# RD Sharma Solutions Class 9 Rationalisation Exercise 3.2

## RD Sharma Solutions Class 9 Chapter 3 Ex 3.2

1. Rationalize the denominator of each of the following:

(i) 3/√5

(ii) 3/2√5

(iii) 1/√12

(iv) √2/√5

(v) (√3+1)/√2

(vi) (√2+√5)/√3

(vii) 3√2/√5

Solution:

(i) Given, 3/√5

Multiply both numerator and denominator to rationalise the denominator, with √5;

= $\frac{3\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}$

= $\frac{3\times \sqrt{5}}{5}$

(ii)Given, 3/2√5

Multiply both numerator and denominator to rationalise the denominator, with √5;

=  $\frac{3\times \sqrt{5}}{2\sqrt{5}\times \sqrt{5}}$

= $\frac{3\sqrt{5}}{2\times \sqrt{5\times 5}}$

= $\frac{3\sqrt{5}}{2\times 5}$

= $\frac{3\sqrt{5}}{10}$

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

Multiply both numerator and denominator to rationalise the denominator, with ${\sqrt{12}}$

$\frac{1\times \sqrt{12}}{\sqrt{12}\times \sqrt{12}}$

= $\frac{ \sqrt{12}}{\sqrt{12\times 12}}$

= $\frac{ \sqrt{12}}{12}$

(iv) Given, $\frac{\sqrt{2}}{\sqrt{3}}$

Multiply both numerator and denominator to rationalise the denominator, with ${\sqrt{3}}$

= $\frac{\sqrt{2}\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}$

= $\frac{\sqrt{2\times 3}}{\sqrt{3\times 3}}$

= $\frac{\sqrt{6}}{3}$

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

Multiply both numerator and denominator to rationalise the denominator, with ${\sqrt{2}}$

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

= $\frac{(\sqrt{3}\times\sqrt{2})+\sqrt{2}}{\sqrt{2\times 2}}$

= $\frac{\sqrt{6}+\sqrt{2}}{2}$

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

Multiply both numerator and denominator to rationalise the denominator, with ${\sqrt{3}}$

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

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

= $\frac{\sqrt{6} + \sqrt{15}}{3}$

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

Multiply both numerator and denominator to rationalise the denominator, with ${\sqrt{5}}$

= $\frac{3\sqrt{2}\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}$

= $\frac{3\sqrt{2\times 5}}{\sqrt{5\times 5}}$

= $\frac{3\sqrt{10}}{5}$

2. Find the value to three places of decimals of each of the following.

It is given that √2 = 1.414, √3 = 1.732, √5 = 2.236, $\sqrt{10}$ = 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) $\frac{\sqrt{2}-1}{\sqrt{5}}$

Solution:

Given here, √2 = 1.414, √3 = 1.732, √5 = 2.236, $\sqrt{10}$ = 3.162.

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

Multiply both numerator and denominator to rationalise the denominator, with ${\sqrt{3}}$

= $\frac{2\sqrt{3}}{\sqrt{3}\times \sqrt{3}}$

= $\frac{2\sqrt{3}}{\sqrt{3\times 3}}$

= $\frac{2\sqrt{3}}{3}$

= $\frac{2\times 1.732}{3}$

= $\frac{3.464}{3}$

= 1.154666666

(ii) Given here, $\frac{3}{\sqrt{10}}$

Multiply both numerator and denominator to rationalise the denominator, with ${\sqrt{10}}$

= $\frac{3\sqrt{10}}{\sqrt{10}\times \sqrt{10}}$

= $\frac{3\sqrt{10}}{\sqrt{10\times10}}$

= $\frac{3\sqrt{10}}{10}$

= $\frac{9.486}{10}$

= 0. 9486

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

Multiply both numerator and denominator to rationalise the denominator, with ${\sqrt{3}}$

= $\frac{(\sqrt{5}\times \sqrt{2}) + \sqrt{2}}{\sqrt{2}\times \sqrt{2}}$

= $\frac{\sqrt{10} + \sqrt{2}}{2}$

= $\frac{4.576}{2}$

= 2.288

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

Multiply both numerator and denominator to rationalise the denominator, with ${\sqrt{2}}$

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

= $\frac{\sqrt{20}+\sqrt{30}}{2}$

= $\frac{(\sqrt{10}\times \sqrt{2})+(\sqrt{10}\times \sqrt{3})}{2}$

= $\frac{(3.162\times 1.414) + (3.162\times 1.732)}{2}$

= $\frac{(4.471068)+(5.476584)}{2}$

= $\frac{9.947652}{2}$

= 4.973826

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

=  $\frac{2 + 1.732}{3}$

= $\frac{3.732}{3}$

= 1.244

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

Multiply both numerator and denominator to rationalise the denominator, with ${\sqrt{5}}$

= $\frac{(\sqrt{2}\times \sqrt{5})-\sqrt{5}}{\sqrt{5}\times \sqrt{5}}$

= $\frac{\sqrt{10}-\sqrt{5}}{5}$

= $\frac{3.162-2.236}{5}$

= $\frac{0.926}{5}$

= 0.1852

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

Solution:

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor ${3-\sqrt{2}}$

= $\frac{3 – \sqrt{2}}{(3 + \sqrt{2}) (3 – \sqrt{2})}$

=  $\frac{3 – \sqrt{2}}{9 – 2}$                          [∵ (a + b) (a – b) = a2 – b2]

=  $\frac{3 – \sqrt{2}}{7}$

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor ${\sqrt{6}+\sqrt{5}}$

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

= $\frac{\sqrt{6}+\sqrt{2}}{6-2}$                 [∵(a + b) (a – b) = a2 – b2]

$\frac{\sqrt{6}+\sqrt{2}}{4}$

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor ${\sqrt{41}+5}$

= $\frac{16\times ({\sqrt{41} + 5)}}{(\sqrt{41} – 5)({\sqrt{41} + 5})}$

= $\frac{16{\sqrt{41} + 80}}{41 – 5}$            [∵ (a + b) (a – b) = a2 – b2]

= $\frac{16{\sqrt{41}+80}}{16}$

= $\frac{16({\sqrt{41}+5)}}{16}$

= ${\sqrt{41}+5}$

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor ${5\sqrt{3}+3\sqrt{5}}$

= $\frac{30\times(5\sqrt{3}+3\sqrt{5}) }{(5\sqrt{3}-3\sqrt{5})(5\sqrt{3}+3\sqrt{5})}$

= $\frac{30\times(5\sqrt{3}+3\sqrt{5}) }{75-45}$                     [∵ (a + b) (a – b) = a2 – b2]

= $\frac{30\times(5\sqrt{3}+3\sqrt{5}) }{30}$

= $5\sqrt{3}+3\sqrt{5}$

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor ${2\sqrt{5}+\sqrt{3}}$

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

= $\frac{{2\sqrt{5}+\sqrt{3}}}{20-3}$                 [∵ (a + b) (a – b) = a2 – b2]

= $\frac{{2\sqrt{5}+\sqrt{3}}}{17}$

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor ${2\sqrt{2}+\sqrt{3}}$

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

= $\frac{(2\sqrt{6}+3+2\sqrt{2}+\sqrt{3})}{8-3}$                 [∵ (a + b) (a – b) = a2 – b2]

= $\frac{(2\sqrt{6}+3+2\sqrt{2}+\sqrt{3})}{5}$

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor ${6-4\sqrt{2}}$

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

= $\frac{(6-4\sqrt{2})^{2}}{36-32}$

∵ [(a + b) (a – b) = a2 – bAnd (a – b)2 = (a2 – 2ab + b2)]

= $\frac{36-48\sqrt{2}+32}{4}$

= $\frac{68-48\sqrt{2}}{4}$

= $\frac{4(17-12\sqrt{2})}{4}$

= $17-12\sqrt{2}$

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $2\sqrt{5}-3$

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

[∵ (a + b) (a – b) = a2 – b2]

= $\frac{6\sqrt{10} – 9\sqrt{2} + 2\sqrt{5} – 3}{(20 – 9)}$

= $\frac{6\sqrt{10}-9\sqrt{2}+2\sqrt{5}-3}{11}$

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $\sqrt{(a^{2}+b^{2})}-a$

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

[∵ (a + b) (a – b) = a2 – b2]

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

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

4. Rationalize the denominator and simplify:

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

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

Solution:

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $\sqrt{3}-\sqrt{2}$

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

[∵ (a + b) (a – b) = a2 – b2]

= $\frac{(\sqrt{3}-\sqrt{2})^{2}}{3-2}$

Since, (a – b)2 = (a2 – 2ab + b2)

= $\frac{3-2\sqrt{3}\sqrt{2}+2}{1}$

= $5-2\sqrt{6}$

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $7-4\sqrt{3}$

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

[∵ (a + b) (a – b) = a2 – b2]

$\frac{(5+2\sqrt{3})(7-4\sqrt{3})}{49-48}$

= 35 – $20\sqrt{3}$ + $14\sqrt{3}$ – 24

= 11 – $6\sqrt{3}$

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $3+2\sqrt{2}$

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

[∵ (a + b) (a – b) = a2 – b2]

= $\frac{(1+\sqrt{2})(3+2\sqrt{2})}{9-8}$

= $3+2\sqrt{2}+3\sqrt{2}+4$

= 7 + $5\sqrt{2}$

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $3\sqrt{5}+2\sqrt{6}$

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

[∵ (a + b) (a – b) = a2 – b2]

= $\frac{(2\sqrt{6}-\sqrt{5})(3\sqrt{5}+2\sqrt{6})}{45-24}$

= $\frac{(2\sqrt{6}-\sqrt{5})(3\sqrt{5}+2\sqrt{6})}{21}$

= $\frac{(2\sqrt{6}-\sqrt{5})(3\sqrt{5}+2\sqrt{6})}{21}$

= $\frac{6\sqrt{30}+24-15-2\sqrt{30}}{21}$

= $\frac{4\sqrt{30}+9}{21}$

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $\sqrt{48} – \sqrt{18}$

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

[∵ (a + b) (a – b) = a2 – b2]

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

= $\frac{48-12\sqrt{6}+20\sqrt{6}-30}{30}$

= $\frac{18+8\sqrt{6}}{30}$

= $\frac{9+4\sqrt{6}}{15}$

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $2\sqrt{2}-3\sqrt{3}$

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

= $\frac{(2\sqrt{3} -\sqrt{5})(2\sqrt{2}-3\sqrt{3})}{8 – 27}$

= $\frac{(4\sqrt{6} – 2\sqrt{10}) – 18 + 3\sqrt{15})}{-19}$

= $\frac{(18 – 4\sqrt{6} + 2\sqrt{10} – 3\sqrt{15})}{19}$

5. Simplify:

(i) $\frac{3\sqrt{2} – 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} + \frac{\sqrt{12}}{\sqrt{3} – \sqrt{2}}$

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

(iii) $\frac{7+3\sqrt{5}}{3+\sqrt{5}} – \frac{7-3\sqrt{5}}{3-\sqrt{5}}$

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

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

Solution:

(i)Given here, $\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}} + \frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}$

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $3\sqrt{2}-2\sqrt{3}$ for $\frac{1}{3\sqrt{2}+2\sqrt{3}}$ and the rationalizing factor $\sqrt{3}+\sqrt{2}$ for $\frac{1}{\sqrt{3}-\sqrt{2}}$

= $\frac{(3\sqrt{2}-2\sqrt{3})(3\sqrt{2} -2\sqrt{3})}{(3\sqrt{2} + 2\sqrt{3})(3\sqrt{2} – 2\sqrt{3})} + \frac{\sqrt{12}({\sqrt{3} + \sqrt{2}})}{(\sqrt{3} – \sqrt{2})({\sqrt{3} + \sqrt{2}})}$

[∵ (a + b) (a – b) = a2 – b2]

= $\frac{(3\sqrt{2}-2\sqrt{3})(3\sqrt{2}-2\sqrt{3})}{18-12} + \frac{\sqrt{12}({\sqrt{3}+\sqrt{2}})}{3-2}$

Since ,(a – b)2 = (a2 – 2ab + b2))

= $\frac{(3\sqrt{2})^{2}-(2\times 3\sqrt{2}\times 2\sqrt{3})+(2\sqrt{3})^{2}}{6} + 2\sqrt{3}({\sqrt{3}+\sqrt{2}})$

= $\frac{(18-12\sqrt{6}+12)}{6} +({6+2\sqrt{6}})$

= $3-2\sqrt{6}+2 +({6+2\sqrt{6}})$

= $5-2\sqrt{6} +({6+2\sqrt{6}})$

= 11

(ii) Given here, $\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $\sqrt{5}+\sqrt{3}$ for $\frac{1}{\sqrt{5}-\sqrt{3}}$ and the rationalizing factor $\sqrt{5}-\sqrt{3}$ for $\frac{1}{\sqrt{5}+\sqrt{3}}$

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

Since, (a + b) (a – b) = a2 – b2, (a – b)2 = (a2 – 2ab + b2) and (a + b)2 = (a2 + 2ab + b2)

= $\frac{5+2\times \sqrt{5}\times \sqrt{3}+3}{5-3}+\frac{5-2\times \sqrt{3}\times \sqrt{5}+3}{5-3}$

= $\frac{8+2\sqrt{15}+8-2\sqrt{15}}{2}$

= $\frac{16}{2}$

= 8

(iii) Given here, $\frac{7+3\sqrt{5}}{3+\sqrt{5}} – \frac{7-3\sqrt{5}}{3-\sqrt{5}}$

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $3-\sqrt{5}$ for $\frac{1}{3+\sqrt{5}}$ and the rationalizing factor $3+\sqrt{5}$ for $\frac{1}{3-\sqrt{5}}$

$\frac{(7+3\sqrt{5})(3-\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})} – \frac{(7-3\sqrt{5})(3+\sqrt{5})}{(3-\sqrt{5})(3-\sqrt{5})}$

Since, (a + b) (a – b) = a2 – b2

= $\frac{(7+3\sqrt{5})(3-\sqrt{5})}{9-5} – \frac{(7-3\sqrt{5})(3+\sqrt{5})}{9-5}$

= $\frac{(21-7\sqrt{5}+9\sqrt{5}-15)}{4} – \frac{(21+7\sqrt{5}-9\sqrt{5}-15)}{4}$

= $\frac{(6+2\sqrt{5})}{4} – \frac{(6-2\sqrt{5})}{4}$

= $\frac{4\sqrt{5}}{4}$

= √5

(iv) Given here, $\frac{1}{2+\sqrt{3}} + \frac{2}{\sqrt{5}-\sqrt{3}} + \frac{1}{2-\sqrt{5}}$

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $2-\sqrt{3}$ for $\frac{1}{2+\sqrt{3}}$, the rationalizing factor $\sqrt{5}+\sqrt{3}$ for $\frac{1}{\sqrt{5}-\sqrt{3}}$, and the rationalizing factor $2+\sqrt{5}$ for $\frac{1}{2-\sqrt{5}}$

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

Since, (a + b) (a – b) = a2 – b2

= $\frac{2-\sqrt{3}}{4 – 3} + \frac{2\times (\sqrt{5}+\sqrt{3})}{5-3} + \frac{2+\sqrt{5}}{4-5}$

= $\frac{2-\sqrt{3}}{1} + \frac{2\sqrt{5}+2\sqrt{3}}{2} + \frac{2+\sqrt{5}}{-1}$

= $\frac{4-2\sqrt{3}+2\sqrt{5}+2\sqrt{3}-4-2\sqrt{5}}{2}$

= $\frac{0}{2}$

= 0

(v) Given here, $\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{3}{\sqrt{5}+\sqrt{2}}$

Multiply both numerator and denominator to rationalise the denominator, with  the rationalizing factor $\sqrt{5}-\sqrt{3}$ for $\frac{1}{\sqrt{5}+\sqrt{3}}$, the rationalizing factor $\sqrt{3}-\sqrt{2}$ for $\frac{1}{\sqrt{3}+\sqrt{2}}$, and the rationalizing factor $\sqrt{5}-\sqrt{2}$ for $\frac{1}{\sqrt{5}+\sqrt{2}}$

= $\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})} + \frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} – \frac{3\times (\sqrt{5}-\sqrt{2})}{\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}$

Since, (a + b) (a – b) = a2 – b2

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

= $\frac{2\sqrt{5}-2\sqrt{3}}{2}+\frac{\sqrt{3}-\sqrt{2}}{1}-\frac{3\times\sqrt{5}-3\sqrt{2}}{3}$

= $\frac{6\sqrt{5}-6\sqrt{3}+6\sqrt{3}-6\sqrt{2}-6\sqrt{5}+6\sqrt{2}}{3}$

= $\frac{0}{3}$

= 0

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) $\frac{4+\sqrt{2}}{2+\sqrt{2}}$ = $a-\sqrt{b}$

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

Solution:

(i) Given here,

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $\sqrt{3}-1$

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

Since, (a + b) (a – b) = a2 – b2

= $\frac{3-2\sqrt{3}+1}{3-1}$

= $\frac{4-2\sqrt{3}}{2}$

= $2-\sqrt{3}$

⇒$2-\sqrt{3}$ = $a-b\sqrt{3}$

If we compare, the rational and irrational parts, in the above equation, we will get,

a = 2 and b = 1

(ii) Given here,

$\frac{4+\sqrt{2}}{2+\sqrt{2}}$ = $a-\sqrt{b}$

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $2-\sqrt{2}$

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

Since, (a + b) (a – b) = a2 – b2

= $\frac{(8-4\sqrt{2}+2\sqrt{2}-2)}{4-2}$

= $\frac{(6-2\sqrt{2})}{2}$

= $3-\sqrt{2}$

⇒$3-\sqrt{2}$ = $a-\sqrt{b}$

If we compare, the rational and irrational parts, in the above equation, we will get,

a = 3 and b = 2

(iii) Given here,

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $3+\sqrt{2}$

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

Since, (a + b) (a – b) = a2 – b2

= $\frac{(9+6\sqrt{2}+2)}{9-2}$

$\frac{(11+6\sqrt{2})}{7}$

= $\frac{11}{7}+\frac{6\sqrt{2}}{7}$

⇒$\frac{11}{7}+\frac{6\sqrt{2}}{7}$ = $a+b\sqrt{2}$

If we compare, the rational and irrational parts, in the above equation, we will get,

a = $\frac{11}{7}+\frac{6\sqrt{2}}{7}$ and

b = $\frac{6}{7}$

(iv)  Given here,

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $7-4\sqrt{3}$

= $\frac{(5+3\sqrt{3})(7-4\sqrt{3})}{(7+4\sqrt{3})(7-4\sqrt{3})}$

Since, (a + b) (a – b) = a2 – b2

= $\frac{(35-20\sqrt{3}+21\sqrt{3}-36)}{49-48}$

= -1 + √3

-1 + √3 = $a + b\sqrt{3}$

If we compare, the rational and irrational parts, in the above equation, we will get,

a = -1 and

b = 1

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $\sqrt{11} – \sqrt{7}$

$\frac{(\sqrt{11}-\sqrt{7})(\sqrt{11} – \sqrt{7})}{(\sqrt{11}+\sqrt{7})(\sqrt{11} – \sqrt{7})}$

Since, (a + b) (a – b) = a2 – b2

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

= $\frac{(18-2\sqrt{77})}{4}$

= $\frac{9}{2}-\frac{\sqrt{77}}{2}$

⇒$\frac{9}{2}-\frac{\sqrt{77}}{2}$ = $a – b\sqrt{77}$

If we compare, the rational and irrational parts, in the above equation, we will get,

a = $\frac{9}{2}$ and

b = $\frac{1}{2}$

(vi) Given here,

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

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $4 + 3\sqrt{5}$

⇒$\frac{(4+3\sqrt{5})(4 + 3\sqrt{5})}{(4-3\sqrt{5})(4 + 3\sqrt{5})}$

Since, (a + b) (a – b) = a2 – b2

⇒$\frac{(16+24\sqrt{5}+45)}{-29}$

⇒$\frac{(61+24\sqrt{5})}{-29}$

$\frac{-61}{29}-\frac{(24\sqrt{5})}{29}$

⇒$\frac{-61}{29}-\frac{(24\sqrt{5})}{29}$ = $a + b\sqrt{5}$

If we compare, the rational and irrational parts, in the above equation, we will get,

a = $\frac{-61}{29}$, and

b = $\frac{-24}{29}$

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

Solution: Given here,

x = 2 + √3,

To find the value of $x^{3}+\frac{1}{x^{3}}$

We have, x = 2 + √3,

⇒$\frac{1}{x}$ = $\frac{1}{2+\sqrt{3}}$

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $2-\sqrt{3}$ for $\frac{1}{2+\sqrt{3}}$

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

Since, (a + b) (a – b) = a2 – b2

⇒$\frac{1}{x}$ = $\frac{2-\sqrt{3}}{4-3}$

x + $\frac{1}{x}$ = 2 + √3 + 2 – √3

x + $\frac{1}{x}$ = 4

We know that, $(a^{3}+b^{3})=(a+b)(a^{2}-ab+b^{2})$

⇒$(x^{3}+\frac{1}{x^{3}}) = (x+\frac{1}{x})(x^{2}-x.\frac{1}{x}+\frac{1}{x^{2}})$

⇒$(x^{3}+\frac{1}{x^{3}})=(x+\frac{1}{x})(x^{2}+\frac{1}{x}^{2}-1)$

⇒$(x^{3}+\frac{1}{x^{3}})=(x+\frac{1}{x})(x^{2}+\frac{1}{x^{2}}+2-2-1)$

⇒$(x^{3}+\frac{1}{x^{3}})=(x+\frac{1}{x})(x^{2}+\frac{1}{x^{2}}+2(x.\frac{1}{x})-2-1)$

⇒$(x^{3}+\frac{1}{x^{3}})=(x+\frac{1}{x})((x+\frac{1}{x})^{2}-3)$

Substituting the value of $x+\frac{1}{x}$ in the above equation, we get,

⇒$(x^{3}+\frac{1}{x^{3}})=(4)(4^{2}-3)$

⇒$(x^{3}+\frac{1}{x^{3}})=52$

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

Solution: Given here,

x = 3 + $\sqrt{8}$,

We need to find the value of $(x^{2}+\frac{1}{x^{2}})$

We already have, x = 3 + $\sqrt{8}$,

⇒ $\frac{1}{x} = \frac{1}{3+\sqrt{8}}$

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $3-\sqrt{8}$ for $\frac{1}{3+\sqrt{8}}$

⇒ $\frac{1}{x} = \frac{3-\sqrt{8}}{(3+\sqrt{8})(3-\sqrt{8})}$

Since, (a + b) (a – b) = a2 – b2

⇒ $\frac{1}{x} = \frac{3-\sqrt{8}}{9-8}$

⇒ $\frac{1}{x} = 3-\sqrt{8}$

⇒ $(x^{2}+\frac{1}{x^{2}}) = ((3+\sqrt{8})^{2}(3-\sqrt{8})^{2})$

⇒ $(x^{2}+\frac{1}{x^{2}})= ((9+8+6\sqrt{8})+(9+8-6\sqrt{8}))$

⇒ 34

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

Solution: Given here,

$\frac{6}{\sqrt{5}-\sqrt{3}}$

Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $\sqrt{5}+\sqrt{3}$ for $\frac{1}{\sqrt{5}-\sqrt{3}}$

= $\frac{6(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}$

Since, (a + b) (a – b) = a2 – b2

= $\frac{6\sqrt{5}+6\sqrt{3}}{5-3}$

= $\frac{6\sqrt{5}+6\sqrt{3}}{2}$

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

= 3(2.236+1.732)

= 3(3.968)

= 11.904

10. Find the values of each of the following correct to three places of decimals, it being given that

√2 = 1.414, √3 = 1.732,  √5 = 2.236, √6 = 2.4495, √10 = 3.162

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

Solution: Multiply both numerator and denominator to rationalise the denominator,with the rationalizing factor $3-2\sqrt{5}$

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

Since, (a + b) (a – b) = a2 – b2

= $\frac{(3-\sqrt{5})(3-2\sqrt{5})}{9-20}$

= $\frac{(9-6\sqrt{5}-3\sqrt{5}+10)}{-11}$

= $\frac{(19-9\sqrt{5})}{-11}$

= $\frac{(9\sqrt{5}-19)}{11}$

= $\frac{(9(2.236))-19)}{11}$

= $\frac{(20.124-19)}{11}$

= $\frac{1.124}{11}$

= 0.102

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

Solution: Multiply both numerator and denominator to rationalise the denominator, with the rationalizing factor $3+2\sqrt{2}$

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

Since, (a + b) (a – b) = a2 – b2

= $\frac{(1+\sqrt{2})(3+2\sqrt{2})}{9-8}$

= $3+2\sqrt{2}+3\sqrt{2}+4$

= 7 + $5\sqrt{2}$

= 7 + 7.07

= 14.07

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

Solution: Given here,

x = $\frac{\sqrt{3}+1}{2}$ and given to find the value of $4x^{3}+2x^{2}-8x+7$

⇒ 2x = $\sqrt{3}+1$

2x – 1 = √3

Squaring both the sides, we will get,

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

⇒ $4x^{2}-4x+1=3$

⇒ $4x^{2}-4x+1-3=0$

⇒ $4x^{2}-4x-2=0$

⇒ $2x^{2}-2x-1=0$

Now, let us take, $4x^{3}+2x^{2}-8x+7$

⇒ 2x($2x^{2}-2x-1$)+ $4x^{2}+2x+2x^{2}-8x+7$

⇒ 2x($2x^{2}-2x-1$)+ $6x^{2}-6x+7$

Since, $2x^{2}-2x-1=0$

⇒ 2x(0)+3($2x^{2}-2x-1$))+7+3

⇒ 0+3(0)+10

⇒ 10