RD Sharma Solutions for Class 9 Maths Chapter 3 Rationalisation - Free PDF Download

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In this article, we have listed RD Sharma Solutions for Class 9 Maths Chapter 3 in PDF that is prepared by Kota’s top IITian’s Faculties by keeping Simplicity in mind.

If you want to learn and understand class 9 Maths Chapter 3 "Rationalisation" in an easy way then you can use these solutions PDF.

Chapter 3 of RD Sharma Class 9 deals with different algebraic identities and rationalisation of the denominator. Rationalisation is a process by which radicals in the denominator of a fraction are eliminated. If you want to improve your basic fundamentals of Rationalisation then you can use this.

RD Sharma Solutions helps students to Practice important concepts of subjects easily. RD Sharma class 9 solutions provide detailed explanations of all the  exercise questions that students can use to clear their doubts instantly.

If you want to become good at Math then it is very important for you to have a good knowledge of all the important topics of class 9 math, so to learn and practice those topics you can use eSaral RD Sharma Solutions.

In this article, we have listed RD Sharma Solutions for Class 9 Maths Chapter 3 that you can download to start your preparations anytime.

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RD Sharma Solutions for Class 9 Maths Chapter 3 Rationalisation - Free PDF Download

Question 1: Simplify each of the following:

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

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

Solution: (i) Using: $\sqrt[n]{a} \times \sqrt[n]{b}=\sqrt[n]{a \times b}$

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

$=\sqrt[3]{64}$

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

$=4$

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

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

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

$=\sqrt[4]{625}$

$=\sqrt[4]{5^{4}}$

$=5^{\left(4 \times \frac{1}{4}\right)}$

$=5$

Question 2: Simplify the following expressions:

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

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

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

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

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

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

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

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

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

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

Question 3: Simplify the following expressions:

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

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

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

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

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

Solution: Using Identity: $(a-b)(a+b)=a^{2}-b^{2}$

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

$=11^{2}-(\sqrt{11})^{2}$

$=121-11$

$=110$

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

$=\left(5^{2}-(\sqrt{7})^{2}\right)$

$=25-7=18$

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

$=(\sqrt{8})^{2}-(\sqrt{2})^{2}$

$=8-2$

$=6$

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

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

$=9-3$

$=6$

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

$=(\sqrt{5})^{2}-(\sqrt{2})^{2}$

$=5-2$

$=3$

Question 4: Simplify the following expressions:

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

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

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

Solution: Using identities: $(a-b)^{2}=a^{2}+b^{2}-2 a b$ and $(a+b)^{2}=a^{2}+b^{2}+2 a b$

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

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

$=3+7+2 \sqrt{21}$

$=10+2 \sqrt{21}$

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

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

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

$=8-2 \sqrt{15}$

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

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

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

$=38+12 \sqrt{10}$

Exercise $3.2$

Question 1: Rationalise the denominators of each of the following (i-vii):

(i) $3 / \sqrt{5}$

(ii) $3 /(2 \sqrt{5})$

(iii) $1 / \sqrt{12}$

(iv) $\sqrt{2} / \sqrt{5}$

$(v)(\sqrt{3}+1) / \sqrt{2}$

(vi) $(\sqrt{2}+\sqrt{5}) / \sqrt{3}$

(vii) $3 \sqrt{2} / \sqrt{5}$

Solution: (i) Multiply both numerator and denominator to with same number to rationalise the denominator.

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

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

$=3 \sqrt{5 / 5}$ (ii) Multiply both numerator and denominator to with same number to rationalise the denominator.

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

$=\frac{3 \sqrt{5}}{2 \times 5}=\frac{3 \sqrt{5}}{10}=\frac{3}{10} \sqrt{5}$

(iii) Multiply both numerator and denominator to with same number to rationalise the denominator.

$\frac{1}{\sqrt{12}}=\frac{1}{\sqrt{4 \times 3}}=\frac{1}{2 \sqrt{3}}$

$=\frac{1 \times \sqrt{3}}{2 \sqrt{3} \times \sqrt{3}}=\frac{\sqrt{3}}{2 \times 3}=\frac{\sqrt{3}}{6}$

(iv) Multiply both numerator and denominator to with same number to rationalise the denominator.

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

$=\frac{\sqrt{10}}{5}=\frac{1}{5} \sqrt{10}$

(v) Multiply both numerator and denominator to with same number to rationalise the denominator.

$\frac{\sqrt{3}+1}{\sqrt{2}}=\frac{(\sqrt{3}+1) \sqrt{2}}{\sqrt{2} \times \sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{2}$

(vi) Multiply both numerator and denominator to with same number to rationalise the denominator.

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

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

(vii) Multiply both numerator and denominator to with same number to rationalise the denominator.

$\frac{3 \sqrt{2}}{\sqrt{5}}=\frac{3 \sqrt{2} \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}}=\frac{3 \times \sqrt{10}}{5}$

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

Question 2: Find the value to three places of decimals of each of the following. It is given that $\sqrt{2}=1.414, \sqrt{3}=1.732, \sqrt{5}=2.236$ and $\sqrt{10}=3.162$

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

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

$=\frac{3.464}{3}=1.154$

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

$=\frac{3(3.162)}{10}=\frac{9.486}{10}=0.9486$

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

$=\frac{\sqrt{10}+\sqrt{2}}{2}=\frac{3.162+1.414}{2}$

$=\frac{4.576}{2}=2.288$ (iv) $\frac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}=\frac{(\sqrt{10}+\sqrt{15}) \sqrt{2}}{\sqrt{2} \times \sqrt{2}}$

$=\frac{\sqrt{20}+\sqrt{30}}{2}=\frac{2 \sqrt{5}+\sqrt{10} \times \sqrt{3}}{2}$

$=\frac{2(2.236)+3.162 \times 1.732}{2}=4.974$

(v) $\frac{2+\sqrt{3}}{3}=\frac{2+1.732}{3}=\frac{3.732}{3}=1.244$

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

$=\frac{\sqrt{10}-\sqrt{5}}{5}=\frac{3.162-2.236}{5}$

$=\frac{0.926}{5}=0.185$

Question 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: Using identity: $(a+b)(a-b)=a^{2}-b^{2}$

(i) Multiply and divide given number by $3-\sqrt{2}$

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

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

$=\frac{3-\sqrt{2}}{9-2}$

$=\frac{3-\sqrt{2}}{7}$

(ii) Multiply and divide given number by $\sqrt{6}+\sqrt{5}$

$\frac{1}{\sqrt{6}-\sqrt{5}}$

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

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

$=\sqrt{6}+\sqrt{5}$

(iii) Multiply and divide given number by $\sqrt{41}+5$

$\frac{16}{\sqrt{41}-5}$

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

$=\frac{16 \sqrt{41}+80}{41-25}$

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

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

$=\sqrt{41}+5$

(iv) Multiply and divide given number by $5 \sqrt{3}+3 \sqrt{5}$

$\frac{30}{5 \sqrt{3}-3 \sqrt{5}}$

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

$=\frac{30 \times(5 \sqrt{3}+3 \sqrt{5})}{75-45}$

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

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

(v) Multiply and divide given number by $2 \sqrt{5}+\sqrt{3}$

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

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

$=\frac{2 \sqrt{5}+\sqrt{3}}{20-3}$

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

(vi) Multiply and divide given number by $2 \sqrt{2}+\sqrt{3}$

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

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

$=\frac{(2 \sqrt{6}+3+2 \sqrt{2}+\sqrt{3})}{8-3}$

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

(vii) Multiply and divide given number by $6-4 \sqrt{2}$

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

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

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

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

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

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

$=17-12 \sqrt{2}$

(viii) Multiply and divide given number by $2 \sqrt{5}+3$

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

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

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

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

(ix) Multiply and divide given number by $\sqrt{\left(a^{2}+b^{2}\right)}-a$

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

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

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

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

Question 4: Rationales 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:

[Use identities: $(a+b)(a-b)=a^{2}-b^{2} ;(a+b)^{2}=\left(a^{2}+2 a b+b^{2}\right.$ and $(a-b)^{2}=\left(a^{2}-2 a b+b^{2}\right]$

(i) Multiply both numerator and denominator by $\sqrt{3}-\sqrt{2}$ to rationalise the denominator.

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

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

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

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

$=5-2 \sqrt{6}$

(ii) Multiply both numerator and denominator by $7-4 \sqrt{3}$ to rationalise the denominator.

$\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}$

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

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

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

$=11-6 \sqrt{3}$ (iii) Multiply both numerator and denominator by $3+2 \sqrt{2}$ to rationalise the denominator.

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

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

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

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

$=7+5 \sqrt{2}$

(iv) Multiply both numerator and denominator by $3 \sqrt{5}+2 \sqrt{6}$ to rationalise the denominator.

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

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

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

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

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

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

(v) Multiply both numerator and denominator by $\sqrt{48}-\sqrt{18}$ to rationalise the denominator.

$\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$

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

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

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

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

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

(vi) Multiply both numerator and denominator by $2 \sqrt{2}-3 \sqrt{3}$ to rationalise the denominator.

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

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

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

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

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

Exercise VSAQs

Question 1: Write the value of $(2+\sqrt{3})(2-\sqrt{3})$.

Solution: $(2+\sqrt{3})(2-\sqrt{3})$

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

[Using identity : $\left.(a+b)(a-b)=a^{2}-b^{2}\right]$

$=4-3$

$=1$

Question 2: Write the reciprocal of $5+\sqrt{2}$

Solution: Reciprocal of $5+\sqrt{2}=\frac{1}{5+\sqrt{2}}$

Rationalisation of fraction

Multiply and divide given fraction by $5-\sqrt{2}$

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

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

$=\frac{5-\sqrt{2}}{25-2}$

$=\frac{5-\sqrt{2}}{23}$

Question 3: Write the rationalisation factor of $7-3 \sqrt{5}$

Solution: Rationalisation factor of $7-3 \sqrt{5}$ is $7+3 \sqrt{5}$

Question 4: If $\frac{\sqrt{3}-1}{\sqrt{3}+1}=x+y \sqrt{3}$

Find the values of $\mathbf{x}$ and $\mathrm{y}$

Solution: [Using identities: $(a+b)(a-b)=a^{2}-b^{2}$ and $\left.(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$

Rationalising Denominator

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

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

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

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

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

$=2-\sqrt{3}$

Now,

$2-\sqrt{3}=x+y \sqrt{3}$

On comparing,

$x=2, y=-1$

Question 5: If $\mathbf{x}=\sqrt{\mathbf{2}}-1$, then write the value of $1 / \mathrm{x}$

Solution:

$x=\sqrt{2}-1$

or $1 / \mathrm{x}=1 /(\sqrt{2}-1)$

Rationalising denominator, we have

$=1 /(\sqrt{2}-1) \times(\sqrt{2}+1) /(\sqrt{2}+1)$

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

$=\sqrt{2}+1$

Question 6: Simplify

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

Solution: $\sqrt{3+2 \sqrt{2}}$

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

$=\sqrt{(\sqrt{2}+1)^{2}}=\sqrt{2}+1$

[ Because: $\left.(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

Question 7: Simplify

$\sqrt{3-2 \sqrt{2}}$

Solution: $\sqrt{3-2 \sqrt{2}}$

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

$=\sqrt{(\sqrt{2})^{2}+(1)^{2}-2 \times \sqrt{2} \times 1}$

$=\sqrt{(\sqrt{2}-1)^{2}}=\sqrt{2}-1$

$\left[\right.$ Because: $\left.(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$

Question 8: If $a=\sqrt{2}+1$, then find the value of a $-1 / a$.

Solution: Given: $a=\sqrt{2}+1$

$1 / a=1 /(\sqrt{2}+1)$

$=1 /(\sqrt{2}+1) \times(\sqrt{2}-1) /(\sqrt{2}-1)$

$=(\sqrt{2}-1) /\left((\sqrt{2})^{2}-(1)^{2}\right)$

$=(\sqrt{2}-1) / 1$

$=\sqrt{2}-1$

Now, $a-1 / a=(\sqrt{2}+1)-(\sqrt{2}-1)$

$=2$

Question 9: If $x=2+\sqrt{3}$, find the value of $x+1 / x$

Solution: Given: $x=2+\sqrt{3}$

$1 / x=1 /(2+\sqrt{3})$

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

$=(2-\sqrt{3}) /\left((2)^{2}-(\sqrt{3})^{2}\right)$

$=(2-\sqrt{3}) /(4-3)$

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

Now, $x+1 / x=(2+\sqrt{3})+(2-\sqrt{3})$

$=4$

Question 10: Write the rationalisation factor of $\sqrt{5}-2$

Solution: Rationalisation factor of $\sqrt{5}-2$ is $\sqrt{5}+2$

Question 11: If $\mathrm{x}=3+2 \sqrt{2}$, then find the value of $\sqrt{\mathbf{x}}-1 / \sqrt{\mathbf{x}}$

Solution: $x=3+2 v 2$

$\frac{1}{x}$

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

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

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

$=\frac{3-2 \sqrt{2}}{9-8}$

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

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

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

$=6$

Now, $\left(v x-\frac{1}{v x}\right)^{2}$

$=x+\frac{1}{x}-2$

$=6-2=4=(2)^{2}$

$\left(v x-\frac{1}{v x}\right)=2$

Also Read,

Download NCERT Class 9 Maths Chapterwise Book Free

Download NCERT Class 10 Maths Chapterwise Book Free

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