# RD Sharma Solutions Class 9 Rationalisation Exercise 3.1

## RD Sharma Solutions Class 9 Chapter 3 Ex 3.1

1. Simplify each of the following:

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

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

Sol:

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

(Note: $\sqrt[n]{a}\times \sqrt[n]{b}$ = $\sqrt[n]{a \times b}$)

= $\sqrt[3]{4\times 16}$

= $\sqrt[3]{64}$

= $\sqrt[3]{4^{3}}$

= $(4^{3})^{\frac{1}{3}}$

= $4^({3\times \frac{1}{3}})$

= $4^{1}$

= 4

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

(Note:  $\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$=  $\sqrt[n]{\frac{a}{b}}$)

= $\sqrt[4]{\frac{1250}{2}}$

= $\sqrt[4]{\frac{2\times 625}{2}}$

= $\sqrt[4]{625}$

= $\sqrt[4]{15^{4}}$

= $15^({4\times \frac{1}{4}})$

= 15

2. Simplify the following expressions:

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

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

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

Solution:

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

= 12 + $4\sqrt{2}$ + $3\sqrt{7}$ + $\sqrt{7\times 2}$

= 12 + $4\sqrt{2}$ + $3\sqrt{7}$ + $\sqrt{14}$

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

= 15 – $3\sqrt{2}$ + $5\sqrt{3}$$\sqrt{3\times 2}$

= 15 – $3\sqrt{2}$ + $5\sqrt{3}$$\sqrt{6}$

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

= $\sqrt{15} – \sqrt{25} – 2\sqrt{3} + 2\sqrt{5}$

= $\sqrt{15} – \sqrt{5\times 5} – 2\sqrt{3} + 2\sqrt{5}$

= $\sqrt{15} – 5 – 2\sqrt{3} + 2\sqrt{5}$

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) ($\sqrt{5} – \sqrt{2}$)( $\sqrt{5} + \sqrt{2}$)

Solution:

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

As we know, (a + b) (a – b) = ($a^{2} – b^{2}$)

So, $11^{2}$ – 11

• 121 – 11 =110

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

As we know, (a + b) (a – b) = ($a^{2} – b^{2}$)

So, $5^{2}$ – 7

• 25 – 7 = 18

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

As we know, (a + b) (a – b) = ($a^{2} – b^{2}$)

• $\sqrt{8\times 8} – \sqrt{2\times 2}$ = 8 – 2

= 6

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

As we know, (a + b) (a – b) = ($a^{2} – b^{2}$)

= 9 – $\sqrt{3\times 3}$

= 6

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

As we know, (a + b) (a – b) = ($a^{2} – b^{2}$)

= $\sqrt{5\times 5} – \sqrt{2\times 2}$

= 5 – 2

= 3

4. Simplify the following expressions:

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

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

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

Solution:

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

As we know , (a + b) $^{2}$ = ($a^{2} + 2\times a\times b + b^{2}$)

= $\sqrt{3^{2}} + 2\times \sqrt{3}\times \sqrt{7} + \sqrt{7^{2}}$

= 3 + $2\times \sqrt{3\times 7}$ + 7

= 10 + $2\times \sqrt{21}$

(ii) $(\sqrt{5} – \sqrt{3})^{2}$

As we know , (a – b) $^{2}$ = ($a^{2} – 2\times a\times b + b^{2}$)

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

As we know , (a + b) $^{2}$ = ($a^{2} + 2\times a\times b + b^{2}$)

= $4\sqrt{5\times 5}+2\times 2\sqrt{5}\times 3\sqrt{2}+9\sqrt{2\times 2}$

= 20 + $12\sqrt{10}$ + 18

= 28 + $12\sqrt{10}$

#### Practise This Question

212+213+214110×25×1256  is equal to: