Rd Sharma 2021 for Class 9 Maths Chapter 3 - Rationalisation

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Rd Sharma 2021 Solutions for Class 9 Maths Chapter 3 Rationalisation are provided here with simple step-by-step explanations. These solutions for Rationalisation are extremely popular among Class 9 students for Maths Rationalisation Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2021 Book of Class 9 Maths Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2021 Solutions. All Rd Sharma 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

Page No 3.14:

Question 1:

Rationalise the denominator of each of the following (i-vii):

(i) $\frac{3}{\sqrt{5}}$

(ii) $\frac{3}{2 \sqrt{5}}$

(iii) $\frac{1}{\sqrt{12}}$

(iv) $\frac{\sqrt{2}}{\sqrt{5}}$

(v) $\frac{\sqrt{3} + 1}{\sqrt{2}}$

(vi) $\frac{\sqrt{2} + \sqrt{5}}{3}$

(vii) $\frac{3 \sqrt{2}}{\sqrt{5}}$

Answer:

(i) We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(ii) We know that rationalization factor foris. We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(iii) We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(iv) We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(v) We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(vi) We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(vii) We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

Page No 3.14:

Question 2:

Find the value to three places of decimals of each of the following. It is given that
$\sqrt{2} = 1.414, \sqrt{3} = 1.732, \sqrt{5} = 2.236$ and $\sqrt{10} = 3.162$ .

(i) $\frac{2}{\sqrt{3}}$

(ii) $\frac{3}{\sqrt{10}}$

(iii) $\frac{\sqrt{5} + 1}{\sqrt{2}}$

(iv) $\frac{\sqrt{10} + \sqrt{15}}{\sqrt{2}}$

(v) $\frac{2 + \sqrt{3}}{3}$

(vi) $\frac{\sqrt{2} - 1}{\sqrt{5}}$

Answer:

(i) We know that rationalization factor of the denominator is. We will multiply numerator and denominator of the given expression by, to get

The value of expression can be round off to three decimal places as.

Hence the given expression is simplified to.

(ii) We know that rationalization factor of the denominator is . We will multiply numerator and denominator of the given expression by , to get

The value of expression can be round off to three decimal places as.

Hence the given expression is simplified to.

(iii) We know that rationalization factor of the denominator is. We will multiply numerator and denominator of the given expression by, to get

The value of expression can be round off to three decimal places as.

Hence the given expression is simplified to.

(iv) We know that rationalization factor of the denominator is. We will multiply numerator and denominator of the given expression by, to get

The value of expression can be round off to three decimal places as.

Hence the given expression is simplified to.

(v) Given that

Putting the value of, we get

The value of expression can be round off to three decimal places as.

Hence the given expression is simplified to.

(vi) We know that rationalization factor of the denominator is. We will multiply numerator and denominator of the given expression by, to get

Putting the value of and, we get

The value of expression can be round off to three decimal places as.

Hence the given expression is simplified to.

Page No 3.14:

Question 3:

Express each one of the following with rational denominator:

(i) $\frac{1}{3 + \sqrt{2}}$

(ii) $\frac{1}{\sqrt{6} - \sqrt{5}}$

(iii) $\frac{16}{\sqrt{41} - 5}$

(iv) $\frac{30}{5 \sqrt{3} - 3 \sqrt{5}}$

(v) $\frac{1}{2 \sqrt{5} - \sqrt{3}}$

(vi) $\frac{\sqrt{3} + 1}{2 \sqrt{2} - \sqrt{3}}$

(vii) $\frac{6 - 4 \sqrt{2}}{6 + 4 \sqrt{2}}$

(viii) $\frac{3 \sqrt{2} + 1}{2 \sqrt{5} - 3}$

(ix) $\frac{b^{2}}{\sqrt{a^{2} + b^{2}} + a}$

Answer:

(i) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified with rational denominator to.

(ii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified with rational denominator to.

(iii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified with rational denominator to.

(iv) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified with rational denominator to.

(v) We know that rationalization factor for is.We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified with rational denominator to.

(vi) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified with rational denominator to.

(vii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified with rational denominator to.

(viii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified with rational denominator to.

(ix) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified with rational denominator to .

Page No 3.14:

Question 4:

Rationales the denominator and simplify:

(i) $\frac{3 - \sqrt{2}}{3 + \sqrt{2}}$

(ii) $\frac{5 + 2 \sqrt{3}}{7 + 4 \sqrt{3}}$

(iii) $\frac{1 + \sqrt{2}}{3 - 2 \sqrt{2}}$

(iv) $\frac{2 \sqrt{6} - \sqrt{5}}{3 \sqrt{5} - 2 \sqrt{6}}$

(v) $\frac{4 \sqrt{3} + 5 \sqrt{2}}{\sqrt{48} + \sqrt{18}}$

(vi) $\frac{2 \sqrt{3} - \sqrt{5}}{2 \sqrt{2} + 3 \sqrt{3}}$

Answer:

(i) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(ii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(iii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(iv) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(v) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

(vi) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence the given expression is simplified to.

Page No 3.14:

Question 5:

Simplify:

(i) $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$

(ii) $\frac{1}{2 + \sqrt{3}} + \frac{2}{\sqrt{5} - \sqrt{3}} + \frac{1}{2 - \sqrt{5}}$

(iii) $\frac{2}{\sqrt{5} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{2}} + \frac{3}{\sqrt{5} + \sqrt{2}}$

Answer:

(i) We know that rationalization factor forand areand respectively. We will multiply numerator and denominator of the given expression and by and respectively, to get

Hence the given expression is simplified to.

(ii) We know that rationalization factor forand areand respectively. We will multiply numerator and denominator of the given expression and by and respectively, to get

Hence the given expression is simplified to.

(iii) We know that rationalization factor forand areand respectively. We will multiply numerator and denominator of the given expression and by and respectively, to get

Hence the given expression is simplified to.

Page No 3.14:

Question 6:

In each of the following determine rational numbers a and b:

(i) $\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = a - b \sqrt{3}$

(ii) $\frac{4 + \sqrt{2}}{2 + \sqrt{2}} = n - \sqrt{b}$

(iii) $\frac{3 + \sqrt{2}}{3 - \sqrt{2}} = a + b \sqrt{2}$

(iv) $\frac{5 + 3 \sqrt{3}}{7 + 4 \sqrt{3}} = a + b \sqrt{3}$

(v) $\frac{\sqrt{11} - \sqrt{7}}{\sqrt{11} + \sqrt{7}} = a - b \sqrt{77}$

(vi) $\frac{4 + 3 \sqrt{5}}{4 - 3 \sqrt{5}} = a + b \sqrt{5}$

Answer:

(i) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

On equating rational and irrational terms, we get

Hence, we get.

(ii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

On equating rational and irrational terms, we get

Hence, we get.

(iii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

On equating rational and irrational terms, we get

Hence, we get.

(iv) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

On equating rational and irrational terms, we get

Hence, we get.

(v) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

On equating rational and irrational terms, we get

Hence, we get.

(vi) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

On equating rational and irrational terms, we get

Hence, we get.

Page No 3.14:

Question 7:

Find the value of $\frac{6}{\sqrt{5} - \sqrt{3}}$ , it being given that $\sqrt{3} = 1.732$ and $\sqrt{5} = 2.236$

Answer:

We know that rationalization factor for is . We will multiply denominator and numerator of the given expression by , to get

Putting the values of and, we get

Hence value of the given expression is.

Page No 3.15:

Question 8:

Find the values of each of the following correct to three places of decimals, it being given that $\sqrt{2} = 1.4142,$ $\sqrt{3} = 1.732$ , $\sqrt{5} = 2.2360,$ $\sqrt{6} = 2.4495$ and $\sqrt{10} = 3.162$ ,

(i) $\frac{3 - \sqrt{5}}{3 + 2 \sqrt{5}}$

(ii) $\frac{1 + \sqrt{2}}{3 - 2 \sqrt{2}}$

Answer:

(i) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Putting the values of, we get

Hence the given expression is simplified to.

(ii) We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Putting the value of, we get

Hence the given expression is simplified to.

Page No 3.15:

Question 9:

Simplify:

(i) $\frac{3 \sqrt{2} - 2 \sqrt{3}}{3 \sqrt{2} + 2 \sqrt{3}} + \frac{\sqrt{12}}{\sqrt{3} - \sqrt{2}}$
(ii) $\frac{7 + 3 \sqrt{5}}{3 + \sqrt{5}} - \frac{7 - 3 \sqrt{5}}{3 - \sqrt{5}}$

Answer:

(i) We know that rationalization factor forand areand respectively. We will multiply numerator and denominator of the given expression and by and respectively, to get

Hence the given expression is simplified to.

(ii) We know that rationalization factor forand areand respectively. We will multiply numerator and denominator of the given expression and by and respectively, to get

Hence the given expression is simplified to.

Page No 3.15:

Question 10:

If x = 2+ $\sqrt{3}$ , find the value of $x^{3} + \frac{1}{x^{3}}$

Answer:

We know that. We have to find the value of.

As therefore,

We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Putting the value of and , we get

Hence the value of the given expression

Page No 3.15:

Question 11:

If x = 3+ $\sqrt{8}$ , find the value of $x^{2} + \frac{1}{x^{2}}$

Answer:

We know that. We have to find the value of . As therefore,

We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Putting the value of x and , we get

Hence the given expression is simplified to.

Page No 3.15:

Question 12:

If $x = \frac{\sqrt{3} + 1}{2},$ find the value of $4 x^{3} + 2 x^{2} - 8 x + 7$ .

Answer:

We have,

It can be simplified as

On squaring both sides, we get

The given equation can be rewritten as.

Therefore, we have

Hence, the value of given expression is.

Page No 3.16:

Question 1:

$\sqrt{10} \times \sqrt{15}$ is equal to

(a) 5 $\sqrt{6}$

(b) 6 $\sqrt{5}$

(c) $\sqrt{30}$

(d) $\sqrt{25}$

Answer:

Given that, it can be simplified as

Therefore given expression is simplified and correct choice is

Page No 3.16:

Question 2:

$\sqrt[5]{6} \times \sqrt[5]{6}$ is equal to

(a) $\sqrt[5]{36}$

(b) $\sqrt[5]{6 \times 0}$

(c) $\sqrt[5]{6}$

(d) $\sqrt[5]{12}$

Answer:

Given that, it can be simplified as

Therefore given expression is simplified and correct choice is.

Page No 3.16:

Question 3:

The rationalisation factor of $\sqrt{3}$ is

(a) $- \sqrt{3}$

(b) $\frac{1}{\sqrt{3}}$

(c) $2 \sqrt{3}$

(d) $- 2 \sqrt{3}$

Answer:

We know that rationalization factor for is. Hence rationalization factor of is.Hence the correct option is.

Page No 3.16:

Question 4:

The rationalisation factor of $2 + \sqrt{3}$ , is

(a) $2 - \sqrt{3}$

(b) $2 + \sqrt{3}$

(c) $\sqrt{2} - 3$

(d) $\sqrt{3} - 2$

Answer:

We know that rationalization factor for is. Hence rationalization factor of is.Hence correct option is

Page No 3.16:

Question 5:

If x = $\sqrt{5} + 2$ , then $x - \frac{1}{x}$ equals

(a) $2 \sqrt{5}$

(b) 4

(c) 2

(d) $\sqrt{5}$

Answer:

Given that.Hence is given as

.We need to find

We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Therefore,

Hence the correct option is.

Page No 3.16:

Question 6:

If $\frac{\sqrt{3 - 1}}{\sqrt{3} + 1}$ = $a - b \sqrt{3}$ , then

(a) a = 2, b =1

(b) a = 2, b =−1

(c) a = −2, b = 1

(d) a = b = 1

Answer:

Given that:

We are asked to find a and b

We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

On equating rational and irrational terms, we get

Comparing rational and irrational part we get

Hence, the correct choice is.

Page No 3.16:

Question 7:

The simplest rationalising factor of $\sqrt[3]{500}$ is

(a) $\sqrt[3]{2}$

(b) $\sqrt[3]{5}$

(c) $\sqrt{3}$

(d) none of these

Answer:

Given that:.To find simplest rationalizing factor of the given expression we will factorize it as

The rationalizing factor of is, since when we multiply given expression with this factor we get rid of irrational term.

Therefore, rationalizing factor of the given expression is

Hence correct option is.

Page No 3.16:

Question 8:

The simplest rationalising factor of $\sqrt{3} + \sqrt{5}$ , is

(a) $\sqrt{3} - 5$

(b) $3 - \sqrt{5}$

(c) $\sqrt{3} - \sqrt{5}$

(d) $\sqrt{3} + \sqrt{5}$

Answer:

We know that rationalization factor for is. Hence rationalization factor of is.

Page No 3.16:

Question 9:

The simplest rationalising factor of $2 \sqrt{5}$ − $\sqrt{3}$ is

(a) $2 \sqrt{5} + 3$

(b) $2 \sqrt{5} + \sqrt{3}$

(c) $\sqrt{5} + \sqrt{3}$

(d) $\sqrt{5} - \sqrt{3}$

Answer:

We know that rationalization factor for is. Hence rationalization factor of is.

Page No 3.16:

Question 10:

If x = $\frac{2}{3 + \sqrt{7}}$ , then (x−3)² =

(a) 1
(b) 3
(c) 6
(d) 7

Answer:

Given that:

We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Therefore,

On squaring both sides, we get

Hence the value of the given expression is.

Page No 3.16:

Question 11:

If $x = 7 + 4 \sqrt{3}$ and xy =1, then $\frac{1}{x^{2}} + \frac{1}{y^{2}} =$

(a) 64

(b) 134

(c) 194

(d) $\frac{1}{49}$

Answer:

Given that,

Hence is given as

We need to find

We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Since so we have

Therefore,

Hence the value of the given expression is.

Page No 3.16:

Question 12:

If $x + \sqrt{15} = 4,$ then $x + \frac{1}{x}$ =

(a) 2
(b) 4
(c) 8
(d) 1

Answer:

Given that .It can be simplified as

We need to find

We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Therefore,

Hence the value of the given expression is 8.Hence correct option is .

Page No 3.17:

Question 13:

If $x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$ and $y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$ , then x + y +xy=

(a) 9
(b) 5
(c) 17
(d) 7

Answer:

Given that and.

We are asked to find

Now we will rationalize x. We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Similarly, we can rationalize y. We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Therefore,

Hence the value of the given expression is.

Page No 3.17:

Question 14:

If x= $\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$ and y = $\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$ , then x² + y +y² =

(a) 101
(b) 99
(c) 98
(d) 102

Answer:

Given that and.

We need to find

Now we will rationalize x. We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Similarly, we can rationalize y. We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Therefore,

Hence the value of the given expression is.

Page No 3.17:

Question 15:

$\frac{1}{\sqrt{9} - \sqrt{8}}$ is equal to

(a) $3 + 2 \sqrt{2}$

(b) $\frac{1}{3 + 2 \sqrt{2}}$

(c) $3 - 2 \sqrt{2}$

(d) $\frac{3}{2} - \sqrt{2}$

Answer:

Given that

We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Hence the correct option is.

Page No 3.17:

Question 16:

The value of $\frac{\sqrt{48} + \sqrt{32}}{\sqrt{27} + \sqrt{18}}$ is

(a) $\frac{4}{3}$

(b) 4

(c) 3

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

Answer:

Given that

We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

We can factor irrational terms as

Hence the value of given expression is.

Page No 3.17:

Question 17:

If $\frac{5 - \sqrt{3}}{2 + \sqrt{3}} = x + y \sqrt{3}$ , then

(a) x = 13, y = −7

(b) x = −13, y = 7

(c) x = −13, y = −7

(d) x = 13, y = 7

Answer:

Given that: .We need to find x and y

We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Since

On equating rational and irrational terms, we get

Hence, the correct choice is.

Page No 3.17:

Question 18:

If x = $\sqrt[3]{2 + \sqrt{3}}$ , then $x^{3} + \frac{1}{x^{3}} =$

(a) 2

(b) 4

(c) 8

(d) 9

Answer:

Given that .It can be simplified as

We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

Therefore,

Hence the value of the given expression is.

Page No 3.17:

Question 19:

The value of $\sqrt{3 - 2 \sqrt{2}}$ is

(a) $\sqrt{2} - 1$

(b) $\sqrt{2} + 1$

(c) $\sqrt{3} - \sqrt{2}$

(d) $\sqrt{3} + \sqrt{2}$

Answer:

Given that:.It can be written in the form as

Hence the value of the given expression is.

Page No 3.17:

Question 20:

The value of $\sqrt{5 + 2 \sqrt{6}}$ ,is

(a) $\sqrt{3} - \sqrt{2}$

(b) $\sqrt{3} + \sqrt{2}$

(c) $\sqrt{5} + \sqrt{6}$

(d) none of these

Answer:

Given that:.It can be written in the form as

Hence the value of the given expression is.

Page No 3.17:

Question 21:

If $\sqrt{2} = 1.4142$ then $\sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}}$ is equal to

(a) 0.1718

(b) 5.8282

(c) 0.4142

(d) 2.4142

Answer:

Given that , we need to find the value of .

We can rationalize the denominator of the given expression. We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

$\sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}} = \frac{\sqrt{2} - 1}{1}$

Putting the value of , we get

Hence the value of the given expression is 0.14142 and correct choice is.

Page No 3.17:

Question 22:

If $\sqrt{2} = 1.414,$ then the value of $\sqrt{6} - \sqrt{3}$ upto three places of decimal is

(a) 0.235

(b) 0.707

(c) 1.414

(d) 0.471

Answer:

Given that.We need to find.

We can factor out from the given expression, to get

Putting the value of, we get

Hence the value of expression must closely resemble be

The correct option is.

Page No 3.17:

Question 23:

The positive square root of $7 + \sqrt{48}$ is

(a) $7 + 2 \sqrt{3}$

(b) $7 + \sqrt{3}$

(c) $2 + \sqrt{3}$

(d) $3 + \sqrt{2}$

Answer:

Given that:.To find square root of the given expression we need to rewrite the expression in the form

Hence the square root of the given expression is

Hence the correct option is.

Page No 3.17:

Question 24:

If $x = \sqrt{6} + \sqrt{5}$ , then $x^{2} + \frac{1}{x^{2}} - 2 =$

(a) $2 \sqrt{6}$

(b) $2 \sqrt{5}$

(c) 24

(d) 20

Answer:

Given that.Hence is given as

We need to find

We know that rationalization factor for is. We will multiply numerator and denominator of the given expression by, to get

We know that therefore,

Hence the value of the given expression is 20 and correct option is (d).

Page No 3.17:

Question 25:

If $\sqrt{13 - a \sqrt{10}} = \sqrt{8} + \sqrt{5}, then a =$

(a) −5

(b) −6

(c) −4

(d) −2

Answer:

Given that:

We need to find a

The given expression can be simplified by taking square on both sides

The irrational terms on right side can be factorized such that it of the same form as left side terms.

Hence,

On comparing rational and irrational terms, we get.Therefore, correct choice is .

Page No 3.18:

Question 1:

The number obtained by rationalizing the denominator of $\frac{1}{\sqrt{7} + 2}$ is __________.

Answer:

$\frac{1}{\sqrt{7} + 2} Multiply and divide by (\sqrt{7} - 2), we get = \frac{1}{\sqrt{7} + 2} \times \frac{\sqrt{7} - 2}{\sqrt{7} - 2} = \frac{\sqrt{7} - 2}{{(\sqrt{7})}^{2} - 2^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) = \frac{\sqrt{7} - 2}{7 - 4} = \frac{\sqrt{7} - 2}{3}$

Hence, the number obtained by rationalizing the denominator of $\frac{1}{\sqrt{7} + 2}$ is $\frac{\sqrt{7} - 2}{3} .$ .

Page No 3.18:

Question 2:

If $\frac{1}{\sqrt{9} - \sqrt{8}} = A + B \sqrt{2}$ , then A = ____________ and B = ____________.

Answer:

$\frac{1}{\sqrt{9} - \sqrt{8}} Multiply and divide by (\sqrt{9} + \sqrt{8}), we get = \frac{1}{\sqrt{9} - \sqrt{8}} \times \frac{\sqrt{9} + \sqrt{8}}{\sqrt{9} + \sqrt{8}} = \frac{\sqrt{9} + \sqrt{8}}{{(\sqrt{9})}^{2} - {(\sqrt{8})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) = \frac{\sqrt{9} + \sqrt{8}}{9 - 8} = \frac{\sqrt{9} + \sqrt{8}}{1} = \sqrt{9} + \sqrt{8} = 3 + 2 \sqrt{2} Now, it is given that \frac{1}{\sqrt{9} - \sqrt{8}} = A + B \sqrt{2} \Rightarrow 3 + 2 \sqrt{2} = A + B \sqrt{2} \Rightarrow A = 3 and B = 2$

Hence, if $\frac{1}{\sqrt{9} - \sqrt{8}} = A + B \sqrt{2}$ , then A = 3 and B = 2.

Page No 3.18:

Question 3:

After rationalizing the denominator of $\frac{7}{3 \sqrt{3} - 2 \sqrt{2}}$ , we get the denominator as __________.

Answer:

$\frac{7}{3 \sqrt{3} - 2 \sqrt{2}} Multiply and divide by (3 \sqrt{3} + 2 \sqrt{2}), we get = \frac{7}{3 \sqrt{3} - 2 \sqrt{2}} \times \frac{3 \sqrt{3} + 2 \sqrt{2}}{3 \sqrt{3} + 2 \sqrt{2}} = \frac{7 (3 \sqrt{3} + 2 \sqrt{2})}{{(3 \sqrt{3})}^{2} - {(2 \sqrt{2})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) = \frac{21 \sqrt{3} + 14 \sqrt{2}}{27 - 8} = \frac{21 \sqrt{3} + 14 \sqrt{2}}{19}$

Hence, after rationalizing the denominator of $\frac{7}{3 \sqrt{3} - 2 \sqrt{2}}$ , we get the denominator as $\frac{21 \sqrt{3} + 14 \sqrt{2}}{19} .$

Page No 3.18:

Question 4:

If $a = 2 + \sqrt{3}, then a - \frac{1}{a} =___________.$

Answer:

$Given : a = 2 + \sqrt{3} Now, \frac{1}{a} = \frac{1}{2 + \sqrt{3}} Multiply and divide RHS by (2 - \sqrt{3}), we get \Rightarrow \frac{1}{a} = \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} \Rightarrow \frac{1}{a} = \frac{2 - \sqrt{3}}{{(2)}^{2} - {(\sqrt{3})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow \frac{1}{a} = \frac{2 - \sqrt{3}}{4 - 3} \Rightarrow \frac{1}{a} = \frac{2 - \sqrt{3}}{1} \Rightarrow \frac{1}{a} = 2 - \sqrt{3} Thus, a - \frac{1}{a} = (2 + \sqrt{3}) - (2 - \sqrt{3}) = 2 + \sqrt{3} - 2 + \sqrt{3} = 2 \sqrt{3}$

Hence, if $a = 2 + \sqrt{3}, then a - \frac{1}{a} = 2 \sqrt{3} .$

Page No 3.18:

Question 5:

If $a = 5 + 2 \sqrt{6}, then a + \frac{1}{a} =_________.$

Answer:

$Given : a = 5 + 2 \sqrt{6} Now, \frac{1}{a} = \frac{1}{5 + 2 \sqrt{6}} Multiply and divide RHS by (5 - 2 \sqrt{6}), we get \Rightarrow \frac{1}{a} = \frac{1}{5 + 2 \sqrt{6}} \times \frac{5 - 2 \sqrt{6}}{5 - 2 \sqrt{6}} \Rightarrow \frac{1}{a} = \frac{5 - 2 \sqrt{6}}{{(5)}^{2} - {(2 \sqrt{6})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow \frac{1}{a} = \frac{5 - 2 \sqrt{6}}{25 - 24} \Rightarrow \frac{1}{a} = \frac{5 - 2 \sqrt{6}}{1} \Rightarrow \frac{1}{a} = 5 - 2 \sqrt{6} Thus, a + \frac{1}{a} = (5 + 2 \sqrt{6}) + (5 - 2 \sqrt{6}) = 5 + 2 \sqrt{6} + 5 - 2 \sqrt{6} = 10$

Hence, if $a = 5 + 2 \sqrt{6}, then a + \frac{1}{a} = 10 .$

Page No 3.18:

Question 6:

If $x = \sqrt{6} + \sqrt{5}, then x^{2} + \frac{1}{x^{2}} - 2 =_________.$

Answer:

$Given : x = \sqrt{6} + \sqrt{5} Now, \frac{1}{x} = \frac{1}{\sqrt{6} + \sqrt{5}} Multiply and divide RHS by (\sqrt{6} - \sqrt{5}), we get \Rightarrow \frac{1}{x} = \frac{1}{\sqrt{6} + \sqrt{5}} \times \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} - \sqrt{5}} \Rightarrow \frac{1}{x} = \frac{\sqrt{6} - \sqrt{5}}{{(\sqrt{6})}^{2} - {(\sqrt{5})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow \frac{1}{x} = \frac{\sqrt{6} - \sqrt{5}}{6 - 5} \Rightarrow \frac{1}{x} = \frac{\sqrt{6} - \sqrt{5}}{1} \Rightarrow \frac{1}{x} = \sqrt{6} - \sqrt{5} Thus, x^{2} + \frac{1}{x^{2}} - 2 = {(x - \frac{1}{x})}^{2} = {((\sqrt{6} + \sqrt{5}) - (\sqrt{6} - \sqrt{5}))}^{2} = {(\sqrt{6} + \sqrt{5} - \sqrt{6} + \sqrt{5})}^{2} = {(2 \sqrt{5})}^{2} = 20$

Hence, if $x = \sqrt{6} + \sqrt{5}, then x^{2} + \frac{1}{x^{2}} - 2 = 20 .$

Page No 3.18:

Question 7:

If $x = 3 - \sqrt{8}, then {(x - \frac{1}{x})}^{2} =___________.$

Answer:

$Given : x = 3 - \sqrt{8} Now, \frac{1}{x} = \frac{1}{3 - \sqrt{8}} Multiply and divide RHS by (3 + \sqrt{8}), we get \Rightarrow \frac{1}{x} = \frac{1}{3 - \sqrt{8}} \times \frac{3 + \sqrt{8}}{3 + \sqrt{8}} \Rightarrow \frac{1}{x} = \frac{3 + \sqrt{8}}{{(3)}^{2} - {(\sqrt{8})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow \frac{1}{x} = \frac{3 + \sqrt{8}}{9 - 8} \Rightarrow \frac{1}{x} = \frac{3 + \sqrt{8}}{1} \Rightarrow \frac{1}{x} = 3 + \sqrt{8} Thus, {(x - \frac{1}{x})}^{2} = {((3 - \sqrt{8}) - (3 + \sqrt{8}))}^{2} = {(3 - \sqrt{8} - 3 - \sqrt{8})}^{2} = {(- 2 \sqrt{8})}^{2} = 32$

Hence, if $x = 3 - \sqrt{8}, then {(x - \frac{1}{x})}^{2} = 32 .$

Page No 3.18:

Question 8:

If $x = \frac{2}{\sqrt{3} - \sqrt{5}} and y = \frac{2}{\sqrt{3} + \sqrt{5}}$ , then x + y = __________.

Answer:

$Given : x = \frac{2}{\sqrt{3} - \sqrt{5}} y = \frac{2}{\sqrt{3} + \sqrt{5}} Now, x = \frac{2}{\sqrt{3} - \sqrt{5}} Multiply and divide RHS by (\sqrt{3} + \sqrt{5}), we get \Rightarrow x = \frac{2}{\sqrt{3} - \sqrt{5}} \times \frac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}} \Rightarrow x = \frac{2 (\sqrt{3} + \sqrt{5})}{{(\sqrt{3})}^{2} - {(\sqrt{5})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow x = \frac{2 (\sqrt{3} + \sqrt{5})}{3 - 5} \Rightarrow x = \frac{2 (\sqrt{3} + \sqrt{5})}{- 2} \Rightarrow x = \frac{- (\sqrt{3} + \sqrt{5})}{1} \Rightarrow x = - \sqrt{3} - \sqrt{5} y = \frac{2}{\sqrt{3} + \sqrt{5}} Multiply and divide RHS by (\sqrt{3} - \sqrt{5}), we get \Rightarrow y = \frac{2}{\sqrt{3} + \sqrt{5}} \times \frac{\sqrt{3} - \sqrt{5}}{\sqrt{3} - \sqrt{5}} \Rightarrow y = \frac{2 (\sqrt{3} - \sqrt{5})}{{(\sqrt{3})}^{2} - {(\sqrt{5})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow y = \frac{2 (\sqrt{3} - \sqrt{5})}{3 - 5} \Rightarrow y = \frac{2 (\sqrt{3} - \sqrt{5})}{- 2} \Rightarrow y = \frac{- (\sqrt{3} - \sqrt{5})}{1} \Rightarrow y = - \sqrt{3} + \sqrt{5} Thus, x + y = (- \sqrt{3} - \sqrt{5}) + (- \sqrt{3} + \sqrt{5}) = (- \sqrt{3} - \sqrt{5} - \sqrt{3} + \sqrt{5}) = (- 2 \sqrt{3}) = - 2 \sqrt{3}$

Hence, x + y = $- 2 \sqrt{3}$ .

Page No 3.18:

Question 9:

If $\sqrt{5 + 2 \sqrt{6}} = A + \sqrt{3}, then A =_________.$

Answer:

$Given : \sqrt{5 + 2 \sqrt{6}} = A + \sqrt{3} Now, \sqrt{5 + 2 \sqrt{6}} = A + \sqrt{3} \Rightarrow \sqrt{2 + 3 + 2 \sqrt{6}} = A + \sqrt{3} \Rightarrow \sqrt{{(\sqrt{2})}^{2} + {(\sqrt{3})}^{2} + 2 \sqrt{6}} = A + \sqrt{3} \Rightarrow \sqrt{{(\sqrt{2} + \sqrt{3})}^{2}} = A + \sqrt{3} (Using the identity : {(a + b)}^{2} = a^{2} + b^{2} + 2 a b) \Rightarrow \sqrt{2} + \sqrt{3} = A + \sqrt{3} \Rightarrow A = \sqrt{2}$

Hence, if $\sqrt{5 + 2 \sqrt{6}} = A + \sqrt{3}, then A = \sqrt{2} .$

Page No 3.18:

Question 10:

$\sqrt{7 + 2 \sqrt{6}} - \sqrt{7 - 2 \sqrt{6}} =__________.$

Answer:

$\sqrt{7 + 2 \sqrt{6}} = \sqrt{6 + 1 + 2 \sqrt{6}} = \sqrt{{(\sqrt{6})}^{2} + {(1)}^{2} + 2 \sqrt{6}} = \sqrt{{(\sqrt{6} + 1)}^{2}} (Using the identity : {(a + b)}^{2} = a^{2} + b^{2} + 2 a b) = \sqrt{6} + 1 . . . . (1) \sqrt{7 - 2 \sqrt{6}} = \sqrt{6 + 1 - 2 \sqrt{6}} = \sqrt{{(\sqrt{6})}^{2} + {(1)}^{2} - 2 \sqrt{6}} = \sqrt{{(\sqrt{6} - 1)}^{2}} (Using the identity : {(a - b)}^{2} = a^{2} + b^{2} - 2 a b) = \sqrt{6} - 1 . . . . (2) From (1) and (2) \sqrt{7 + 2 \sqrt{6}} - \sqrt{7 - 2 \sqrt{6}} = (\sqrt{6} + 1) - (\sqrt{6} - 1) = \sqrt{6} + 1 - \sqrt{6} + 1 = 2$

Hence, $\sqrt{7 + 2 \sqrt{6}} - \sqrt{7 - 2 \sqrt{6}} = 2 .$

Page No 3.18:

Question 11:

If $x = \frac{2}{\sqrt{10} - \sqrt{8}} and y = \frac{2}{\sqrt{10} + 2 \sqrt{2}}, then {(x - y)}^{2} =____________.$

Answer:

$Given : x = \frac{2}{\sqrt{10} - \sqrt{8}} y = \frac{2}{\sqrt{10} + 2 \sqrt{2}} Now, x = \frac{2}{\sqrt{10} - \sqrt{8}} Multiply and divide RHS by (\sqrt{10} + \sqrt{8}), we get \Rightarrow x = \frac{2}{\sqrt{10} - \sqrt{8}} \times \frac{\sqrt{10} + \sqrt{8}}{\sqrt{10} + \sqrt{8}} \Rightarrow x = \frac{2 (\sqrt{10} + \sqrt{8})}{{(\sqrt{10})}^{2} - {(\sqrt{8})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow x = \frac{2 (\sqrt{10} + \sqrt{8})}{10 - 8} \Rightarrow x = \frac{2 (\sqrt{10} + \sqrt{8})}{2} \Rightarrow x = \frac{(\sqrt{10} + \sqrt{8})}{1} \Rightarrow x = \sqrt{10} + \sqrt{8} \Rightarrow x = \sqrt{10} + 2 \sqrt{2} y = \frac{2}{\sqrt{10} + 2 \sqrt{2}} Multiply and divide RHS by (\sqrt{10} - 2 \sqrt{2}), we get \Rightarrow y = \frac{2}{\sqrt{10} + 2 \sqrt{2}} \times \frac{\sqrt{10} - 2 \sqrt{2}}{\sqrt{10} - 2 \sqrt{2}} \Rightarrow y = \frac{2 (\sqrt{10} - 2 \sqrt{2})}{{(\sqrt{10})}^{2} - {(2 \sqrt{2})}^{2}} (Using the identity : (a + b) (a - b) = a^{2} - b^{2}) \Rightarrow y = \frac{2 (\sqrt{10} - 2 \sqrt{2})}{10 - 8} \Rightarrow y = \frac{2 (\sqrt{10} - 2 \sqrt{2})}{2} \Rightarrow y = \frac{(\sqrt{10} - 2 \sqrt{2})}{1} \Rightarrow y = \sqrt{10} - 2 \sqrt{2} Thus, {(x - y)}^{2} = {((\sqrt{10} + 2 \sqrt{2}) - (\sqrt{10} - 2 \sqrt{2}))}^{2} = {(\sqrt{10} + 2 \sqrt{2} - \sqrt{10} + 2 \sqrt{2})}^{2} = {(4 \sqrt{2})}^{2} = 32$

Hence, if $x = \frac{2}{\sqrt{10} - \sqrt{8}} and y = \frac{2}{\sqrt{10} + 2 \sqrt{2}}, then {(x - y)}^{2} = 32 .$

Page No 3.18:

Question 12:

If $\sqrt{13 - x \sqrt{10}} = \sqrt{8} + \sqrt{5}, then x =__________.$

Answer:

$Given : \sqrt{13 - x \sqrt{10}} = \sqrt{8} + \sqrt{5} Now, \sqrt{13 - x \sqrt{10}} = \sqrt{8} + \sqrt{5} \Rightarrow \sqrt{13 - x \sqrt{10}} = \sqrt{{(\sqrt{8} + \sqrt{5})}^{2}} \Rightarrow \sqrt{13 - x \sqrt{10}} = \sqrt{{(\sqrt{8})}^{2} + {(\sqrt{5})}^{2} + 2 (\sqrt{8}) (\sqrt{5})} (Using the identity : {(a + b)}^{2} = a^{2} + b^{2} + 2 a b) \Rightarrow \sqrt{13 - x \sqrt{10}} = \sqrt{8 + 5 + 2 (2 \sqrt{2}) (\sqrt{5})} \Rightarrow \sqrt{13 - x \sqrt{10}} = \sqrt{13 + 4 \sqrt{10}} \Rightarrow - x = 4 \Rightarrow x = - 4$

Hence, if $\sqrt{13 - x \sqrt{10}} = \sqrt{8} + \sqrt{5}, then x = - 4 .$

Page No 3.19:

Question 1:

Write the value of $(2 + \sqrt{3}) (2 - \sqrt{3}) .$

Answer:

Given that

It can be simplified as

Hence the value of the given expression is.

Page No 3.19:

Question 2:

Write the reciprocal of $5 + \sqrt{2}$ .

Answer:

Given that, it’s reciprocal is given as

It can be simplified by rationalizing the denominator. The rationalizing factor of is, we will multiply numerator and denominator of the given expression by, to get

Hence reciprocal of the given expression is.

Page No 3.19:

Question 3:

Write the rationalisation factor of $7 - 3 \sqrt{5}$ .

Answer:

The rationalizing factor of is. Hence the rationalizing factor of is .

Page No 3.19:

Question 4:

If $\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = x + y \sqrt{3},$ find the values of x and y.

Answer:

It is given that;

.we need to find x and y

We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

On equating rational and irrational terms, we get

Hence, we get.

Page No 3.19:

Question 5:

If x= $\sqrt{2} - 1$ , then write the value of $\frac{1}{x} .$

Answer:

Given that.Hence is given as

We know that rationalization factor for is . We will multiply each side of the given expression by, to get

Hence the value of the given expression is.

Page No 3.19:

Question 6:

If $a = \sqrt{2} + 1$ , then find the value of $a - \frac{1}{a}$ .

Answer:

Given that, hence is given as

.we are asked to find

We know that rationalization factor for is . We will multiply each side of the given expression by, to get

Therefore,

Hence value of the given expression is.

Page No 3.19:

Question 7:

If $x = 2 + \sqrt{3}$ , find the value of $x + \frac{1}{x}$ .

Answer:

Given that, hence $\frac{1}{x}$ is given as

.We are asked to find

We know that rationalization factor for is . We will multiply each side of the given expression by, to get

Therefore,

Hence value of the given expression is.

Page No 3.19:

Question 8:

Write the rationalisation factor of $\sqrt{5} - 2$ .

Answer:

Given that, we know that rationalization factor of is

So the rationalization factor of is.

Page No 3.19:

Question 9:

Simplify $\sqrt{3 + 2 \sqrt{2}}$ .

Answer:

We are asked to simplify. It can be written in the form as

Hence the value of given expression is.

Page No 3.19:

Question 10:

Simplify $\sqrt{3 - 2 \sqrt{2}}$ .

Answer:

We are asked to simplify. It can be written in the form as

Hence the value of the given expression is.

Page No 3.19:

Question 11:

If $x = 3 + 2 \sqrt{2}$ , then find the value of $\sqrt{x} - \frac{1}{\sqrt{x}}$ .

Answer:

Given that:.It can be written in the form as

Therefore,

We know that rationalization factor for is . We will multiply numerator and denominator of the given expression by, to get

Hence,

Therefore, value of the given expression is.

Page No 3.2:

Question 1:

Simplify each of the following:

(i) $\sqrt[3]{4} \times \sqrt[3]{16}$

(ii) $\frac{\sqrt[4]{1250}}{\sqrt[4]{2}}$

Answer:

(i) We know that. We will use this property to simplify the expression.

Hence the value of the given expression is .

(ii) We know that. We will use this property to simplify the expression.

Hence the value of the given expression is.

Page No 3.2:

Question 2:

Simplify the following expressions:

(i) $(4 + \sqrt{7}) (3 + \sqrt{2})$

(ii) $(3 + \sqrt{3}) (5 - \sqrt{2})$

(iii) $(\sqrt{5} - 2) (\sqrt{3} - \sqrt{5})$

Answer:

(i) We can simplify the expression as

Hence the value of the expression is

(ii) We can simplify the expression as

Hence the value of the expression is

(iii) We can simplify the expression as

Hence the value of the expression is .

Page No 3.2:

Question 3:

Simplify the following expressions:

(i) $(11 + \sqrt{11}) (11 - \sqrt{11})$

(ii) $(5 + \sqrt{7}) (5 - \sqrt{7})$

(iii) $(\sqrt{8} - \sqrt{2}) (\sqrt{8} + \sqrt{2})$

(iv) $(3 + \sqrt{3}) (3 - \sqrt{3})$

(v) $(\sqrt{5} - \sqrt{2}) (\sqrt{5} + \sqrt{2})$

Answer:

(i) We know that. We will use this property to simplify the expression.

Hence the value of expression is 110.

(ii) We know that. We will use this property to simplify the expression.

Hence the value of expression is 18.

(iii) We know that. We will use this property to simplify the expression.

Hence the value of expression is 6

(iv) We know that. We will use this property to simplify the expression.

Hence the value of expression is 6.

(v) We know that. We will use this property to simplify the expression.

Hence the value of expression is 3.

Page No 3.3:

Question 4:

Simplify the following expressions:

(i) ${(\sqrt{3} + \sqrt{7})}^{2}$

(ii) ${(\sqrt{5} - \sqrt{3})}^{2}$

(iii) ${(2 \sqrt{5} + 3 \sqrt{2})}^{2}$

Answer:

(i) We know that. We will use this property to simplify the expression.

Hence the value of expression is

(ii) We know that. We will use this property to simplify the expression.

Hence the value of expression is

(iii) We know that. We will use this property to simplify the expression.

Hence the value of expression is.

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