180 . . . (i i)$
Put the value of x in (ii) we get
$(\frac{180}{x} - 3) (x + 3) = 180 \Rightarrow (\frac{180 - 3 x}{x}) (x + 3) = 180 \Rightarrow (180 - 3 x) (x + 3) = 180 x \Rightarrow 180 x + 540 - 3 x^{2} - 9 x = 180 x \Rightarrow 3 x^{2} + 9 x - 540 = 0 \Rightarrow x^{2} + 3 x - 180 = 0 \Rightarrow x^{2} + 15 x - 12 x - 180 = 0 \Rightarrow x (x + 15) - 12 (x + 15) = 0 \Rightarrow (x + 15) (x - 12) = 0 \Rightarrow x + 15 = 0 or x - 12 = 0 \Rightarrow x = - 15 or x = 12$
We ignore the negative value.
$\Rightarrow x = 12$
Therefore, number of pens is 12.

Page No 226:

Answer:

Let the cost price of the article be ₹x.

∴ Gain percent = x%

According to the given condition,
₹x + ₹ $(\frac{x}{100} \times x)$ = ₹75 (Cost price + Gain = Selling price)
$\Rightarrow \frac{100 x + x^{2}}{100} = 75 \Rightarrow x^{2} + 100 x = 7500 \Rightarrow x^{2} + 100 x - 7500 = 0 \Rightarrow x^{2} + 150 x - 50 x - 7500 = 0$
$\Rightarrow x (x + 150) - 50 (x + 150) = 0 \Rightarrow (x - 50) (x + 150) = 0 \Rightarrow x - 50 = 0 or x + 150 = 0 \Rightarrow x = 50 or x = - 150$
∴ x = 50 (Cost price cannot be negative)

Hence, the cost price of the article is ₹50.

Page No 226:

Answer:

(i)
Let the present age of the son be x years.

∴ Present age of the man = x² years

One year ago,

Age of the son = (x − 1) years

Age of the man = (x² − 1) years

According to the given condition,

Age of the man = 8 × Age of the son

$∴ x^{2} - 1 = 8 (x - 1) \Rightarrow x^{2} - 1 = 8 x - 8 \Rightarrow x^{2} - 8 x + 7 = 0 \Rightarrow x^{2} - 7 x - x + 7 = 0$
$\Rightarrow x (x - 7) - 1 (x - 7) = 0 \Rightarrow (x - 1) (x - 7) = 0 \Rightarrow x - 1 = 0 or x - 7 = 0 \Rightarrow x = 1 or x = 7$
∴ x = 7 (Man's age cannot be 1 year)

Present age of the son = 7 years

Present age of the man = 7² years = 49 years

(ii)
Let the age of man be m and the age of son be s
It is given that man is $3 \frac{1}{2}$ times as old as his son.
$\Rightarrow m = 3 \frac{1}{2} s \Rightarrow m = \frac{7}{2} s . . . (i)$
Also given that
$m^{2} + s^{2} = 1325 . . . . . (ii)$
Put value of (i) in (ii), we get
${(\frac{7}{2} s)}^{2} + s^{2} = 1325 \frac{\Rightarrow 49 s^{2} + 4 s^{2}}{4} = 1325 53 s^{2} = 5300 \Rightarrow s^{2} = 100 \Rightarrow s = \pm 10$
Ignore the negative value
So, the age of son = s = 10 years
Also, from (i) we have
$m = \frac{7}{2} s \Rightarrow m = \frac{7}{2} \times 10 \Rightarrow m = 35$
So, age of man = 35 years
Age of son = 10 years

Page No 226:

Answer:

Let the present age of Meena be x years.

Meena's age 3 years ago = (x − 3) years

Meena's age 5 years hence = (x + 5) years

According to the given condition,
$\frac{1}{x - 3} + \frac{1}{x + 5} = \frac{1}{3} \Rightarrow \frac{x + 5 + x - 3}{(x - 3) (x + 5)} = \frac{1}{3} \Rightarrow \frac{2 x + 2}{x^{2} + 2 x - 15} = \frac{1}{3} \Rightarrow x^{2} + 2 x - 15 = 6 x + 6$
$\Rightarrow x^{2} - 4 x - 21 = 0 \Rightarrow x^{2} - 7 x + 3 x - 21 = 0 \Rightarrow x (x - 7) + 3 (x - 7) = 0 \Rightarrow (x - 7) (x + 3) = 0$
$\Rightarrow x - 7 = 0 or x + 3 = 0 \Rightarrow x = 7 or x = - 3$
∴ x = 7 (Age cannot be negative)

Hence, the present age of Meena is 7 years.

Page No 226:

Answer:

$Let the present ages of the boy and his brother be x years and (25 - x) years . According to the question : x (25 - x) = 126 \Rightarrow 25 x - x^{2} = 126 \Rightarrow x^{2} - (18 + 7) x + 126 = 0 \Rightarrow x^{2} - 18 x - 7 x + 126 = 0 \Rightarrow x (x - 18) - 7 (x - 18) = 0 \Rightarrow (x - 18) (x - 7) = 0 \Rightarrow x - 18 = 0 or x - 7 = 0 \Rightarrow x = 18 or x = 7 \Rightarrow x = 18 (∵ P resent age of the boy cannot be less than his brother) If x = 18, we have : Present ages of the boy = 18 years Present age of his brother = (25 - 18) years = 7 years Thus, the present ages of the boy and his brother are 18 years and 7 years, respectively.$

Page No 226:

Answer:

$Let the present age of Meena be x years . According to the question : (x - 5) (x + 8) = 30 \Rightarrow x^{2} + 3 x - 40 = 30 \Rightarrow x^{2} + 3 x - 70 = 0 \Rightarrow x^{2} + (10 - 7) x - 70 = 0 \Rightarrow x^{2} + 10 x - 7 x - 70 = 0 \Rightarrow x (x + 10) - 7 (x + 10) = 0 \Rightarrow (x + 10) (x - 7) = 0 \Rightarrow x + 10 = 0 or x - 7 = 0 \Rightarrow x = - 10 or x = 7 \Rightarrow x = 7 (∵ Age cannot be negative) Thus, the present age of Meena is 7 years .$

Page No 227:

Answer:

Let son's age 2 years ago be x years. Then,

Man's age 2 years ago = 3x²years

∴ Son's present age = (x + 2) years

Man's present age = (3x² + 2) years

In three years time,

Son's age = (x + 2 + 3) years = (x + 5) years

Man's age = (3x² + 2 + 3) years = (3x² + 5) years

According to the given condition,

Man's age = 4 × Son's age

∴ 3x² + 5 = 4(x + 5)

$\Rightarrow 3 x^{2} + 5 = 4 x + 20 \Rightarrow 3 x^{2} - 4 x - 15 = 0 \Rightarrow 3 x^{2} - 9 x + 5 x - 15 = 0 \Rightarrow 3 x (x - 3) + 5 (x - 3) = 0$
$\Rightarrow (x - 3) (3 x + 5) = 0 \Rightarrow x - 3 = 0 or 3 x + 5 = 0 \Rightarrow x = 3 or x = - \frac{5}{3}$
∴ x = 3 (Age cannot be negative)

Son's present age = (x + 2) years = (3 + 2) years = 5 years

Man's present age = (3x² + 2) years = (3× 9 + 2) years = 29 years

Page No 227:

Answer:

Let the first speed of the truck be x km/h.

∴ Time taken to cover 150 km = $\frac{150}{x}$ h $(Time = \frac{Distance}{Speed})$

New speed of the truck = (x + 20) km/h

∴ Time taken to cover 200 km = $\frac{200}{x + 20}$ h

According to the given condition,

Time taken to cover 150 km + Time taken to cover 200 km = 5 h
$∴ \frac{150}{x} + \frac{200}{x + 20} = 5 \Rightarrow \frac{150 x + 3000 + 200 x}{x (x + 20)} = 5 \Rightarrow 350 x + 3000 = 5 (x^{2} + 20 x) \Rightarrow 350 x + 3000 = 5 x^{2} + 100 x$
$\Rightarrow 5 x^{2} - 250 x - 3000 = 0 \Rightarrow x^{2} - 50 x - 600 = 0 \Rightarrow x^{2} - 60 x + 10 x - 600 = 0 \Rightarrow x (x - 60) + 10 (x - 60) = 0$
$\Rightarrow (x - 60) (x + 10) = 0 \Rightarrow x - 60 = 0 or x + 10 = 0 \Rightarrow x = 60 or x = - 10$
∴ x = 60 (Speed cannot be negative)

Hence, the first speed of the truck is 60 km/h.

Page No 227:

Answer:

Let the original speed of the plane be x km/h.

∴ Actual speed of the plane = (x + 100) km/h

Distance of the journey = 1500 km

Time taken to reach the destination at original speed = $\frac{1500}{x}$ h $(Time = \frac{Distance}{Speed})$

Time taken to reach the destination at actual speed = $\frac{1500}{x + 100}$ h

According to the given condition,

Time taken to reach the destination at original speed = Time taken to reach the destination at actual speed + 30 min

$∴ \frac{1500}{x} = \frac{1500}{x + 100} + \frac{1}{2} (30 \min = \frac{30}{60} h = \frac{1}{2} h) \Rightarrow \frac{1500}{x} - \frac{1500}{x + 100} = \frac{1}{2} \Rightarrow \frac{1500 x + 150000 - 1500 x}{x (x + 100)} = \frac{1}{2} \Rightarrow \frac{150000}{x^{2} + 100 x} = \frac{1}{2}$
$\Rightarrow x^{2} + 100 x = 300000 \Rightarrow x^{2} + 100 x - 300000 = 0 \Rightarrow x^{2} + 600 x - 500 x - 300000 = 0 \Rightarrow x (x + 600) - 500 (x + 600) = 0$
$\Rightarrow (x + 600) (x - 500) = 0 \Rightarrow x + 600 = 0 or x - 500 = 0 \Rightarrow x = - 600 or x = 500$
∴ x = 500 (Speed cannot be negative)

Hence, the original speed of the plane is 500 km/h.

Yes, we appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time. This reflects the caring nature of the pilot and his dedication to the work.

Page No 227:

Answer:

Let the usual speed of the train be x km/h.

∴ Reduced speed of the train = (x − 8) km/h

Total distance to be covered = 480 km

Time taken by the train to cover the distance at usual speed = $\frac{480}{x}$ h $(Time = \frac{Distance}{Speed})$

Time taken by the train to cover the distance at reduced speed = $\frac{480}{x - 8}$ h

According to the given condition,

Time taken by the train to cover the distance at reduced speed = Time taken by the train to cover the distance at usual speed + 3 h

$∴ \frac{480}{x - 8} = \frac{480}{x} + 3 \Rightarrow \frac{480}{x - 8} - \frac{480}{x} = 3 \Rightarrow \frac{480 x - 480 x + 3840}{x (x - 8)} = 3 \Rightarrow \frac{3840}{x^{2} - 8 x} = 3$
$\Rightarrow x^{2} - 8 x = 1280 \Rightarrow x^{2} - 8 x - 1280 = 0 \Rightarrow x^{2} - 40 x + 32 x - 1280 = 0 \Rightarrow x (x - 40) + 32 (x - 40) = 0$
$\Rightarrow (x - 40) (x + 32) = 0 \Rightarrow x - 40 = 0 or x + 32 = 0 \Rightarrow x = 40 or x = - 32$
∴ x = 40 (Speed cannot be negative)

Hence, the usual speed of the train is 40 km/h.

Page No 227:

Answer:

Let the first speed of the train be x km/h.

∴ Time taken to cover 54 km = $\frac{54}{x}$ h $(Time = \frac{Distance}{Speed})$

New speed of the train = (x + 6) km/h

∴ Time taken to cover 63 km = $\frac{63}{x + 6}$ h

According to the given condition,

Time taken to cover 54 km + Time taken to cover 63 km = 3 h
$∴ \frac{54}{x} + \frac{63}{x + 6} = 3 \Rightarrow \frac{54 x + 324 + 63 x}{x (x + 6)} = 3 \Rightarrow 117 x + 324 = 3 (x^{2} + 6 x) \Rightarrow 117 x + 324 = 3 x^{2} + 18 x$
$\Rightarrow 3 x^{2} - 99 x - 324 = 0 \Rightarrow x^{2} - 33 x - 108 = 0 \Rightarrow x^{2} - 36 x + 3 x - 108 = 0 \Rightarrow x (x - 36) + 3 (x - 36) = 0$
$\Rightarrow (x - 36) (x + 3) = 0 \Rightarrow x - 36 = 0 or x + 3 = 0 \Rightarrow x = 36 or x = - 3$
∴ x = 36 (Speed cannot be negative)

Hence, the first speed of the train is 36 km/h.

Page No 227:

Answer:

Distance covered by the train = 180 km
We know that distance covered $(d) = speed (s) \times time (t)$
$\Rightarrow s \times t = 180 \Rightarrow s = \frac{180}{t} . . . (i)$
Also, given that if the speed is increased by 9km/h, time of travel gets reduced by 1 hour.
$\Rightarrow (s + 9) (t - 1) = 180 . . . (ii)$
Put the value of (i) in (ii), we get
$(\frac{180}{t} + 9) (t - 1) = 180 (180 + 9 t) (t - 1) = 180 t 180 t - 180 + 9 t^{2} - 9 t = 180 t 9 t^{2} - 9 t - 180 = 0 t^{2} - t - 20 = 0 t^{2} - 5 t + 4 t - 20 = 0 t (t - 5) + 4 (t - 5) = 0 (t + 4) (t - 5) = 0 (t + 4) = 0 or (t - 5) = 0 t = - 4 or t = 5$
Ignore the negative value
So, time taken = 5 hours
From (i)
$s = \frac{180}{t} = \frac{180}{5} = 36$
Hence, the speed is 36 km/h.

Page No 227:

Answer:

Distance covered by the train = 300 km
We know that distance covered $(d) = speed (s) \times time (t)$
$\Rightarrow s \times t = 300 \Rightarrow t = \frac{300}{s} . . . (i)$
Also, given that if the speed is increased by 5 km/h, time of travel gets reduced by 2 hours.
$\Rightarrow (s + 5) (t - 2) = 300 . . . . . (ii)$
Put the value of (i) in (ii), we get
$(s + 5) (\frac{300}{s} - 2) = 300 (s + 5) (\frac{300 - 2 s}{s}) = 300 (s + 5) (300 - 2 s) = 300 s 300 s - 2 s^{2} + 1500 - 10 s = 300 s - 2 s^{2} + 1500 - 10 s = 0 - s^{2} + 750 - 5 s = 0 s^{2} + 5 s - 750 = 0 s^{2} + 30 s - 25 s - 750 = 0 s (s + 30) - 25 (s + 30) = 0 (s - 25) (s + 30) = 0 (s - 25) = 0 or (s + 30) = 0 s = 25 or s = - 30$
Ignore the negative value
Therefore, the original speed = 25 km/h

Page No 227:

Answer:

$\begin{array}{l} Let the usual speed be x km/hr . \end{array} According to the question : \frac{300}{x} - \frac{300}{(x + 5)} = 2 \Rightarrow \frac{300 (x + 5) - 300 x}{x (x + 5)} = 2 \Rightarrow \frac{300 x + 1500 - 300 x}{x^{2} + 5 x} = 2 \Rightarrow 1500 = 2 (x^{2} + 5 x) \Rightarrow 1500 = 2 x^{2} + 10 x \Rightarrow x^{2} + 5 x - 750 = 0 \Rightarrow x^{2} + (30 - 25) x - 750 = 0 \Rightarrow x^{2} + 30 x - 25 x - 750 = 0 \Rightarrow x (x + 30) - 25 (x + 30) = 0 \Rightarrow (x + 30) (x - 25) = 0 \Rightarrow x = - 30 or x = 25 T he usual speed cannot be negative; therefore, the speed is 25 km/hr.$

Page No 227:

Answer:

$\begin{array}{l} Let the speed of the Deccan Queen be x km/hr . \end{array} According to the question : Speed of another train = (x - 20) km / h ∴ \frac{192}{x - 20} - \frac{192}{x} = \frac{48}{60} \Rightarrow \frac{4}{x - 20} - \frac{4}{x} = \frac{1}{60} \Rightarrow \frac{4 x - 4 (x - 20)}{(x - 20) x} = \frac{1}{60} \Rightarrow \frac{4 x - 4 x + 80}{x^{2} - 20 x} = \frac{1}{60} \Rightarrow \frac{80}{x^{2} - 20 x} = \frac{1}{60} \Rightarrow x^{2} - 20 x = 4800 \Rightarrow x^{2} - 20 x - 4800 = 0 \Rightarrow x^{2} - (80 - 60) x - 4800 = 0 \Rightarrow x^{2} - 80 x + 60 x - 4800 = 0 \Rightarrow x (x - 80) + 60 (x - 80) = 0 \Rightarrow (x - 80) (x + 60) = 0 \Rightarrow x = 80 or x = - 60 The value of x cannot be negative; therefore, the original speed of Deccan Queen is 80 km/hr.$

Page No 228:

Answer:

$\begin{array}{l} Let the speed of the stream be x km/hr. \end{array} Given : Speed of the boat = 18 km / hr ∴ Speed downstream = (18 + x) km/hr S peed upstream = (18 - x) km/hr ∴ \frac{24}{(18 - x)} - \frac{24}{(18 + x)} = 1 \Rightarrow \frac{1}{(18 - x)} - \frac{1}{(18 + x)} = \frac{1}{24} \Rightarrow \frac{18 + x - 18 + x}{(18 - x) (18 + x)} = \frac{1}{24} \Rightarrow \frac{2 x}{18^{2} - x^{2}} = \frac{1}{24} \Rightarrow 324 - x^{2} = 48 x \Rightarrow 324 - x^{2} - 48 x = 0 \Rightarrow x^{2} + 48 x - 324 = 0 \Rightarrow x^{2} + (54 - 6) x - 324 = 0 \Rightarrow x^{2} + 54 x - 6 x - 324 = 0 \Rightarrow x (x + 54) - 6 (x + 54) = 0 \Rightarrow (x + 54) (x - 6) = 0 \Rightarrow x = - 54 or x = 6 The value of x cannot be negative; thereore, the speed of the stream is 6 km/hr.$

Page No 228:

Answer:

Let, speed of stream be x km/h
Speed of boat = 15 km/h
Distance from each side = 30 km
We know that time taken $= \frac{distance covered}{Speed}$
Total speed of the boat while going upstream $= 15 - x$ km/h
Time taken to go upstream $= \frac{30}{15 - x}$ hrs
Total speed of boat while going downstream $= 15 + x$ km/h
Time taken to go downstream $= \frac{30}{15 + x}$ hrs
Total time of the journey $= 4 \frac{1}{2}$ hrs = 4.5 hrs
$\Rightarrow \frac{30}{15 - x} + \frac{30}{15 + x} = 4.5 \Rightarrow \frac{30 [(15 + x) + (15 - x)]}{(15)^{2} - x^{2}} = 4.5 \Rightarrow 30 [15 + x + 15 - x] = 4.5 [(15)^{2} - x^{2}] \Rightarrow 30 [30] = 4.5 [225 - x^{2}] \Rightarrow \frac{30 \times 30}{4.5} = 225 - x^{2} \Rightarrow 225 - x^{2} = 200 \Rightarrow x^{2} = 25 \Rightarrow x = \pm 5$
Ignore the negative value.
So, the speed of the stream = x = 5 km/h

Page No 228:

Answer:

$\begin{array}{l} Let the speed of the stream be x km / hr . \end{array} ∴ D ownstream speed= (9 + x) km/hr Upstream speed= (9 - x) km/hr Distance covered downstream = Distance covered upstream = 15 km Total time taken = 3 hours 45 minutes = (3 + \frac{45}{60}) minutes = \frac{225}{60} minutes = \frac{15}{4} minutes ∴ \frac{15}{(9 + x)} + \frac{15}{(9 - x)} = \frac{15}{4} \Rightarrow \frac{1}{(9 + x)} + \frac{1}{(9 - x)} = \frac{1}{4} \Rightarrow \frac{9 - x + 9 + x}{(9 + x) (9 - x)} = \frac{1}{4} \Rightarrow \frac{18}{9^{2} - x^{2}} = \frac{1}{4} \Rightarrow \frac{18}{81 - x^{2}} = \frac{1}{4} \Rightarrow 81 - x^{2} = 72 \Rightarrow 81 - x^{2} - 72 = 0 \Rightarrow - x^{2} + 9 = 0 \Rightarrow x^{2} = 9 \Rightarrow x = 3 or x = - 3 The value of x cannot be negative; therefore, the speed of the stream is 3 km/hr.$

Page No 228:

Answer:

$\begin{array}{l} Let B takes x days to complete the work. \end{array} Therefore, A will take (x - 10) days. ∴ \frac{1}{x} + \frac{1}{(x - 10)} = \frac{1}{12} \Rightarrow \frac{(x - 10) + x}{x (x - 10)} = \frac{1}{12} \Rightarrow \frac{2 x - 10}{x^{2} - 10 x} = \frac{1}{12} \Rightarrow x^{2} - 10 x = 12 (2 x - 10) \Rightarrow x^{2} - 10 x = 24 x - 120 \Rightarrow x^{2} - 34 x + 120 = 0 \Rightarrow x^{2} - (30 + 4) x + 120 = 0 \Rightarrow x^{2} - 30 x - 4 x + 120 = 0 \Rightarrow x (x - 30) - 4 (x - 30) = 0 \Rightarrow (x - 30) (x - 4) = 0 \Rightarrow x = 30 or x = 4 Number of days to complete the work by B cannot be less than that by A; therefore, we get: x = 30 Thus, B completes the work in 30 days .$

Page No 228:

Answer:

$Let one tap fills the tank in x hours. Therefore, the o ther tap will fill the tank in (x + 3) hours. Time taken by both taps, running together, to fill the tank= 3 \frac{1}{13} hours = \frac{40}{13} hours Part filled by one tap in 1 hour = \frac{1}{x} Part filled by the other tap in 1 hour = \frac{1}{x + 3} Part filled by both taps, running together, in 1 hour = \frac{1}{x} + \frac{1}{x + 3} ∴ \frac{1}{x} + \frac{1}{x + 3} = \frac{1}{\frac{40}{13}} \Rightarrow \frac{(x + 3) + x}{x (x + 3)} = \frac{13}{40} \Rightarrow \frac{2 x + 3}{x^{2} + 3 x} = \frac{13}{40} \Rightarrow 13 x^{2} + 39 x = 80 x + 120 \Rightarrow 13 x^{2} - 41 x - 120 = 0 \Rightarrow 13 x^{2} - (65 - 24) x - 120 = 0 \Rightarrow 13 x^{2} - 65 x + 24 x - 120 = 0 \Rightarrow 13 x (x - 5) + 24 (x - 5) = 0 \Rightarrow (x - 5) (13 x + 24) = 0 \Rightarrow x - 5 = 0 or 13 x + 24 = 0 ⇒ x = 5 or x = \frac{- 24}{13} \Rightarrow x = 5 (∵ time cannot be a negative fraction) Thus, o ne pipe will take 5 hour and the other will take {(5 + 3) = 8} hour s to fill the tank.$

Page No 228:

Answer:

Let the time taken by one pipe to fill the tank be x minutes.

∴ Time taken by the other pipe to fill the tank = (x + 5) min

Suppose the volume of the tank be V.

Volume of the tank filled by one pipe in x minutes = V

∴ Volume of the tank filled by one pipe in 1 minute = $\frac{V}{x}$

⇒ Volume of the tank filled by one pipe in $11 \frac{1}{9}$ minutes = $\frac{V}{x} \times 11 \frac{1}{9} = \frac{V}{x} \times \frac{100}{9}$

Similarly,

Volume of the tank filled by the other pipe in $11 \frac{1}{9}$ minutes = $\frac{V}{(x + 5)} \times 11 \frac{1}{9} = \frac{V}{(x + 5)} \times \frac{100}{9}$

Now,

Volume of the tank filled by one pipe in $11 \frac{1}{9}$ minutes + Volume of the tank filled by the other pipe in $11 \frac{1}{9}$ minutes = V
$∴ V (\frac{1}{x} + \frac{1}{x + 5}) \times \frac{100}{9} = V \Rightarrow \frac{1}{x} + \frac{1}{x + 5} = \frac{9}{100} \Rightarrow \frac{x + 5 + x}{x (x + 5)} = \frac{9}{100} \Rightarrow \frac{2 x + 5}{x^{2} + 5 x} = \frac{9}{100}$
$\Rightarrow 200 x + 500 = 9 x^{2} + 45 x \Rightarrow 9 x^{2} - 155 x - 500 = 0 \Rightarrow 9 x^{2} - 180 x + 25 x - 500 = 0 \Rightarrow 9 x (x - 20) + 25 (x - 20) = 0$
$\Rightarrow (x - 20) (9 x + 25) = 0 \Rightarrow x - 20 = 0 or 9 x + 25 = 0 \Rightarrow x = 20 or x = - \frac{25}{9}$
∴ x = 20 (Time cannot be negative)

Time taken by one pipe to fill the tank = 20 min

Time taken by other pipe to fill the tank = (20 + 5) = 25 min

Page No 228:

Answer:

Let the tap of smaller diameter fill the tank in x hours.

∴ Time taken by the tap of larger diameter to fill the tank = (x − 9) h

Suppose the volume of the tank be V.

Volume of the tank filled by the tap of smaller diameter in x hours = V

∴ Volume of the tank filled by the tap of smaller diameter in 1 hour = $\frac{V}{x}$

⇒ Volume of the tank filled by the tap of smaller diameter in 6 hours = $\frac{V}{x} \times 6$

Similarly,

Volume of the tank filled by the tap of larger diameter in 6 hours = $\frac{V}{(x - 9)} \times 6$

Now,

Volume of the tank filled by the tap of smaller diameter in 6 hours + Volume of the tank filled by the tap of larger diameter in 6 hours = V
$∴ V (\frac{1}{x} + \frac{1}{x - 9}) \times 6 = V \Rightarrow \frac{1}{x} + \frac{1}{x - 9} = \frac{1}{6} \Rightarrow \frac{x - 9 + x}{x (x - 9)} = \frac{1}{6} \Rightarrow \frac{2 x - 9}{x^{2} - 9 x} = \frac{1}{6}$
$\Rightarrow 12 x - 54 = x^{2} - 9 x \Rightarrow x^{2} - 21 x + 54 = 0 \Rightarrow x^{2} - 18 x - 3 x + 54 = 0 \Rightarrow x (x - 18) - 3 (x - 18) = 0$
$\Rightarrow (x - 18) (x - 3) = 0 \Rightarrow x - 18 = 0 or x - 3 = 0 \Rightarrow x = 18 or x = 3$

For x = 3, time taken by the tap of larger diameter to fill the tank is negative which is not possibe.

∴ x = 18

Time taken by the tap of smaller diameter to fill the tank = 18 h

Time taken by the tap of larger diameter to fill the tank = (18 − 9) = 9 h

Hence, the time taken by the taps of smaller and larger diameter to fill the tank is 18 hours and 9 hours, respectively.

Page No 228:

Answer:

$Let the length and breadth of the rectangle be 2 x m and x m, respectively. According to the question : 2 x \times x = 288 \Rightarrow 2 x^{2} = 288 \Rightarrow x^{2} = 144 \Rightarrow x = 12 or x = - 12 \Rightarrow x = 12 (∵ x cannot be negative) ∴ L ength = 2 \times 12 = 24 m Breadth = 12 m$

Page No 228:

Answer:

$Let the length and breadth of the rectangle be 3 x m and x m, respectively. According to the question : 3 x \times x = 147 \Rightarrow 3 x^{2} = 147 \Rightarrow x^{2} = 49 \Rightarrow x = 7 or x = - 7 \Rightarrow x = 7 (∵ x cannot be negative) ∴ Length = 3 \times 7 = 21m Breadth = 7 m$

Page No 228:

Answer:

$\begin{array}{l} Let the breadth of the rectangular hall be x metre. \end{array} Therefore, the length of the rectangular hall will be (x + 3) metre. According to the question : x (x + 3) = 238 \Rightarrow x^{2} + 3 x = 238 \Rightarrow x^{2} + 3 x - 238 = 0 \Rightarrow x^{2} + (17 - 14) x - 238 = 0 \Rightarrow x^{2} + 17 x - 14 x - 238 = 0 \Rightarrow x (x + 17) - 14 (x + 17) = 0 \Rightarrow (x + 17) (x - 14) = 0 \Rightarrow x = - 17 or x = 14 But the value x cannot be negative. Therefore, the breadth of the hall is 14 metre and the length is 17 metre.$

Page No 228:

Answer:

Given:
Perimeter of a rectangular plot = 60 m
Area = 200 m²

Let the length of rectangular plot be x m.
$Perimeter = 60 \Rightarrow 2 (Length + Breadth) = 60 \Rightarrow Length + Breadth = 30 \Rightarrow Breadth = 30 - Length$

Then, the breadth is (30 − x) m.

According to the question,

$Length \times Breadth = Area \Rightarrow x (30 - x) = 200 \Rightarrow 30 x - x^{2} = 200 \Rightarrow x^{2} - 30 x + 200 = 0 \Rightarrow x^{2} - 10 x - 20 x + 200 = 0 \Rightarrow x (x - 10) - 20 (x - 10) = 0 \Rightarrow (x - 10) (x - 20) = 0 \Rightarrow x = 10, 20 For x = 10, 30 - x = 20 For x = 20, 30 - x = 10$

Hence, the dimensions of the plot are 10 m and 20 m.

Page No 228:

Answer:

$Let the width of the path be x m. ∴ Length of the field including the path = 16 + x + x = 16 + 2 x Breadth of the field including the path = 10 + x + x = 10 + 2 x Now, (Area of the field including path) - (A rea of the field excluding path) = Area of the path \Rightarrow (16 + 2 x) (10 + 2 x) - (16 \times 10) = 120 \Rightarrow 160 + 32 x + 20 x + 4 x^{2} - 160 = 120 \Rightarrow 4 x^{2} + 52 x - 120 = 0 \Rightarrow x^{2} + 13 x - 30 = 0 \Rightarrow x^{2} + (15 - 2) x + 30 = 0 \Rightarrow x^{2} + 15 x - 2 x + 30 = 0 \Rightarrow x (x + 15) - 2 (x + 15) = 0 \Rightarrow (x - 2) (x + 15) = 0 \Rightarrow x - 2 = 0 or x + 15 = 0 \Rightarrow x = 2 or x = - 15 \Rightarrow x = 2 (∵ Width cannot be negative) Thus, the width of the path is 2 m.$

Page No 229:

Answer:

Let the length of the side of the first and the second square be $x$ and y, respectively.

$According to the question : x^{2} + y^{2} = 640 . . . (i) Also, 4 x - 4 y = 64 \Rightarrow x - y = 16 \Rightarrow x = 16 + y$

Putting the value of $x$ in (i), we get:

$x^{2} + y^{2} = 640 \Rightarrow {(16 + y)}^{2} + y^{2} = 640 \Rightarrow 256 + 32 y + y^{2} + y^{2} = 640 \Rightarrow 2 y^{2} + 32 y - 384 = 0 \Rightarrow y^{2} + 16 y - 192 = 0 \Rightarrow y^{2} + (24 - 8) y - 192 = 0 \Rightarrow y^{2} + 24 y - 8 y - 192 = 0 \Rightarrow y (y + 24) - 8 (y + 24) = 0 \Rightarrow (y + 24) (y - 8) = 0 \Rightarrow y = - 24 or y = 8 ∴ y = 8 (∵ S ide cannot be negative) ∴ x = 16 + y = 16 + 8 = 24 m Thus, the sides of the squares are 8 m and 24 m.$

Page No 229:

Answer:

$Let the breadth of rectangle be x cm . According to the question : Side of the square = (x + 4) cm Length of the rectangle = {3 (x + 4)} cm It is given that the areas of the rectangle and square are same. ∴ 3 (x + 4) \times x = {(x + 4)}^{2} \Rightarrow 3 x^{2} + 12 x = {(x + 4)}^{2} \Rightarrow 3 x^{2} + 12 x = x^{2} + 8 x + 16 \Rightarrow 2 x^{2} + 4 x - 16 = 0 \Rightarrow x^{2} + 2 x - 8 = 0 \Rightarrow x^{2} + (4 - 2) x - 8 = 0 \Rightarrow x^{2} + 4 x - 2 x - 8 = 0 \Rightarrow x (x + 4) - 2 (x + 4) = 0 \Rightarrow (x + 4) (x - 2) = 0 \Rightarrow x = - 4 or x = 2 ∴ x = 2 (∵ The value of x cannot be negative) Thus, t he breadth of the rectangle is 2 cm and length is {3 (2 + 4) =18} cm. Also, the side of the square is 6 cm.$

Page No 229:

Answer:

$Let the length and breadth of the rectangular garden be x and y metre, respectively. Given: x y = 180 sq m ... (i) and 2 y + x = 39 \Rightarrow x = 39 - 2 y Putting the value of x in (i), we get: (39 - 2 y) y = 180 \Rightarrow 39 y - 2 y^{2} = 180 \Rightarrow 39 y - 2 y^{2} - 180 = 0 \Rightarrow 2 y^{2} - 39 y + 180 = 0 \Rightarrow 2 y^{2} - (24 + 15) y + 180 = 0 \Rightarrow 2 y^{2} - 24 y - 15 y + 180 = 0 \Rightarrow 2 y (y - 12) - 15 (y - 12) = 0 \Rightarrow (y - 12) (2 y - 15) = 0 \Rightarrow y = 12 or y = \frac{15}{2} = 7.5 If y = 12, x = 39 - 24 = 15 If y = 7.5, x = 39 - 15 = 24 Thus, the l ength and breadth of the garden are (15 m and 12 m) or (24 m and 7 .5 m), respectively.$

Page No 229:

Answer:

Let the altitude of the triangle be x cm.
Therefore, the base of the triangle will be (x + 10) cm.
$Area of triangle = \frac{1}{2} x (x + 10) = 600 \Rightarrow x (x + 10) = 1200 \Rightarrow x^{2} + 10 x - 1200 = 0 \Rightarrow x^{2} + (40 - 30) x - 1200 = 0 \Rightarrow x^{2} + 40 x - 30 x - 1200 = 0 \Rightarrow x (x + 40) - 30 (x + 40) = 0 \Rightarrow (x + 40) (x - 30) = 0 \Rightarrow x = - 40 or x = 30 \Rightarrow x = 30 [∵ Altitude cannot be negative] Thus, the altitude and base of the triangle are 30 cm and (30 + 10 = 40) cm, respectively. ∵ {(Hypotenuse)}^{2} = {(Altitude)}^{2} + {(Base)}^{2} \Rightarrow {(Hypotenuse)}^{2} = {(30)}^{2} + {(40)}^{2} \Rightarrow {(Hypotenuse)}^{2} = 900 + 1600 = 2500 \Rightarrow {(Hypotenuse)}^{2} = {(50)}^{2} \Rightarrow (Hypotenuse) = 50 Thus, the dimensions of the triangle are :, Hypotenuse = 50 cm Altitude = 30 cm Base = 40 cm$

Page No 229:

Answer:

Let the altitude of the triangle be x m.
Therefore, the base will be 3x m.

Area of a triangle = $\frac{1}{2} \times Base \times Altitude$

$∴ \frac{1}{2} \times 3 x \times x = 96 (∵ Area = 96 sq m) \Rightarrow \frac{x^{2}}{2} = 32 \Rightarrow x^{2} = 64 \Rightarrow x = \pm 8$

The value of x cannot be negative.
Therefore, the altitude and base of the triangle are 8 m and (3 $\times$ 8 = 24 m), respectively.

Page No 229:

Answer:

Let the base be $x$ m.
Therefore, the altitude will be $(x + 7)$ m.

Area of a triangle = $\frac{1}{2} \times Base \times Altitude$

$∴ \frac{1}{2} \times x \times (x + 7) = 165 \Rightarrow x^{2} + 7 x = 330 \Rightarrow x^{2} + 7 x - 330 = 0 \Rightarrow x^{2} + (22 - 15) x - 330 = 0 \Rightarrow x^{2} + 22 x - 15 x - 330 = 0 \Rightarrow x (x + 22) - 15 (x + 22) = 0 \Rightarrow (x + 22) (x - 15) = 0 \Rightarrow x = - 22 or x = 15$

The value of $x$ cannot be negative.
Therefore, the base is 15 m and the altitude is {(15 + 7) = 22 m}.

Page No 229:

Answer:

Let one side of the right-angled triangle be $x$ m and the other side be $(x + 4)$ m.
On applying Pythagoras theorem, we have:

$20^{2} = {(x + 4)}^{2} + x^{2} \Rightarrow 400 = x^{2} + 8 x + 16 + x^{2} \Rightarrow 2 x^{2} + 8 x - 384 = 0 \Rightarrow x^{2} + 4 x - 192 = 0 \Rightarrow x^{2} + (16 - 12) x - 192 = 0 \Rightarrow x^{2} + 16 x - 12 x - 192 = 0 \Rightarrow x (x + 16) - 12 (x + 16) = 0 \Rightarrow (x + 16) (x - 12) = 0 \Rightarrow x = - 16 or x = 12$

The value of x cannot be negative.
Therefore, the base is 12 m and the other side is {(12 + 4) = 16 m}.

Page No 229:

Answer:

Let the base and altitude of the right-angled triangle be $x$ and y cm, respectively.
Therefore, the hypotenuse will be $(x + 2)$ cm.
$∴ {(x + 2)}^{2} = y^{2} + x^{2} . . . (i) Again, the hypotenuse exceeds twice the length of the altitude by 1 cm. ∴ h = (2 y + 1) \Rightarrow x + 2 = 2 y + 1 \Rightarrow x = 2 y - 1 P utting the value of x in (i), we get : {(2 y - 1 + 2)}^{2} = y^{2} + {(2 y - 1)}^{2} \Rightarrow {(2 y + 1)}^{2} = y^{2} + 4 y^{2} - 4 y + 1 \Rightarrow 4 y^{2} + 4 y + 1 = 5 y^{2} - 4 y + 1 \Rightarrow - y^{2} + 8 y = 0 \Rightarrow y^{2} - 8 y = 0 \Rightarrow y (y - 8) = 0 \Rightarrow y = 8 cm ∴ x = 16 - 1 = 15 cm ∴ h = 16 + 1 = 17 cm$

Thus, the base, altitude and hypotenuse of the triangle are 15 cm, 8 cm and 17 cm, respectively.

Page No 229:

Answer:

Let the shortest side be $x$ m.
Therefore, according to the question:
Hypotenuse = $(2 x - 1)$ m
Third side = $(x + 1)$ m
On applying Pythagoras theorem, we get:

${(2 x - 1)}^{2} = {(x + 1)}^{2} + x^{2} \Rightarrow 4 x^{2} - 4 x + 1 = x^{2} + 2 x + 1 + x^{2} \Rightarrow 2 x^{2} - 6 x = 0 \Rightarrow 2 x (x - 3) = 0 \Rightarrow x = 0 or x = 3$

The length of the side cannot be 0; therefore, the shortest side is 3 m.
Therefore,
Hypotenuse = $(2 \times 3 - 1) =$ 5 m
Third side = (3 + 1) = 4 m

Page No 229:

Answer:

Let the speed of faster train be x km/h.
Then, the speed of slower train is (x − 10) km/h.

Given:
A faster train takes one hour less than a slower train for a journey of 200 km.

$\frac{Distance}{Speed} = Time$

Time taken by faster train to cover 200 km = $\frac{200}{x}$ h
Time taken by slower train to cover 200 km = $\frac{200}{x - 10}$ h

According to the question,

$\frac{200}{x - 10} - \frac{200}{x} = 1 \Rightarrow \frac{200 (x) - 200 (x - 10)}{(x) (x - 10)} = 1 \Rightarrow \frac{200 x - 200 x + 2000}{x^{2} - 10 x} = 1 \Rightarrow 2000 = x^{2} - 10 x \Rightarrow x^{2} - 10 x - 2000 = 0 \Rightarrow x^{2} - 50 x + 40 x - 2000 = 0 \Rightarrow x (x - 50) + 40 (x - 50) = 0 \Rightarrow (x - 50) (x + 40) = 0 \Rightarrow x = 50, - 40 But (x) is the speed of the train, which is always positive . Thus, x = 50$
and x − 10 = 40

Hence, the speed of fast train is 50 km/h and the speed of slow train is 40 km/h.

Page No 229:

Answer:

Let speed of stream be x km/h.

Given:

Speed of boat = 18 km/h
Distance covered upstream = 36 km
Distance covered downstream = 36 km
It takes $\frac{3}{2} hours$ more to go 36 km upstream than to return downstream to the same spot

Now, Speed of boat upstream = 18 − x km/h
Speed of boat downstream = 18 + x km/h

$\frac{Distance}{Speed} = Time$

According to the question,

Time to go upstream = $\frac{3}{2}$ hours + Time to go downstream

Time to go upstream − Time to go downstream = $\frac{3}{2}$

$\frac{36}{18 - x} - \frac{36}{18 + x} = \frac{3}{2} \Rightarrow \frac{36 (18 + x) - 36 (18 - x)}{(18 + x) (18 - x)} = \frac{3}{2} \Rightarrow \frac{648 + 36 x - 648 + 36 x}{324 - x^{2}} = \frac{3}{2} \Rightarrow \frac{72 x}{324 - x^{2}} = \frac{3}{2} \Rightarrow \frac{24 x}{324 - x^{2}} = \frac{1}{2} \Rightarrow 24 x (2) = 324 - x^{2} \Rightarrow x^{2} + 48 x - 324 = 0 \Rightarrow x^{2} + 54 x - 6 x - 324 = 0 \Rightarrow x (x + 54) - 6 (x + 54) = 0 \Rightarrow (x + 54) (x - 6) = 0 \Rightarrow x = - 54, 6 But x is the speed of stream which is always positive . Thus, x = 6 km / h$

Hence, the speed of the stream is 6 km/h.

Page No 229:

Answer:

Let the time required to fill the tank by second pipe be x hours.
Then, the time required to fill the tank by first pipe is (x − 10) hours.

Given:
Two pipes together can fill a tank in 12 hours.

According to the question,

$\frac{1}{x} + \frac{1}{x - 10} = \frac{1}{12} \Rightarrow \frac{(x - 10) + (x)}{(x) (x - 10)} = \frac{1}{12} \Rightarrow \frac{2 x - 10}{x^{2} - 10 x} = \frac{1}{12} \Rightarrow 24 x - 120 = x^{2} - 10 x \Rightarrow x^{2} - 10 x - 24 x + 120 = 0 \Rightarrow x^{2} - 34 x + 120 = 0 \Rightarrow x^{2} - 30 x - 4 x + 120 = 0 \Rightarrow x (x - 30) - 4 (x - 30) = 0 \Rightarrow (x - 30) (x - 4) = 0 \Rightarrow x = 30, 4 But (x - 10) is the time required by the first pipe to fill the tank, which is always positive . Thus, x = 30 and x \neq 4$

Hence, the second pipe will take 30 hours to fill the tank.

Page No 229:

Answer:

Let the smaller tap takes x hours to fill the tank.
Then, the larger one takes (x − 2) hours to fill the tank.

Tank filled in 1 hour by smaller tap = $\frac{1}{x}$
Tank filled in 1 hour by larger tap = $\frac{1}{x - 2}$
Tank filled in 1 hour by both the taps = $\frac{8}{15}$

According to the question,

$\frac{1}{x} + \frac{1}{x - 2} = \frac{8}{15} \Rightarrow \frac{(x - 2) + (x)}{(x) (x - 2)} = \frac{8}{15} \Rightarrow \frac{2 x - 2}{x^{2} - 2 x} = \frac{8}{15} \Rightarrow 15 (2 x - 2) = 8 (x^{2} - 2 x) \Rightarrow 30 x - 30 = 8 x^{2} - 16 x \Rightarrow 8 x^{2} - 46 x + 30 = 0 \Rightarrow 4 x^{2} - 23 x + 15 = 0 \Rightarrow x = \frac{- (- 23) \pm \sqrt{{(- 23)}^{2} - 4 (4) (15)}}{2 (4)} \Rightarrow x = \frac{23 \pm \sqrt{529 - 240}}{8} \Rightarrow x = \frac{23 \pm \sqrt{289}}{8} \Rightarrow x = \frac{23 \pm 17}{8} \Rightarrow x = \frac{23 + 17}{8} or x = \frac{23 - 17}{8} \Rightarrow x = 5 or x = 0.75 But x \neq 0.75 (∵ (x - 2) becomes negative) Thus, x = 5$

Hence, the time in which each tap can fill the tank separately is 5 hours and 3 hours respectively.

Page No 237:

Answer:

(d) $2 x^{2} - 5 x = (x - 1)^{2}$

$A quadratic equation is the equation with degree 2 . ∵ 2 x^{2} - 5 x = {(x - 1)}^{2} \Rightarrow 2 x^{2} - 5 x = x^{2} - 2 x + 1 \Rightarrow 2 x^{2} - 5 x - x^{2} + 2 x - 1 = 0 \Rightarrow x^{2} - 3 x - 1 = 0, which is a quadratic equation$

Page No 237:

Answer:

(b) $x^{3} - x^{2} = (x - 1)^{3}$

$∵ x^{3} - x^{2} = {(x - 1)}^{3} ⇒ x^{3} - x^{2} = x^{3} - 3 x^{2} + 3 x - 1 ⇒ 2 x^{2} - 3 x + 1 = 0, which is a quadratic equation$

Page No 237:

Answer:

(c) $(\sqrt{2} x + 3)^{2} = 2 x^{2} + 6$

$∵ {(\sqrt{2} x + 3)}^{2} = 2 x^{2} + 6 ⇒ 2 x^{2} + 9 + 6 \sqrt{2} x = 2 x^{2} + 6 ⇒ 6 \sqrt{2} x + 3 = 0, which is not a quadratic equation$

Page No 237:

Answer:

(b) −11
$It is given that x = 3 is a solution of 3 x^{2} + (k - 1) x + 9 = 0; therefore, we have : 3 {(3)}^{2} + (k - 1) \times 3 + 9 = 0 \Rightarrow 27 + 3 (k - 1) + 9 = 0 \Rightarrow 3 (k - 1) = - 36 \Rightarrow (k - 1) = - 12 \Rightarrow k = - 11$

Page No 237:

Answer:

(b) −7
$It is given that one root of the equation 2 x^{2} + a x + 6 = 0 is 2. ∴ 2 \times 2^{2} + a \times 2 + 6 = 0 \Rightarrow 2 a + 14 = 0 \Rightarrow a = - 7$

Page No 237:

Answer:

(c) 6
$Sum of the roots of the equation x^{2} - 6 x + 2 = 0 is α + β = \frac{- b}{a} = \frac{- (- 6)}{1} = 6 , where α and β are the roots of the equation.$

Page No 238:

Answer:

(c) 8
$It is given that the product of the roots of the equation x^{2} - 3 x + k = 10 is - 2 . The equation can be rewritten as : x^{2} - 3 x + (k - 10) = 0 Product of the roots of a quadratic equation = \frac{c}{a} \Rightarrow \frac{c}{a} = - 2 \Rightarrow \frac{(k - 10)}{1} = - 2 \Rightarrow k = 8$

Page No 238:

Answer:

(c) 2 : 3

$Given: 7 x^{2} - 12 x + 18 = 0 ∴ α + β = \frac{12}{7} and α β = \frac{18}{7}, where α and β are the roots of the equation ∴ Ratio of the sum and product of the roots = \frac{12}{7} : \frac{18}{7} = 12 : 18 = 2 : 3$

Page No 238:

Answer:

(d) 3
$Given: 3 x^{2} - 10 x + 3 = 0 One root of the equation is \frac{1}{3} . Let the other root be α . We know that : Product of the roots = \frac{c}{a} \Rightarrow \frac{1}{3} \times α = \frac{3}{3} \Rightarrow α = 3$

Page No 238:

Answer:

(d) 5
$Let the roots of the equation (5 x^{2} + 13 x + k = 0) be α and \frac{1}{α} . ∴ Product of the roots = \frac{c}{a} \Rightarrow α \times \frac{1}{α} = \frac{k}{5} \Rightarrow 1 = \frac{k}{5} \Rightarrow k = 5$

Page No 238:

Answer:

(d) $\frac{- 2}{3}$
$Given: k x^{2} + 2 x + 3 k = 0 Sum of the roots = Product of the roots ⇒ \frac{- 2}{k} = \frac{3 k}{k} ⇒ 3 k = - 2 ⇒ k = \frac{- 2}{3}$

Page No 238:

Answer:

(b) $x^{2} - 3 x - 10 = 0$

$It is given that the roots of the quadratic equation are 5 and - 2 . Then, the equation is: x^{2} - (5 - 2) x + 5 \times (- 2) = 0 ⇒ x^{2} - 3 x - 10 = 0$

Page No 238:

Answer:

(a) $x^{2} - 6 x + 6 = 0$

$Given : Sum of roots = 6 Product of roots = 6 Thus, the equation is: x^{2} - 6 x + 6 = 0$

Page No 238:

Answer:

It is given that α and β are the roots of the equation $3 x^{2} + 8 x + 2 = 0$ .

$∴ α + β = - \frac{8}{3}$ and $α β = \frac{2}{3}$
$\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{- \frac{8}{3}}{\frac{2}{3}} = - 4$

Hence, the correct answer is option C.

Page No 238:

Answer:

(c) c = a
$Let the roots of the equation (a x^{2} + b x + c = 0) be α and \frac{1}{α} . ∴ Product of the roots = α \times \frac{1}{α} = 1 \Rightarrow \frac{c}{a} = 1 \Rightarrow c = a$

Page No 238:

Answer:

(d) $\frac{b^{2}}{4 a}$

$It is given that the roots of the equation (a x^{2} + b x + c = 0) are equal. ∴ (b^{2} - 4 a c) = 0 \Rightarrow b^{2} = 4 a c \Rightarrow c = \frac{b^{2}}{4 a}$

Page No 238:

Answer:

(c) 2 or −2
$It is given that the roots of the equation (9 x^{2} + 6 k x + 4 = 0) are equal. ∴ (b^{2} - 4 a c) = 0 \Rightarrow {(6 k)}^{2} - 4 \times 9 \times 4 = 0 \Rightarrow 36 k^{2} = 144 \Rightarrow k^{2} = 4 \Rightarrow k = \pm 2$

Page No 238:

Answer:

(a) 1 or 4

$It is given that the roots of the equation (x^{2} + 2 (k + 2) x + 9 k = 0) are equal . ∴ (b^{2} - 4 a c) = 0 \Rightarrow {2 (k + 2)}^{2} - 4 \times 1 \times 9 k = 0 \Rightarrow 4 (k^{2} + 4 k + 4) - 36 k = 0 \Rightarrow 4 k^{2} + 16 k + 16 - 36 k = 0 \Rightarrow 4 k^{2} - 20 k + 16 = 0 \Rightarrow k^{2} - 5 k + 4 = 0 \Rightarrow k^{2} - 4 k - k + 4 = 0 \Rightarrow k (k - 4) - (k - 4) = 0 \Rightarrow (k - 4) (k - 1) = 0 \Rightarrow k = 4 or k = 1$

Page No 239:

Answer:

(d) $\pm \frac{4}{3}$

$It is given that the roots of the equation (4 x^{2} - 3 k x + 1 = 0) are equal . ∴ (b^{2} - 4 a c) = 0 \Rightarrow {(3 k)}^{2} - 4 \times 4 \times 1 = 0 \Rightarrow 9 k^{2} = 16 \Rightarrow k^{2} = \frac{16}{9} \Rightarrow k = \pm \frac{4}{3}$

Page No 239:

Answer:

(a) > 0
$\begin{array}{l} The roots of the equation are real and unequal when (b^{2} - 4 a c) > 0 . \end{array}$

Page No 239:

Answer:

We know that when discriminant, $D > 0$ , the roots of the given quadratic equation are real and unequal.

Hence, the correct answer is option B.

Page No 239:

Answer:

(d) imaginary

$∵ D = (b^{2} - 4 a c) = {(- 6)}^{2} - 4 \times 2 \times 7 = 36 - 56 = - 20 < 0 Thus, the roots of the equation are imaginary .$

Page No 239:

Answer:

(b) real, unequal and irrational

$∵ D = (b^{2} - 4 a c) = {(- 6)}^{2} - 4 \times 2 \times 3 = 36 - 24 = 12 12 is greater than 0 and it is not a perfect square; therefore, the roots of the equation are real, unequal and irrational .$

Page No 239:

Answer:

(d) either $k > 2 \sqrt{5}$ or $k < - 2 \sqrt{5}$

$It is given that the roots of the equation (5 x^{2} - k x + 1 = 0) are real and distinct . ∴ (b^{2} - 4 a c) > 0 \Rightarrow {(- k)}^{2} - 4 \times 5 \times 1 > 0 \Rightarrow k^{2} - 20 > 0 \Rightarrow k^{2} > 20 \Rightarrow k > \sqrt{20} or k < - \sqrt{20} \Rightarrow k > 2 \sqrt{5} or k < - 2 \sqrt{5}$

Page No 239:

Answer:

(c) $\frac{- 8}{5} < k > \frac{8}{5}$

$It is given that the equation (x^{2} + 5 k x + 16 = 0) has no real roots. ∴ (b^{2} - 4 a c) < 0 \Rightarrow {(5 k)}^{2} - 4 \times 1 \times 16 < 0 \Rightarrow 25 k^{2} - 64 < 0 \Rightarrow k^{2} < \frac{64}{25} \Rightarrow \frac{- 8}{5} < k < \frac{8}{5}$

Page No 239:

Answer:

(c) −2 < k < 2

$It is given that the equation x^{2} - k x + 1 = 0 has no real roots. ∴ (b^{2} - 4 a c) < 0 \Rightarrow {(- k)}^{2} - 4 \times 1 \times 1 < 0 \Rightarrow k^{2} < 4 \Rightarrow - 2 < k < 2$

Page No 239:

Answer:

(b) $k \geq \frac{- 9}{2}$

$It is given that the roots of the equation (k x^{2} - 6 x - 2 = 0) are real. ∴ D \geq 0 \Rightarrow (b^{2} - 4 a c) \geq 0 \Rightarrow {(- 6)}^{2} - 4 \times k \times (- 2) \geq 0 \Rightarrow 36 + 8 k \geq 0 \Rightarrow k \geq \frac{- 36}{8} \Rightarrow k \geq \frac{- 9}{2}$

Page No 239:

Answer:

(a) $\frac{5}{4} or \frac{4}{5}$

$Let the required number be x . According to the question : x + \frac{1}{x} = \frac{41}{20} \Rightarrow \frac{x^{2} + 1}{x} = \frac{41}{20} \Rightarrow 20 x^{2} - 41 x + 20 = 0 \Rightarrow 20 x^{2} - 25 x - 16 x + 20 = 0 \Rightarrow 5 x (4 x - 5) - 4 (4 x - 5) = 0 \Rightarrow (4 x - 5) (5 x - 4) = 0 \Rightarrow x = \frac{5}{4} or x = \frac{4}{5}$

Page No 239:

Answer:

(c) 16 m

$Let the length and breadth of the rectangle be l and b . Perimeter of the rectangle = 82 m \Rightarrow 2 \times (l + b) = 82 \Rightarrow l + b = 41 \Rightarrow l = (41 - b) . . . (i) Area of the rectangle = 400 m^{2} \Rightarrow l \times b = 400 m^{2} \Rightarrow (41 - b) b = 400 (using (i)) \Rightarrow 41 b - b^{2} = 400 \Rightarrow b^{2} - 41 b + 400 = 0 \Rightarrow b^{2} - 25 b - 16 b + 400 = 0 \Rightarrow b (b - 25) - 16 (b - 25) = 0 \Rightarrow (b - 25) (b - 16) = 0 \Rightarrow b = 25 or b = 16 If b = 25, we have : l = 41 - 25 = 16 Since, l cannot be less than b, ∴ b = 16 m$

Page No 240:

Answer:

Let the breadth of the rectangular field be x m.

∴ Length of the rectangular field = (x + 8) m

Area of the rectangular field = 240 m²(Given)

$∴ (x + 8) \times x = 240$ (Area = Length × Breadth)
$\Rightarrow x^{2} + 8 x - 240 = 0 \Rightarrow x^{2} + 20 x - 12 x - 240 = 0 \Rightarrow x (x + 20) - 12 (x + 20) = 0 \Rightarrow (x + 20) (x - 12) = 0$
$\Rightarrow x + 20 = 0 or x - 12 = 0 \Rightarrow x = - 20 or x = 12$
∴ x = 12 (Breadth cannot be negative)

Thus, the breadth of the field is 12 m.

Hence, the correct answer is option C.

Page No 240:

Answer:

The given quadratic equation is $2 x^{2} - x - 6 = 0$ .

$2 x^{2} - x - 6 = 0 \Rightarrow 2 x^{2} - 4 x + 3 x - 6 = 0 \Rightarrow 2 x (x - 2) + 3 (x - 2) = 0 \Rightarrow (x - 2) (2 x + 3) = 0$
$\Rightarrow x - 2 = 0 or 2 x + 3 = 0 \Rightarrow x = 2 or x = \frac{- 3}{2}$
Thus, the roots of the given equation are 2 and $\frac{- 3}{2}$ .

Hence, the correct answer is option B.

Page No 240:

Answer:

Let the required natural numbers be x and (8 − x).

It is given that the product of the two numbers is 15.
$∴ x (8 - x) = 15 \Rightarrow 8 x - x^{2} = 15 \Rightarrow x^{2} - 8 x + 15 = 0 \Rightarrow x^{2} - 5 x - 3 x + 15 = 0$
$\Rightarrow x (x - 5) - 3 (x - 5) = 0 \Rightarrow (x - 5) (x - 3) = 0 \Rightarrow x - 5 = 0 or x - 3 = 0 \Rightarrow x = 5 or x = 3$

Hence, the required numbers are 3 and 5.

Page No 240:

Answer:

The given equation is $x^{2} + 6 x + 9 = 0$ .

Putting x = −3 in the given equation, we get

LHS = ${(- 3)}^{2} + 6 \times (- 3) + 9 = 9 - 18 + 9 = 0$ = RHS

∴ x = −3 is a solution of the given equation.

Page No 240:

Answer:

The given equation is $3 x^{2} + 13 x + 14 = 0$ .

Putting x = −2 in the given equation, we get

LHS = $3 \times {(- 2)}^{2} + 13 \times (- 2) + 14 = 12 - 26 + 14 = 0$ = RHS

∴ x = −2 is a solution of the given equation.

Page No 240:

Answer:

It is given that $x = \frac{- 1}{2}$ is a solution of the quadratic equation $3 x^{2} + 2 k x - 3 = 0$ .

$∴ 3 \times {(\frac{- 1}{2})}^{2} + 2 k \times (\frac{- 1}{2}) - 3 = 0 \Rightarrow \frac{3}{4} - k - 3 = 0 \Rightarrow k = \frac{3 - 12}{4} = - \frac{9}{4}$

Hence, the value of k is $- \frac{9}{4}$ .

Page No 240:

Answer:

The given quadratic equation is $2 x^{2} - x - 6 = 0$ .
$2 x^{2} - x - 6 = 0 \Rightarrow 2 x^{2} - 4 x + 3 x - 6 = 0 \Rightarrow 2 x (x - 2) + 3 (x - 2) = 0 \Rightarrow (x - 2) (2 x + 3) = 0$
$\Rightarrow x - 2 = 0 or 2 x + 3 = 0 \Rightarrow x = 2 or x = - \frac{3}{2}$
Hence, the roots of the given equation are 2 and $- \frac{3}{2}$ .

Page No 240:

Answer:

The given quadratic equation is $3 \sqrt{3} x^{2} + 10 x + \sqrt{3} = 0$ .
$3 \sqrt{3} x^{2} + 10 x + \sqrt{3} = 0 \Rightarrow 3 \sqrt{3} x^{2} + 9 x + x + \sqrt{3} = 0 \Rightarrow 3 \sqrt{3} x (x + \sqrt{3}) + 1 (x + \sqrt{3}) = 0 \Rightarrow (x + \sqrt{3}) (3 \sqrt{3} x + 1) = 0$
$\Rightarrow x + \sqrt{3} = 0 or 3 \sqrt{3} x + 1 = 0 \Rightarrow x = - \sqrt{3} or x = - \frac{1}{3 \sqrt{3}} = - \frac{\sqrt{3}}{9}$
Hence, $- \sqrt{3}$ and $- \frac{\sqrt{3}}{9}$ are the solutions of the given equation.

Page No 240:

Answer:

It is given that the roots of the quadratic equation $2 x^{2} + 8 x + k = 0$ are equal.
$∴ D = 0 \Rightarrow 8^{2} - 4 \times 2 \times k = 0 \Rightarrow 64 - 8 k = 0 \Rightarrow k = 8$
Hence, the value of k is 8.

Page No 240:

Answer:

It is given that the quadratic equation $p x^{2} - 2 \sqrt{5} p x + 15 = 0$ has two equal roots.

$∴ D = 0 \Rightarrow {(- 2 \sqrt{5} p)}^{2} - 4 \times p \times 15 = 0 \Rightarrow 20 p^{2} - 60 p = 0 \Rightarrow 20 p (p - 3) = 0$
$\Rightarrow p = 0 or p - 3 = 0 \Rightarrow p = 0 or p = 3$

For p = 0, we get 15 = 0, which is not true.

∴ p ≠ 0

Hence, the value of p is 3.

Page No 240:

Answer:

It is given that y = 1 is a root of the equation $a y^{2} + a y + 3 = 0$ .
$∴ a \times {(1)}^{2} + a \times 1 + 3 = 0 \Rightarrow a + a + 3 = 0 \Rightarrow 2 a + 3 = 0 \Rightarrow a = - \frac{3}{2}$

Also, y = 1 is a root of the equation $y^{2} + y + b = 0$ .
$∴ {(1)}^{2} + 1 + b = 0 \Rightarrow 1 + 1 + b = 0 \Rightarrow b + 2 = 0 \Rightarrow b = - 2$

$∴ a b = (- \frac{3}{2}) \times (- 2) = 3$

Hence, the value of ab is 3.

Page No 240:

Answer:

Let the other zero of the given polynomial be $α$ .

Now,
Sum of the zeroes of the given polynomial = $\frac{- (- 4)}{1}$ = 4
$∴ α + (2 + \sqrt{3}) = 4 \Rightarrow α = 4 - 2 - \sqrt{3} = 2 - \sqrt{3}$

Hence, the other zero of the given polynomial is $(2 - \sqrt{3})$ .

Page No 240:

Answer:

Let $α$ and $β$ be the roots of the equation $3 x^{2} - 10 x + k = 0$ .
$∴ α = \frac{1}{β}$ (Given)
$\Rightarrow α β = 1 \Rightarrow \frac{k}{3} = 1 (Product of the roots = \frac{c}{a}) \Rightarrow k = 3$

Hence, the value of k is 3.

Page No 240:

Answer:

It is given that the roots of the quadratic equation $p x^{2} - 2 p x + 6 = 0$ are equal.
$∴ D = 0 \Rightarrow {(- 2 p)}^{2} - 4 \times p \times 6 = 0 \Rightarrow 4 p^{2} - 24 p = 0 \Rightarrow 4 p (p - 6) = 0$
$\Rightarrow p = 0 or p - 6 = 0 \Rightarrow p = 0 or p = 6$

For p = 0, we get 6 = 0, which is not true.

∴ p ≠ 0

Hence, the value of p is 6.

Page No 240:

Answer:

It is given that the quadratic equation $x^{2} - 4 k x + k = 0$ has equal roots.
$∴ D = 0 \Rightarrow {(- 4 k)}^{2} - 4 \times 1 \times k = 0 \Rightarrow 16 k^{2} - 4 k = 0 \Rightarrow 4 k (4 k - 1) = 0$
$\Rightarrow k = 0 or 4 k - 1 = 0 \Rightarrow k = 0 or k = \frac{1}{4}$

Hence, 0 and $\frac{1}{4}$ are the required values of k.

Page No 240:

Answer:

It is given that the quadratic equation $9 x^{2} - 3 k x + k = 0$ has equal roots.
$∴ D = 0 \Rightarrow {(- 3 k)}^{2} - 4 \times 9 \times k = 0 \Rightarrow 9 k^{2} - 36 k = 0 \Rightarrow 9 k (k - 4) = 0$
$\Rightarrow k = 0 or k - 4 = 0 \Rightarrow k = 0 or k = 4$

Hence, 0 and 4 are the required values of k.

Page No 240:

Answer:

$x^{2} - (\sqrt{3} + 1) x + \sqrt{3} = 0 \Rightarrow x^{2} - \sqrt{3} x - x + \sqrt{3} = 0 \Rightarrow x (x - \sqrt{3}) - 1 (x - \sqrt{3}) = 0 \Rightarrow (x - \sqrt{3}) (x - 1) = 0$
$\Rightarrow x - \sqrt{3} = 0 or x - 1 = 0 \Rightarrow x = \sqrt{3} or x = 1$

Hence, 1 and $\sqrt{3}$ are the roots of the given equation.

Page No 240:

Answer:

$2 x^{2} + a x - a^{2} = 0 \Rightarrow 2 x^{2} + 2 a x - a x - a^{2} = 0 \Rightarrow 2 x (x + a) - a (x + a) = 0 \Rightarrow (x + a) (2 x - a) = 0$
$\Rightarrow x + a = 0 or 2 x - a = 0 \Rightarrow x = - a or x = \frac{a}{2}$

Hence, −a and $\frac{a}{2}$ are the roots of the given equation.

Page No 241:

Answer:

$3 x^{2} + 5 \sqrt{5} x - 10 = 0 \Rightarrow 3 x^{2} + 6 \sqrt{5} x - \sqrt{5} x - 10 = 0 \Rightarrow 3 x (x + 2 \sqrt{5}) - \sqrt{5} (x + 2 \sqrt{5}) = 0 \Rightarrow (x + 2 \sqrt{5}) (3 x - \sqrt{5}) = 0$
$\Rightarrow x + 2 \sqrt{5} = 0 or 3 x - \sqrt{5} = 0 \Rightarrow x = - 2 \sqrt{5} or x = \frac{\sqrt{5}}{3}$

Hence, $- 2 \sqrt{5}$ and $\frac{\sqrt{5}}{3}$ are the roots of the given equation.

Page No 241:

Answer:

$\sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0 \Rightarrow \sqrt{3} x^{2} + 12 x - 2 x - 8 \sqrt{3} = 0 \Rightarrow \sqrt{3} x (x + 4 \sqrt{3}) - 2 (x + 4 \sqrt{3}) = 0 \Rightarrow (x + 4 \sqrt{3}) (\sqrt{3} x - 2) = 0$
$\Rightarrow x + 4 \sqrt{3} = 0 or \sqrt{3} x - 2 = 0 \Rightarrow x = - 4 \sqrt{3} or x = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$

Hence, $- 4 \sqrt{3}$ and $\frac{2 \sqrt{3}}{3}$ are the roots of the given equation.

Page No 241:

Answer:

$\sqrt{3} x^{2} - 2 \sqrt{2} x - 2 \sqrt{3} = 0 \Rightarrow \sqrt{3} x^{2} - 3 \sqrt{2} x + \sqrt{2} x - 2 \sqrt{3} = 0 \Rightarrow \sqrt{3} x (x - \sqrt{6}) + \sqrt{2} (x - \sqrt{6}) = 0 \Rightarrow (x - \sqrt{6}) (\sqrt{3} x + \sqrt{2}) = 0$
$\Rightarrow x - \sqrt{6} = 0 or \sqrt{3} x + \sqrt{2} = 0 \Rightarrow x = \sqrt{6} or x = - \frac{\sqrt{2}}{\sqrt{3}} = - \frac{\sqrt{6}}{3}$

Hence, $\sqrt{6}$ and $- \frac{\sqrt{6}}{3}$ are the roots of the given equation.

Page No 241:

Answer:

$4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3} = 0 \Rightarrow 4 \sqrt{3} x^{2} + 8 x - 3 x - 2 \sqrt{3} = 0 \Rightarrow 4 x (\sqrt{3} x + 2) - \sqrt{3} (\sqrt{3} x + 2) = 0 \Rightarrow (\sqrt{3} x + 2) (4 x - \sqrt{3}) = 0$
$\Rightarrow \sqrt{3} x + 2 = 0 or 4 x - \sqrt{3} = 0 \Rightarrow x = - \frac{2}{\sqrt{3}} = - \frac{2 \sqrt{3}}{3} or x = \frac{\sqrt{3}}{4}$

Hence, $- \frac{2 \sqrt{3}}{3}$ and $\frac{\sqrt{3}}{4}$ are the roots of the given equation.

Page No 241:

Answer:

$4 x^{2} + 4 b x - (a^{2} - b^{2}) = 0 \Rightarrow 4 x^{2} + 4 b x - (a - b) (a + b) = 0 \Rightarrow 4 x^{2} + 2 [(a + b) - (a - b)] x - (a - b) (a + b) = 0 \Rightarrow 4 x^{2} + 2 (a + b) x - 2 (a - b) x - (a - b) (a + b) = 0$
$\Rightarrow 2 x [2 x + (a + b)] - (a - b) [2 x + (a + b)] = 0 \Rightarrow [2 x + (a + b)] [2 x - (a - b)] = 0 \Rightarrow 2 x + (a + b) = 0 or 2 x - (a - b) = 0 \Rightarrow x = - \frac{a + b}{2} or x = \frac{a - b}{2}$

Hence, $- \frac{a + b}{2}$ and $\frac{a - b}{2}$ are the roots of the given equation.

Page No 241:

Answer:

$x^{2} + 5 x - (a^{2} + a - 6) = 0 \Rightarrow x^{2} + 5 x - (a + 3) (a - 2) = 0 \Rightarrow x^{2} + [(a + 3) - (a - 2)] x - (a + 3) (a - 2) = 0 \Rightarrow x^{2} + (a + 3) x - (a - 2) x - (a + 3) (a - 2) = 0$
$\Rightarrow x [x + (a + 3)] - (a - 2) [x + (a + 3)] = 0 \Rightarrow [x + (a + 3)] [x - (a - 2)] = 0 \Rightarrow x + (a + 3) = 0 or x - (a - 2) = 0 \Rightarrow x = - (a + 3) or x = (a - 2)$

Hence, $- (a + 3)$ and $(a - 2)$ are the roots of the given equation.

Page No 241:

Answer:

$x^{2} + 6 x - (a^{2} + 2 a - 8) = 0 \Rightarrow x^{2} + 6 x - (a + 4) (a - 2) = 0 \Rightarrow x^{2} + [(a + 4) - (a - 2)] x - (a + 4) (a - 2) = 0 \Rightarrow x^{2} + (a + 4) x - (a - 2) x - (a + 4) (a - 2) = 0$
$\Rightarrow x [x + (a + 4)] - (a - 2) [x + (a + 4)] = 0 \Rightarrow [x + (a + 4)] [x - (a - 2)] = 0 \Rightarrow x + (a + 4) = 0 or x - (a - 2) = 0 \Rightarrow x = - (a + 4) or x = (a - 2)$

Hence, $- (a + 4)$ and $(a - 2)$ are the roots of the given equation.

Page No 241:

Answer:

$x^{2} - 4 a x + 4 a^{2} - b^{2} = 0 \Rightarrow x^{2} - 4 a x + (2 a + b) (2 a - b) = 0 \Rightarrow x^{2} - [(2 a + b) + (2 a - b)] x + (2 a + b) (2 a - b) = 0 \Rightarrow x^{2} - (2 a + b) x - (2 a - b) x + (2 a + b) (2 a - b) = 0$
$\Rightarrow x [x - (2 a + b)] - (2 a - b) [x - (2 a + b)] = 0 \Rightarrow [x - (2 a + b)] [x - (2 a - b)] = 0 \Rightarrow x - (2 a + b) = 0 or x - (2 a - b) = 0 \Rightarrow x = (2 a + b) or x = (2 a - b)$

Hence, (2a + b) and (2a − b) are the roots of the given equation.

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