RD Sharma Solutions Class 9 Chapter 3 Ex 3.2
1. Rationalize the denominator of each of the following:
(i) \(\frac{3}{\sqrt{5}}\)
(ii) \(\frac{3}{2\sqrt{5}}\)
(iii) \(\frac{1}{\sqrt{12}}\)
(iv) \(\frac{\sqrt{2}}{\sqrt{3}}\)
(v) \(\frac{\sqrt{2} + \sqrt{5}}{\sqrt{3}}\)
(vi) \(\frac{\sqrt{2} + \sqrt{5}}{\sqrt{3}}\)
 (vii) \(\frac{3\sqrt{2}}{\sqrt{5}}\)
Solution:
(i) \(\frac{3}{\sqrt{5}}\)
For rationalizing the denominator, multiply both numerator and denominator with \({\sqrt{5}}\)
= \(\frac{3\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}\)
= \(\frac{3\times \sqrt{5}}{5}\)
(ii) \(\frac{3}{2\sqrt{5}}\)
For rationalizing the denominator, multiply both numerator and denominator with \({\sqrt{5}}\)
= Â \(\frac{3\times \sqrt{5}}{2\sqrt{5}\times \sqrt{5}}\)
= \(\frac{3\sqrt{5}}{2\times \sqrt{5\times 5}}\)
= \(\frac{3\sqrt{5}}{2\times 5}\)
= \(\frac{3\sqrt{5}}{10}\)
(iii) \(\frac{1}{\sqrt{12}}\)
For rationalizing the denominator, multiply both numerator and denominator with \({\sqrt{12}}\)
=Â \(\frac{1\times \sqrt{12}}{\sqrt{12}\times \sqrt{12}}\)
= \(\frac{ \sqrt{12}}{\sqrt{12\times 12}}\)
= \(\frac{ \sqrt{12}}{12}\)
(iv) \(\frac{\sqrt{2}}{\sqrt{3}}\)
For rationalizing the denominator, multiply both numerator and 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) \(\frac{\sqrt{3}+1}{\sqrt{2}}\)
For rationalizing the denominator, multiply both numerator and 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) \(\frac{\sqrt{2} + \sqrt{5}}{\sqrt{3}}\)
For rationalizing the denominator, multiply both numerator and 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) \(\frac{3\sqrt{2}}{\sqrt{5}}\)
For rationalizing the denominator, multiply both numerator and 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 \(\sqrt{2}\)
(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) Â \(\frac{2+\sqrt{3}}{3}\)
(vi) \(\frac{\sqrt{2}-1}{\sqrt{5}}\)
Solution:
Given, \(\sqrt{2}\)
(i) \(\frac{2}{\sqrt{3}}\)
Rationalizing the denominator by multiplying both numerator and 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) Â \(\frac{3}{\sqrt{10}}\)
Rationalizing the denominator by multiplying both numerator and 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) \(\frac{\sqrt{5} + 1}{\sqrt{2}}\)
Rationalizing the denominator by multiplying both numerator and 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) \(\frac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}\)
Rationalizing the denominator by multiplying both numerator and denominator with \({\sqrt{2}}\)
= Â \(\frac{(\sqrt{10}\times \sqrt{2}) + (\sqrt{15}\times \sqrt{2})}{\sqrt{2}\times \sqrt{2}}\)
= \(\frac{\sqrt{20}+\sqrt{30}}{2}\)
= \(\frac{(\sqrt{10}\times \sqrt{2})+(\sqrt{10}\times \sqrt{3})}{2}\)
= \(\frac{(3.162\times 1.414) + (3.162\times 1.732)}{2}\)
= \(\frac{(4.471068)+(5.476584)}{2}\)
= \(\frac{9.947652}{2}\)
= 4.973826
(v) \(\frac{2 + \sqrt{3}}{3}\)
= Â \(\frac{2 + 1.732}{3}\)
= \(\frac{3.732}{3}\)
= 1.244
(vi) \(\frac{\sqrt{2}-1}{\sqrt{5}}\)
Rationalizing the denominator by multiplying both numerator and 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) \(\frac{1}{3+\sqrt{2}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \({3-\sqrt{2}}\)
= \(\frac{3 – \sqrt{2}}{(3 + \sqrt{2}) (3 – \sqrt{2})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= Â \(\frac{3 – \sqrt{2}}{9 – 2}\)
= Â \(\frac{3 – \sqrt{2}}{7}\)
(ii) \(\frac{1}{\sqrt{6}-\sqrt{5}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \({\sqrt{6}+\sqrt{5}}\)
= \(\frac{\sqrt{6} + \sqrt{2}}{(\sqrt{6} -\sqrt{2})(\sqrt{6} + \sqrt{2})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{\sqrt{6}+\sqrt{2}}{6-2}\)
=Â \(\frac{\sqrt{6}+\sqrt{2}}{4}\)
(iii) Â \(\frac{16}{\sqrt{41}-5}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \({\sqrt{41}+5}\)
= \(\frac{16\times ({\sqrt{41} + 5)}}{(\sqrt{41} – 5)({\sqrt{41} + 5})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{16{\sqrt{41} + 80}}{41 – 5}\)
= \(\frac{16{\sqrt{41}+80}}{16}\)
= \(\frac{16({\sqrt{41}+5)}}{16}\)
= \({\sqrt{41}+5}\)
(iv) \(\frac{30}{5\sqrt{3}-3\sqrt{5}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \({5\sqrt{3}+3\sqrt{5}}\)
= \(\frac{30\times(5\sqrt{3}+3\sqrt{5}) }{(5\sqrt{3}-3\sqrt{5})(5\sqrt{3}+3\sqrt{5})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{30\times(5\sqrt{3}+3\sqrt{5}) }{75-45}\)
= \(\frac{30\times(5\sqrt{3}+3\sqrt{5}) }{30}\)
= \(5\sqrt{3}+3\sqrt{5}\)
(v)Â \(\frac{1}{2\sqrt{5}-\sqrt{3}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \({2\sqrt{5}+\sqrt{3}}\)
= \(\frac{{2\sqrt{5}+\sqrt{3}}}{(2\sqrt{5}-\sqrt{3})({2\sqrt{5}+\sqrt{3}})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{{2\sqrt{5}+\sqrt{3}}}{20-3}\)
= \(\frac{{2\sqrt{5}+\sqrt{3}}}{17}\)
(vi) \(\frac{\sqrt{3}+1}{2\sqrt{2}-\sqrt{3}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \({2\sqrt{2}+\sqrt{3}}\)
= \(\frac{(\sqrt{3} + 1)(2\sqrt{2} + \sqrt{3})}{(2\sqrt{2} + \sqrt{3})(2\sqrt{2} – \sqrt{3})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{(2\sqrt{6}+3+2\sqrt{2}+\sqrt{3})}{8-3}\)
= \(\frac{(2\sqrt{6}+3+2\sqrt{2}+\sqrt{3})}{5}\)
(vii) \(\frac{6-4\sqrt{2}}{6+4\sqrt{2}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \({6-4\sqrt{2}}\)
= \(\frac{(6-4\sqrt{2})(6-4\sqrt{2})}{(6+4\sqrt{2})(6-4\sqrt{2})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{(6-4\sqrt{2})^{2}}{36-32}\)
As we know , (a – b) \(^{2}\)
= \(\frac{36-48\sqrt{2}+32}{4}\)
= \(\frac{68-48\sqrt{2}}{4}\)
= \(\frac{4(17-12\sqrt{2})}{4}\)
= \(17-12\sqrt{2}\)
(viii) \(\frac{3\sqrt{2}+1}{2\sqrt{5}-3}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(2\sqrt{5}-3\)
= \(\frac{(3\sqrt{2}+1)\times( 2\sqrt{5}-3)}{(2\sqrt{5}-3)(2\sqrt{5}-3)}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{6\sqrt{10} – 9\sqrt{2} + 2\sqrt{5} – 3}{(20 – 9)}\)
= \(\frac{6\sqrt{10}-9\sqrt{2}+2\sqrt{5}-3}{11}\)
(ix) Â \(\frac{b^{2}}{\sqrt{(a^{2}+b^{2})}+a}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(\sqrt{(a^{2}+b^{2})}-a\)
= \(\frac{b^{2}(\sqrt{(a^{2} + b^{2})} – a)}{(\sqrt{(a^{2} + b^{2})} + a)(\sqrt{(a^{2} + b^{2})} – a)}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
=Â \(\frac{b^{2}(\sqrt{(a^{2}+b^{2})}-a)}{(a^{2}+b^{2})-a^{2})}\)
= \(\frac{b^{2}(\sqrt{(a^{2}+b^{2})}-a)}{b^{2}}\)
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) \(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(\sqrt{3}-\sqrt{2}\)
= \(\frac{(\sqrt{3} – \sqrt{2})(\sqrt{3} -\sqrt{2})}{(\sqrt{3} + \sqrt{2})(\sqrt{3} – \sqrt{2})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{(\sqrt{3}-\sqrt{2})^{2}}{3-2}\)
As we know , (a – b) \(^{2}\)
= \(\frac{3-2\sqrt{3}\sqrt{2}+2}{1}\)
= \(5-2\sqrt{6}\)
(ii) \(\frac{5+2\sqrt{3}}{7+4\sqrt{3}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(7-4\sqrt{3}\)
= \(\frac{(5+2\sqrt{3})(7-4\sqrt{3})}{(7+4\sqrt{3})(7-4\sqrt{3})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
=Â \(\frac{(5+2\sqrt{3})(7-4\sqrt{3})}{49-48}\)
= 35 – \(20\sqrt{3}\)
= 11 – \(6\sqrt{3}\)
(iii) \(\frac{1+\sqrt{2}}{3-2\sqrt{2}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(3+2\sqrt{2}\)
= Â \(\frac{(1+\sqrt{2})(3+2\sqrt{2})}{(3-2\sqrt{2})(3+2\sqrt{2})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{(1+\sqrt{2})(3+2\sqrt{2})}{9-8}\)
= \(3+2\sqrt{2}+3\sqrt{2}+4\)
= 7 + \(5\sqrt{2}\)
(iv) Â \(\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(3\sqrt{5}+2\sqrt{6}\)
= \(\frac{(2\sqrt{6}-\sqrt{5})(3\sqrt{5}+2\sqrt{6})}{(3\sqrt{5}-2\sqrt{6})(3\sqrt{5}+2\sqrt{6})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{(2\sqrt{6}-\sqrt{5})(3\sqrt{5}+2\sqrt{6})}{45-24}\)
= \(\frac{(2\sqrt{6}-\sqrt{5})(3\sqrt{5}+2\sqrt{6})}{21}\)
= \(\frac{(2\sqrt{6}-\sqrt{5})(3\sqrt{5}+2\sqrt{6})}{21}\)
= \(\frac{6\sqrt{30}+24-15-2\sqrt{30}}{21}\)
= \(\frac{4\sqrt{30}+9}{21}\)
(v) \(\frac{4\sqrt{3}+5\sqrt{2}}{\sqrt{48}+\sqrt{18}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(\sqrt{48} – \sqrt{18}\)
= Â \(\frac{(4\sqrt{3}+5\sqrt{2})(\sqrt{48}-\sqrt{18})}{(\sqrt{48}+\sqrt{18})(\sqrt{48}-\sqrt{18})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{(4\sqrt{3}+5\sqrt{2})(\sqrt{48}-\sqrt{18})}{48-18}\)
= \(\frac{48-12\sqrt{6}+20\sqrt{6}-30}{30}\)
= \(\frac{18+8\sqrt{6}}{30}\)
= \(\frac{9+4\sqrt{6}}{15}\)
(vi) \(\frac{2\sqrt{3}-\sqrt{5}}{2\sqrt{2}+3\sqrt{3}}\)
Rationalizing the denominator by multiplying both numerator and 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) \(\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}} + \frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(3\sqrt{2}-2\sqrt{3}\)
= \(\frac{(3\sqrt{2}-2\sqrt{3})(3\sqrt{2} -2\sqrt{3})}{(3\sqrt{2} + 2\sqrt{3})(3\sqrt{2} – 2\sqrt{3})} + \frac{\sqrt{12}({\sqrt{3} + \sqrt{2}})}{(\sqrt{3} – \sqrt{2})({\sqrt{3} + \sqrt{2}})}\)
Now, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{(3\sqrt{2}-2\sqrt{3})(3\sqrt{2}-2\sqrt{3})}{18-12} + \frac{\sqrt{12}({\sqrt{3}+\sqrt{2}})}{3-2}\)
As we know , (a – b) \(^{2}\)
= \(\frac{(3\sqrt{2})^{2}-(2\times 3\sqrt{2}\times 2\sqrt{3})+(2\sqrt{3})^{2}}{6} + 2\sqrt{3}({\sqrt{3}+\sqrt{2}})\)
= \(\frac{(18-12\sqrt{6}+12)}{6} +({6+2\sqrt{6}})\)
= \(3-2\sqrt{6}+2 +({6+2\sqrt{6}})\)
= \(5-2\sqrt{6} +({6+2\sqrt{6}})\)
= 11
(ii) \(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(\sqrt{5}+\sqrt{3}\)
= \(\frac{(\sqrt{5}+\sqrt{3})(\sqrt{5} + \sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5} + \sqrt{3})}+\frac{(\sqrt{5}-\sqrt{3})(\sqrt{5} – \sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5} – \sqrt{3})}\)
Now as we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{5+2\times \sqrt{5}\times \sqrt{3}+3}{5-3}+\frac{5-2\times \sqrt{3}\times \sqrt{5}+3}{5-3}\)
= \(\frac{8+2\sqrt{15}+8-2\sqrt{15}}{2}\)
= \(\frac{16}{2}\)
= 8
(iii) \(\frac{7+3\sqrt{5}}{3+\sqrt{5}} – \frac{7-3\sqrt{5}}{3-\sqrt{5}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(3-\sqrt{5}\)
\(\frac{(7+3\sqrt{5})(3-\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})} – \frac{(7-3\sqrt{5})(3+\sqrt{5})}{(3-\sqrt{5})(3-\sqrt{5})}\)
Now as we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{(7+3\sqrt{5})(3-\sqrt{5})}{9-5} – \frac{(7-3\sqrt{5})(3+\sqrt{5})}{9-5}\)
= \(\frac{(21-7\sqrt{5}+9\sqrt{5}-15)}{4} – \frac{(21+7\sqrt{5}-9\sqrt{5}-15)}{4}\)
= \(\frac{(6+2\sqrt{5})}{4} – \frac{(6-2\sqrt{5})}{4}\)
= \(\frac{4\sqrt{5}}{4}\)
= \(\sqrt{5}\)
(iv) \(\frac{1}{2+\sqrt{3}} + \frac{2}{\sqrt{5}-\sqrt{3}} + \frac{1}{2-\sqrt{5}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(2-\sqrt{3}\)
= \(\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})} + \frac{2\times (\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})} + \frac{2+\sqrt{5}}{(2-\sqrt{5})(2+\sqrt{5})}\)
Since, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{2-\sqrt{3}}{4 – 3} + \frac{2\times (\sqrt{5}+\sqrt{3})}{5-3} + \frac{2+\sqrt{5}}{4-5}\)
= \(\frac{2-\sqrt{3}}{1} + \frac{2\sqrt{5}+2\sqrt{3}}{2} + \frac{2+\sqrt{5}}{-1}\)
= \(\frac{4-2\sqrt{3}+2\sqrt{5}+2\sqrt{3}-4-2\sqrt{5}}{2}\)
= \(\frac{0}{2}\)
= 0
(v) \(\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{3}{\sqrt{5}+\sqrt{2}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(\sqrt{5}-\sqrt{3}\)
= \(\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})} + \frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} – \frac{3\times (\sqrt{5}-\sqrt{2})}{\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}\)
Since, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{2(\sqrt{5}-\sqrt{3})}{5-3}+\frac{\sqrt{3}-\sqrt{2}}{3-2}-\frac{3\times (\sqrt{5}-\sqrt{2})}{5-2}\)
= \(\frac{2\sqrt{5}-2\sqrt{3}}{2}+\frac{\sqrt{3}-\sqrt{2}}{1}-\frac{3\times\sqrt{5}-3\sqrt{2}}{3}\)
= \(\frac{6\sqrt{5}-6\sqrt{3}+6\sqrt{3}-6\sqrt{2}-6\sqrt{5}+6\sqrt{2}}{3}\)
= \(\frac{0}{3}\)
= 0
- In each of the following determine rational numbers a and b:
(i) \(\frac{\sqrt{3}-1}{\sqrt{3}+1}\)
(ii) \(\frac{4+\sqrt{2}}{2+\sqrt{2}}\)
(iii) \(\frac{3+\sqrt{2}}{3-\sqrt{2}}\)
(iv) \(\frac{5+3\sqrt{3}}{7+4\sqrt{3}}\)
(v) \(\frac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}\)
(vi) \(\frac{4+3\sqrt{5}}{4-3\sqrt{5}}\)
Solution:
(i) Given,
\(\frac{\sqrt{3}-1}{\sqrt{3}+1}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(\sqrt{3}-1\)
= \(\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{3-2\sqrt{3}+1}{3-1}\)
= \(\frac{4-2\sqrt{3}}{2}\)
= \(2-\sqrt{3}\)
- \(2-\sqrt{3}\)
= \(a-b\sqrt{3}\)
On comparing the rational and irrational parts of the above equation, we get,
a = 2 and b = 1
(ii) Given;
\(\frac{4+\sqrt{2}}{2+\sqrt{2}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(2-\sqrt{2}\)
= \(\frac{(4+\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{(8-4\sqrt{2}+2\sqrt{2}-2)}{4-2}\)
= \(\frac{(6-2\sqrt{2})}{2}\)
= \(3-\sqrt{2}\)
- \(3-\sqrt{2}\)
= \(a-\sqrt{b}\)
On comparing the rational and irrational parts of the above equation, we get,
- a = 3 and b = 2
(iii) Given,
\(\frac{3+\sqrt{2}}{3-\sqrt{2}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(3+\sqrt{2}\)
= \(\frac{(3+\sqrt{2})(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{(9+6\sqrt{2}+2)}{9-2}\)
=Â \(\frac{(11+6\sqrt{2})}{7}\)
= \(\frac{11}{7}+\frac{6\sqrt{2}}{7}\)
- \(\frac{11}{7}+\frac{6\sqrt{2}}{7}\)
= \(a+b\sqrt{2}\)
On comparing the rational and irrational parts of the above equation, we get,
- a = \(\frac{11}{7}+\frac{6\sqrt{2}}{7}\)
and - b = \(\frac{6}{7}\)
(iv)Â Given,
\(\frac{5+3\sqrt{3}}{7+4\sqrt{3}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(7-4\sqrt{3}\)
= \(\frac{(5+3\sqrt{3})(7-4\sqrt{3})}{(7+4\sqrt{3})(7-4\sqrt{3})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{(35-20\sqrt{3}+21\sqrt{3}-36)}{49-48}\)
= -1 + \(\sqrt{3}\)
- -1 + \(\sqrt{3}\)
= \(a + b\sqrt{3}\)
On comparing the rational and irrational parts of the above equation, we get,
- a = -1 and
- b = 1
(v) \(\frac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(\sqrt{11} – \sqrt{7}\)
=Â \(\frac{(\sqrt{11}-\sqrt{7})(\sqrt{11} – \sqrt{7})}{(\sqrt{11}+\sqrt{7})(\sqrt{11} – \sqrt{7})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{(11-\sqrt{77}-\sqrt{77}+7)}{11-7}\)
= \(\frac{(18-2\sqrt{77})}{4}\)
= \(\frac{9}{2}-\frac{\sqrt{77}}{2}\)
- \(\frac{9}{2}-\frac{\sqrt{77}}{2}\)
= \(a – b\sqrt{77}\)
On comparing the rational and irrational parts of the above equation, we get,
- a = \(\frac{9}{2}\)
and - b = \(\frac{1}{2}\)
(vi) Given,
= \(\frac{4+3\sqrt{5}}{4-3\sqrt{5}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(4 + 3\sqrt{5}\)
= \(\frac{(4+3\sqrt{5})(4 + 3\sqrt{5})}{(4-3\sqrt{5})(4 + 3\sqrt{5})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{(16+24\sqrt{5}+45)}{-29}\)
= \(\frac{(61+24\sqrt{5})}{-29}\)
=Â \(\frac{-61}{29}-\frac{(24\sqrt{5})}{29}\)
- \(\frac{-61}{29}-\frac{(24\sqrt{5})}{29}\)
= \(a + b\sqrt{5}\)
On comparing the rational and irrational parts of the above equation, we get,
- a = \(\frac{-61}{29}\)
, and - b = \(\frac{-24}{29}\)
7. If x = 2 + \(\sqrt{3}\)
Solution:
Given,
x = 2 + \(\sqrt{3}\)
To find the value of \(x^{3}+\frac{1}{x^{3}}\)
We have, x = 2 + \(\sqrt{3}\)
- \(\frac{1}{x}\)
= \(\frac{1}{2+\sqrt{3}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(2-\sqrt{3}\)
= \(\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}\)
Since, (a + b) (a – b) = (\(a^{2} – b^{2}\)
- \(\frac{1}{x}\)
= \(\frac{2-\sqrt{3}}{4-3}\) - x + \(\frac{1}{x}\)
= 2 + \(\sqrt{3}\) + 2 – \(\sqrt{3}\) - x + \(\frac{1}{x}\)
= 4
We know that, \((a^{3}+b^{3})=(a+b)(a^{2}-ab+b^{2})\)
- \((x^{3}+\frac{1}{x^{3}}) = (x+\frac{1}{x})(x^{2}-x.\frac{1}{x}+\frac{1}{x^{2}})\)
- \((x^{3}+\frac{1}{x^{3}})=(x+\frac{1}{x})(x^{2}+\frac{1}{x}^{2}-1)\)
- \((x^{3}+\frac{1}{x^{3}})=(x+\frac{1}{x})(x^{2}+\frac{1}{x^{2}}+2-2-1)\)
- \((x^{3}+\frac{1}{x^{3}})=(x+\frac{1}{x})(x^{2}+\frac{1}{x^{2}}+2(x.\frac{1}{x})-2-1)\)
- \((x^{3}+\frac{1}{x^{3}})=(x+\frac{1}{x})((x+\frac{1}{x})^{2}-3)\)
Putting the value of \(x+\frac{1}{x}\)
- \((x^{3}+\frac{1}{x^{3}})=(4)(4^{2}-3)\)
- \((x^{3}+\frac{1}{x^{3}})=52\)
8. If x = 3 + \(\sqrt{8}\)
Solution:
Given,
x = 3 + \(\sqrt{8}\)
To find the value of \((x^{2}+\frac{1}{x^{2}})\)
We have, x = 3 + \(\sqrt{8}\)
- \(\frac{1}{x} = \frac{1}{3+\sqrt{8}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(3-\sqrt{8}\)
- \(\frac{1}{x} = \frac{3-\sqrt{8}}{(3+\sqrt{8})(3-\sqrt{8})}\)
Since, (a + b) (a – b) = (\(a^{2} – b^{2}\)
- \(\frac{1}{x} = \frac{3-\sqrt{8}}{9-8}\)
- \(\frac{1}{x} = 3-\sqrt{8}\)
- \((x^{2}+\frac{1}{x^{2}}) = ((3+\sqrt{8})^{2}(3-\sqrt{8})^{2})\)
- \((x^{2}+\frac{1}{x^{2}})= ((9+8+6\sqrt{8})+(9+8-6\sqrt{8}))\)
- 34
9. Find the value of \(\frac{6}{\sqrt{5}-\sqrt{3}}\)
Given,
\(\frac{6}{\sqrt{5}-\sqrt{3}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(\sqrt{5}+\sqrt{3}\)
= \(\frac{6(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}\)
Since, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{6\sqrt{5}+6\sqrt{3}}{5-3}\)
= \(\frac{6\sqrt{5}+6\sqrt{3}}{2}\)
= \(3(\sqrt{5}+\sqrt{3})\)
= 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
\(\sqrt{2}\)
(i) \(\frac{3-\sqrt{5}}{3+2\sqrt{5}}\)
Rationalizing the denominator by multiplying both numerator and 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) = (\(a^{2} – b^{2}\)
= \(\frac{(3-\sqrt{5})(3-2\sqrt{5})}{9-20}\)
= \(\frac{(9-6\sqrt{5}-3\sqrt{5}+10)}{-11}\)
= \(\frac{(19-9\sqrt{5})}{-11}\)
= \(\frac{(9\sqrt{5}-19)}{11}\)
= \(\frac{(9(2.236))-19)}{11}\)
= \(\frac{(20.124-19)}{11}\)
= \(\frac{1.124}{11}\)
= 0.102
 Â
(ii) \(\frac{1+\sqrt{2}}{3-2\sqrt{2}}\)
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor \(3+2\sqrt{2}\)
= Â \(\frac{(1+\sqrt{2})(3+2\sqrt{2})}{(3-2\sqrt{2})(3+2\sqrt{2})}\)
As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\)
= \(\frac{(1+\sqrt{2})(3+2\sqrt{2})}{9-8}\)
= \(3+2\sqrt{2}+3\sqrt{2}+4\)
= 7 + \(5\sqrt{2}\)
= 7 + 7.07
= 14.07
11. If x = \(\frac{\sqrt{3}+1}{2}\)
Solution:
Given,
x = \(\frac{\sqrt{3}+1}{2}\)
- 2x = \(\sqrt{3}+1\)
- 2x – 1 = \(\sqrt{3}\)
Now, squaring on both the sides, we 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 taking \(4x^{3}+2x^{2}-8x+7\)
- 2x(\(2x^{2}-2x-1\)
)+ \(4x^{2}+2x+2x^{2}-8x+7\) - 2x(\(2x^{2}-2x-1\)
)+ \(6x^{2}-6x+7\)
As, \(2x^{2}-2x-1=0\)
- 2x(0)+3(\(2x^{2}-2x-1\)
))+7+3 - 0+3(0)+10
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