RD Sharma Solutions Class 9 Rationalisation Exercise 3.2

RD Sharma Class 9 Solutions Chapter 3 Ex 3.2 Free Download

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

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