# RD Sharma Class 9 Rationalization Exercise 3.2 Solutions

On this page you will find Maths RD Sharma Class 9 Rationalization Exercise 3.2 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 3 for class 9 deals with the topic of triangles and it is one of the most important chapters.

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Students can use these solutions to overcome the fear of maths and the solutions have been designed in such a way that it enables them to discover easy ways to solve different problems. These solutions can help students in refining their maths fluency and problem-solving skills. Students can go through the RD Sharma solutions for class 9 chapter 3 Rationalization Exercise 3.2 below and it will be beneficial for them.

## RD Sharma Class 9 Rationalization Exercise 3.2 Solutions

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 $\sqrt{5}$
$=\frac{3 \times \sqrt{5}}{2 \sqrt{5} \times \sqrt{5}}$
$=\frac{3 \sqrt{5}}{2 \times \sqrt{5 \times 5}}$
$=\frac{3 \sqrt{5}}{2 \times 5}$
$=\frac{3 \sqrt{5}}{10}$
(iii) $\frac{1}{\sqrt{12}}$
For rationalizing the denominator, multiply both numerator and denominator with $\sqrt{12}$
$=\frac{1 \times \sqrt{12}}{\sqrt{12} \times \sqrt{12}}$
$=\frac{\sqrt{12}}{\sqrt{12 \times 12}}$
$=\frac{\sqrt{12}}{12}$
(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 $\sqrt{2}=1.414, \sqrt{3}=1.732, \sqrt{5}=2.236, \sqrt{10}=3.162$.
(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, $\sqrt{2}=1.414, \sqrt{3}=1.732, \sqrt{5}=2.236, \sqrt{10}=3.162$
(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) $\frac{3}{\sqrt{10}}$
Rationalizing the denominator by multiplying both numerator and denominator with $\sqrt{10}$
$=\frac{3 \sqrt{10}}{\sqrt{10} \times \sqrt{10}}$
$=\frac{3 \sqrt{10}}{\sqrt{10 \times 10}}$
$=\frac{3 \sqrt{10}}{10}$
$=\frac{9.486}{10}$
$=0.9486$
(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 $\sqrt{2}$
$=\frac{(\sqrt{10} \times \sqrt{2})+(\sqrt{15} \times \sqrt{2})}{\sqrt{2} \times \sqrt{2}}$
$=\frac{\sqrt{20}+\sqrt{30}}{2}$
$=\frac{(\sqrt{10} \times \sqrt{2})+(\sqrt{10} \times \sqrt{3})}{2}$
$=\frac{(3.162 \times 1.414)+(3.162 \times 1.732)}{2}$
$=\frac{(4.471068)+(5.476584)}{2}$
$=\frac{9.947652}{2}$
$=4.973826$
(v) $\frac{2+\sqrt{3}}{3}$
$=\frac{2+1.732}{3}$
$=\frac{3.732}{3}$
$=1.244$
(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) $\frac{b^{2}}{\sqrt{\left(a^{2}+b^{2}\right)+a}}$
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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{3-\sqrt{2}}{9-2}$
$=\frac{3-\sqrt{2}}{7}$
(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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{\sqrt{6}+\sqrt{2}}{6-2}$
$=\frac{\sqrt{6}+\sqrt{2}}{4}$
(iii) $\frac{16}{\sqrt{41}-5}$
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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{16 \sqrt{41}+80}{41-5}$
$=\frac{16 \sqrt{41}+80}{16}$
$=\frac{16(\sqrt{41}+5)}{16}$
$=\sqrt{41}+5$
(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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{30 \times(5 \sqrt{3}+3 \sqrt{5})}{75-45}$
$=\frac{30 \times(5 \sqrt{3}+3 \sqrt{5})}{30}$
$=5 \sqrt{3}+3 \sqrt{5}$
(v) $\frac{1}{2 \sqrt{5}-\sqrt{3}}$
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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{2 \sqrt{5}+\sqrt{3}}{20-3}$
$=\frac{2 \sqrt{5}+\sqrt{3}}{17}$
(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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(2 \sqrt{6}+3+2 \sqrt{2}+\sqrt{3})}{8-3}$
$=\frac{(2 \sqrt{6}+3+2 \sqrt{2}+\sqrt{3})}{5}$
(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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(6-4 \sqrt{2})^{2}}{36-32}$
As we know,$(a-b)^{2}=\left(a^{2}-2 \times a \times b+b^{2}\right)$
$=\frac{36-48 \sqrt{2}+32}{4}$
$=\frac{68-48 \sqrt{2}}{4}$
$=\frac{4(17-12 \sqrt{2})}{4}$
$=17-12 \sqrt{2}$
(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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{6 \sqrt{10}-9 \sqrt{2}+2 \sqrt{5}-3}{(20-9)}$
$=\frac{6 \sqrt{10}-9 \sqrt{2}+2 \sqrt{5}-3}{11}$
(ix) $\frac{\mathrm{b}^{2}}{\sqrt{\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)}+\mathrm{a}}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $\sqrt{\left(a^{2}+b^{2}\right)}-a$
$=\frac{b^{2}\left(\sqrt{\left(a^{2}+b^{2}\right)}-a\right)}{\left(\sqrt{\left(a^{2}+b^{2}\right)}+a\right)\left(\sqrt{\left(a^{2}+b^{2}\right)}-a\right)}$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{b^{2}\left(\sqrt{\left(a^{2}+b^{2}\right)}-a\right)}{ \left.\left(a^{2}+b^{2}\right)-a^{2}\right)}$
$=\frac{b^{2}\left(\sqrt{\left(a^{2}+b^{2}\right)}-a\right)}{b^{2}}$

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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(\sqrt{3}-\sqrt{2})^{2}}{3-2}$
As we know,$(a-b)^{2}=\left(a^{2}-2 \times a \times b+b^{2}\right)$
$=\frac{3-2 \sqrt{3} \sqrt{2}+2}{1}$
$=5-2 \sqrt{6}$
(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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(5+2 \sqrt{3})(7-4 \sqrt{3})}{49-48}$
$=35-20 \sqrt{3}+14 \sqrt{3}-24$
$=11-6 \sqrt{3}$
(iii) $\frac{1+\sqrt{2}}{3-2 \sqrt{2}}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $3+2 \sqrt{2}$
$=\frac{(1+\sqrt{2})(3+2 \sqrt{2})}{(3-2 \sqrt{2})(3+2 \sqrt{2})}$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(1+\sqrt{2})(3+2 \sqrt{2})}{9-8}$
$=3+2 \sqrt{2}+3 \sqrt{2}+4$
$=7+5 \sqrt{2}$
(iv) $\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}$
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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(2 \sqrt{6}-\sqrt{5})(3 \sqrt{5}+2 \sqrt{6})}{45-24}$
$=\frac{(2 \sqrt{6}-\sqrt{5})(3 \sqrt{5}+2 \sqrt{6})}{21}$
$=\frac{(2 \sqrt{6}-\sqrt{5})(3 \sqrt{5}+2 \sqrt{6})}{21}$
$=\frac{6 \sqrt{30}+24-15-2 \sqrt{30}}{21}$
$=\frac{4 \sqrt{30}+9}{21}$
(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 $\sqrt{48}-\sqrt{18}$
$=\frac{(4 \sqrt{3}+5 \sqrt{2})(\sqrt{48}-\sqrt{18})}{(\sqrt{48}+\sqrt{18})(\sqrt{48}-\sqrt{18})}$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(4 \sqrt{3}+5 \sqrt{2})(\sqrt{48}-\sqrt{18})}{48-18}$
$=\frac{48-12 \sqrt{6}+20 \sqrt{6}-30}{30}$
$=\frac{18+8 \sqrt{6}}{30}$
$=\frac{9+4 \sqrt{6}}{15}$
(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 $3 \sqrt{2}-2 \sqrt{3}$ for $\frac{1}{3 \sqrt{2}+2 \sqrt{3}}$ and the rationalizing factor $\sqrt{3}+\sqrt{2}$ for $\frac{1}{\sqrt{3}-\sqrt{2}}$
$=\frac{(3 \sqrt{2}-2 \sqrt{3})(3 \sqrt{2}-2 \sqrt{3})}{(3 \sqrt{2}+2 \sqrt{3})(3 \sqrt{2}-2 \sqrt{3})}+\frac{\sqrt{12}(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$
Now, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(3 \sqrt{2}-2 \sqrt{3})(3 \sqrt{2}-2 \sqrt{3})}{18-12}+\frac{\sqrt{12}(\sqrt{3}+\sqrt{2})}{3-2}$
As we know,$(a-b)^{2}=\left(a^{2}-2 \times a \times b+b^{2}\right)$
$=\frac{(3 \sqrt{2})^{2}-(2 \times 3 \sqrt{2} \times 2 \sqrt{3})+(2 \sqrt{3})^{2}}{6}+2 \sqrt{3}(\sqrt{3}+\sqrt{2})$
$=\frac{(18-12 \sqrt{6}+12)}{6}+(6+2 \sqrt{6})$
$=3-2 \sqrt{6}+2+(6+2 \sqrt{6})$
$=5-2 \sqrt{6}+(6+2 \sqrt{6})$
$=11$
(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 $\sqrt{5}+\sqrt{3}$ for $\frac{1}{\sqrt{5}-\sqrt{3}}$ and the rationalizing factor $\sqrt{5}-\sqrt{3}$ for $\frac{1}{\sqrt{5}+\sqrt{3}}$
$=\frac{(\sqrt{5}+\sqrt{3})(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}+\frac{(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}$
Now as we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right),(a-b)^{2}=\left(a^{2}-2 \times a \times b+b^{2}\right)$ and
$(a+b)^{2}=\left(a^{2}+2 \times a \times b+b^{2}\right)$
$=\frac{5+2 \times \sqrt{5} \times \sqrt{3}+3}{5-3}+\frac{5-2 \times \sqrt{3} \times \sqrt{5}+3}{5-3}$
$=\frac{8+2 \sqrt{15}+8-2 \sqrt{15}}{2}$
$=\frac{16}{2}$
$=8$
(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 $3-\sqrt{5}$ for $\frac{1}{3+\sqrt{5}}$ and the rationalizing factor $3+\sqrt{5}$ for $\frac{1}{3-\sqrt{5}}$
$\frac{(7+3 \sqrt{5})(3-\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}-\frac{(7-3 \sqrt{5})(3+\sqrt{5})}{(3-\sqrt{5})(3-\sqrt{5})}$
Now as we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$ $=\frac{(7+3 \sqrt{5})(3-\sqrt{5})}{9-5}-\frac{(7-3 \sqrt{5})(3+\sqrt{5})}{9-5}$
$=\frac{(21-7 \sqrt{5}+9 \sqrt{5}-15)}{4}-\frac{(21+7 \sqrt{5}-9 \sqrt{5}-15)}{4}$
$=\frac{(6+2 \sqrt{5})}{4}-\frac{(6-2 \sqrt{5})}{4}$
$=\frac{4 \sqrt{5}}{4}$
$=\sqrt{5}$
(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 $2-\sqrt{3}$ 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 $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{2-\sqrt{3}}{4-3}+\frac{2 \times(\sqrt{5}+\sqrt{3})}{5-3}+\frac{2+\sqrt{5}}{4-5}$
$=\frac{2-\sqrt{3}}{1}+\frac{2 \sqrt{5}+2 \sqrt{3}}{2}+\frac{2+\sqrt{5}}{-1}$
$=\frac{4-2 \sqrt{3}+2 \sqrt{5}+2 \sqrt{3}-4-2 \sqrt{5}}{2}$
$=\frac{0}{2}$
$=0$
(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 $\sqrt{5}-\sqrt{3}$ 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 $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{2(\sqrt{5}-\sqrt{3})}{5-3}+\frac{\sqrt{3}-\sqrt{2}}{3-2}-\frac{3 \times(\sqrt{5}-\sqrt{2})}{5-2}$
$=\frac{2 \sqrt{5}-2 \sqrt{3}}{2}+\frac{\sqrt{3}-\sqrt{2}}{1}-\frac{3 \times \sqrt{5}-3 \sqrt{2}}{3}$
$=\frac{6 \sqrt{5}-6 \sqrt{3}+6 \sqrt{3}-6 \sqrt{2}-6 \sqrt{5}+6 \sqrt{2}}{3}$
$=\frac{0}{3}$
$=0$

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) $\frac{4+\sqrt{2}}{2+\sqrt{2}}=\mathrm{a}-\sqrt{\mathrm{b}}$
(iii) $\frac{3+\sqrt{2}}{3-\sqrt{2}}=\mathrm{a}+\mathrm{b} \sqrt{2}$
(iv) $\frac{5+3 \sqrt{3}}{7+4 \sqrt{3}}=\mathrm{a}+\mathrm{b} \sqrt{3}$
(v) $\frac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}=\mathrm{a}-\mathrm{b} \sqrt{77}$
(vi) $\frac{4+3 \sqrt{5}}{4-3 \sqrt{5}}=\mathrm{a}+\mathrm{b} \sqrt{5}$
Solution:
(i) Given,
$\frac{\sqrt{3}-1}{\sqrt{3}+1}=\mathrm{a}-\mathrm{b} \sqrt{3}$
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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{3-2 \sqrt{3}+1}{3-1}$
$=\frac{4-2 \sqrt{3}}{2}$
\begin{aligned}=& 2-\sqrt{3} \\ & 2-\sqrt{3}=a-b \sqrt{3} \end{aligned}
On comparing the rational and irrational parts of the above equation, we get,
$a=2$ and $b=1$
(ii) Given;
$\frac{4+\sqrt{2}}{2+\sqrt{2}}=a-\sqrt{b}$
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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(8-4 \sqrt{2}+2 \sqrt{2}-2)}{4-2}$
$=\frac{(6-2 \sqrt{2})}{2}$
$=3-\sqrt{2}$
$3-\sqrt{2}=\mathrm{a}-\sqrt{\mathrm{b}}$
On comparing the rational and irrational parts of the above equation, we get,
$a=3$ and $b=2$
(iii) Given,
$\frac{3+\sqrt{2}}{3-\sqrt{2}}=\mathrm{a}+\mathrm{b} \sqrt{2}$
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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(9+6 \sqrt{2}+2)}{9-2}$
$=\frac{(11+6 \sqrt{2})}{7}$
$=\frac{11}{7}+\frac{6 \sqrt{2}}{7}$
$\frac{11}{7}+\frac{6 \sqrt{2}}{7}=\mathrm{a}+\mathrm{b} \sqrt{2}$
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) Given,
$\frac{5+3 \sqrt{3}}{7+4 \sqrt{3}}=\mathrm{a}+\mathrm{b} \sqrt{3}$9Rationalizing 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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(35-20 \sqrt{3}+21 \sqrt{3}-36)}{49-48}$
$=-1+\sqrt{3}$
$-1+\sqrt{3}=a+b \sqrt{3}$
On comparing the rational and irrational parts of the above equation, we get,
$a=-1$ and
$b=1$
(v) $\frac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}=\mathrm{a}-\mathrm{b} \sqrt{77}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $\sqrt{11}-\sqrt{7}$
$=\frac{(\sqrt{11}-\sqrt{7})(\sqrt{11}-\sqrt{7})}{(\sqrt{11}+\sqrt{7})(\sqrt{11}-\sqrt{7})}$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(11-\sqrt{77}-\sqrt{77}+7)}{11-7}$
$=\frac{(18-2 \sqrt{77})}{4}$
$=\frac{9}{2}-\frac{\sqrt{77}}{2}$
On comparing the rational and irrational parts of the above equation, we get,
$a=\frac{9}{2}$ and
$b=\frac{1}{2}$
(vi) Given,
$=\frac{4+3 \sqrt{5}}{4-3 \sqrt{5}}=\mathrm{a}+\mathrm{b} \sqrt{5}$
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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(16+24 \sqrt{5}+45)}{-29}$
$=\frac{(61+24 \sqrt{5})}{-29}$
$=\frac{-61}{29}-\frac{(24 \sqrt{5})}{29}$
$\frac{-61}{29}-\frac{(24 \sqrt{5})}{29}=\mathrm{a}+\mathrm{b} \sqrt{5}$
On comparing the rational and irrational parts of the above equation, we get,
$a=\frac{-61}{29}$, and
$b=\frac{-24}{29}$

7. If $x=2+\sqrt{3}$, find the value of $x^{3}+\frac{1}{x^{3}}$
Solution:
Given,
$x=2+\sqrt{3}$
To find the value of $\mathrm{x}^{3}+\frac{1}{\mathrm{x}^{3}}$
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 $2-\sqrt{3}$ for $\frac{1}{2+\sqrt{3}}$
$=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}$
Since, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$\frac{1}{x}=\frac{2-\sqrt{3}}{4-3}$
$x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}$
$x+\frac{1}{x}=4$
We know that, $\left(\mathrm{a}^{3}+\mathrm{b}^{3}\right)=(\mathrm{a}+\mathrm{b})\left(\mathrm{a}^{2}-\mathrm{ab}+\mathrm{b}^{2}\right)$
$\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x+\frac{1}{x}\right)\left(x^{2}-x \cdot \frac{1}{x}+\frac{1}{x^{2}}\right)$
$\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x}^{2}-1\right)$
$\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}+2-2-1\right)$
$\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}+2\left(x \cdot \frac{1}{x}\right)-2-1\right)$
$\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x+\frac{1}{x}\right)\left(\left(x+\frac{1}{x}\right)^{2}-3\right)$
Putting the value of $\mathrm{x}+\frac{1}{\mathrm{x}}$ in the above equation, we get,
$\left(x^{3}+\frac{1}{x^{3}}\right)=(4)\left(4^{2}-3\right)$
$\left(x^{3}+\frac{1}{x^{3}}\right)=52$

8. If $x=3+\sqrt{8}$, find the value of $\left(x^{2}+\frac{1}{x^{2}}\right)$
Solution:
Given,
$x=3+\sqrt{8}$
To find the value of $\left(\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}\right)$
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 $3-\sqrt{8}$ for $\frac{1}{3+\sqrt{8}}$
$\frac{1}{x}=\frac{3-\sqrt{8}}{(3+\sqrt{8})(3-\sqrt{8})}$
Since $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$\frac{1}{x}=\frac{3-\sqrt{8}}{9-8}$
$\frac{1}{x}=3-\sqrt{8}$
$\left(\mathrm{x}^{2}+\frac{1}{x^{2}}\right)=\left((3+\sqrt{8})^{2}(3-\sqrt{8})^{2}\right)$
$\left(\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}\right)=((9+8+6 \sqrt{8})+(9+8-6 \sqrt{8}))$
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9. Find the value of $\frac{6}{\sqrt{5}-\sqrt{3}}$, it being given that $\sqrt{3}=1.732$ and $\sqrt{5}=2.236$.
Solution:
Given,
$\frac{6}{\sqrt{5}-\sqrt{3}}$
Rationalizing the denominator by multiplying both numerator and 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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{6 \sqrt{5}+6 \sqrt{3}}{5-3}$
$=\frac{6 \sqrt{5}+6 \sqrt{3}}{2}$
$=3(\sqrt{5}+\sqrt{3})$
$=3(2.236+1.732)$
$=3(3.968)$
$=11.904$

10. Find the values of each of the following correct to three places of decimals, it being given that
$\sqrt{2}=1.414, \sqrt{3}=1.732, \sqrt{5}=2.236, \sqrt{6}=2.4495, \sqrt{10}=3.162$
(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, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(3-\sqrt{5})(3-2 \sqrt{5})}{9-20}$
$=\frac{(9-6 \sqrt{5}-3 \sqrt{5}+10)}{-11}$
$=\frac{(19-9 \sqrt{5})}{-11}$
$=\frac{(9 \sqrt{5}-19)}{11}$
$=\frac{(9(2.236))-19)}{11}$
$=\frac{(20.124-19)}{11}$
$=\frac{1.124}{11}$
$=0.102$
(ii) $\frac{1+\sqrt{2}}{3-2 \sqrt{2}}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $3+2 \sqrt{2}$
$=\frac{(1+\sqrt{2})(3+2 \sqrt{2})}{(3-2 \sqrt{2})(3+2 \sqrt{2})}$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(1+\sqrt{2})(3+2 \sqrt{2})}{9-8}$
$=3+2 \sqrt{2}+3 \sqrt{2}+4$
$=7+5 \sqrt{2}$
$=7+7.07$
$=14.07$

11. If $x=\frac{\sqrt{3}+1}{2}$, find the value of $4 x^{3}+2 x^{2}-8 x+7$.
Solution:
Given,
$x=\frac{\sqrt{3}+1}{2}$ and given to find the value of $4 x^{3}+2 x^{2}-8 x+7$
$2 x=\sqrt{3}+1$
$2 x-1=\sqrt{3}$
Now, squaring on both the sides, we get,
$(2 x-1)^{2}=3$
$4 x^{2}-4 x+1=3$
$4 x^{2}-4 x+1-3=0$
$4 x^{2}-4 x-2=0$
$2 x^{2}-2 x-1=0$
Now taking $4 \mathrm{x}^{3}+2 \mathrm{x}^{2}-8 \mathrm{x}+7$
$2 x\left(2 x^{2}-2 x-1\right)+4 x^{2}+2 x+2 x^{2}-8 x+7$
$2 x\left(2 x^{2}-2 x-1\right)+6 x^{2}-6 x+7$
As, $2 \mathrm{x}^{2}-2 \mathrm{x}-1=0$
$\left.2 x(0)+3\left(2 x^{2}-2 x-1\right)\right)+7+3$
$0+3(0)+10$
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