Class 9 RD Sharma Solutions - Chapter 3 Rationalisation- Exercise 3.2 | Set 2

Chapter 3 of RD Sharma's Class 9 mathematics textbook focuses on the concept of rationalization, a crucial algebraic technique used to simplify expressions containing surds or irrational numbers. Exercise 3.2 specifically deals with the rationalization of denominators, which is an essential skill for simplifying complex fractions and solving equations involving irrational numbers. This set of problems aims to reinforce students' understanding of rationalization techniques and their application in various mathematical contexts.

Question 6. In each of the following determine rational numbers a and b:

(i)\frac{\sqrt3-1}{\sqrt3+1}=a-b\sqrt3

(ii)\frac{4+\sqrt2}{2+\sqrt2}=n-\sqrt{b}

(iii)\frac{3+\sqrt2}{3-\sqrt2}=a+b\sqrt2

(iv)\frac{5+3\sqrt3}{7+4\sqrt3}=a+b\sqrt3

(v)\frac{\sqrt{11}-\sqrt7}{\sqrt{11}+\sqrt7}=a-b\sqrt{77}

(vi)\frac{4+3\sqrt5}{4-3\sqrt5}=a+b\sqrt5

Solution:

(i) We know that rationalisation factor for\sqrt3+1\ is\ \sqrt3-1 . We will multiply numerator and denominator of the given expression\frac{\sqrt3-1}{\sqrt3+1}\ by\ \sqrt3-1 , to get

\frac{\sqrt3-1}{\sqrt3+1}\times\frac{\sqrt3-1}{\sqrt3-1}=\frac{\sqrt3^2+1^2-2\times\sqrt3\times1}{\sqrt3^2-1^2}\\ =\frac{3+1-2\sqrt3}{3-1}\\ =\frac{4-2\sqrt3}{2}\\ =2-\sqrt3

On equating rational and irrational terms, we get

a-b\sqrt3=2-\sqrt3\\ =2-1\sqrt3

Hence, we get a = 2, b = 1

(ii) We know that rationalisation factor for2+\sqrt2\ is\ 2-\sqrt2 . We will multiply numerator and denominator of the given expression\frac{4+\sqrt2}{2+\sqrt2}\ by\ 2-\sqrt2 , to get

\frac{4+\sqrt2}{2+\sqrt2}\times\frac{2-\sqrt2}{2-\sqrt2}=\frac{4\times2-4\times\sqrt2+2\times\sqrt2-\sqrt2^2}{2^2-\sqrt2^2}\\ =\frac{8-4\sqrt2+2\sqrt2-2}{4-2}\\ =\frac{6-2\sqrt2}{2}\\ =3-\sqrt2

On equating rational and irrational terms, we get

a-\sqrt{b}=3-\sqrt2

Hence, we get a = 3, b = 2

(iii) We know that rationalisation factor for3-\sqrt2\ is\ 3+\sqrt2 . We will multiply numerator and denominator of the given expression, to get

\frac{3+\sqrt2}{3-\sqrt2}\times\frac{3+\sqrt2}{3+\sqrt2}=\frac{3^2+\sqrt2^2+2\times3\times\sqrt2}{3^2-\sqrt2^2}\\ =\frac{9+2+6\sqrt2}{9-2}\\ =\frac{11+6\sqrt2}{7}\\ =\frac{11}{7}+\frac{6}{7}\sqrt2

On equating rational and irrational terms, we get

a+b\sqrt2=\frac{11}{7}+\frac{6}{7}\sqrt2

Hence, we get a =\frac{11}7{} , b =\frac{6}{7}

(iv) We know that rationalisation factor for7+4\sqrt3\ is\ 7-4\sqrt3 . We will multiply numerator and denominator of the given expression\frac{5+3\sqrt3}{7+4\sqrt3}\ by\ 7-4\sqrt3 , to get

\frac{5+3\sqrt3}{7+4\sqrt3}\times\frac{7-4\sqrt3}{7-4\sqrt3}=\frac{5\times7-5\times4\times\sqrt3+3\times7\times\sqrt3-3\times4\times\sqrt3^2}{7^2-(4\sqrt3)^2}\\ =\frac{35-20\sqrt3+21\sqrt3-36}{49-48}\\ =\frac{\sqrt3-1}{1}\\ =\sqrt3-1

On equating rational and irrational terms, we get

a+b\sqrt3=\sqrt3-1\\ =-1+1\sqrt3

Hence, we get a = -1, b = 1

(v) We know that rationalisation factor for\sqrt{11}+\sqrt7\ is\ \sqrt{11}-\sqrt7 . We will multiply numerator and denominator of the given expression\frac{\sqrt{11}-\sqrt7}{\sqrt{11}+\sqrt7}\ by\ \sqrt{11}-\sqrt7 , to get

\frac{\sqrt{11}-\sqrt7}{\sqrt{11}+\sqrt7}\times\frac{\sqrt{11}-\sqrt7}{\sqrt{11}-\sqrt7}=\frac{\sqrt{11}^2+\sqrt7^2-2\times\sqrt{11}\times\sqrt7}{\sqrt{11}^2-\sqrt7^2}\\ =\frac{11+7-2\sqrt{77}}{11-7}\\ =\frac{18-2\sqrt{77}}{4}\\ =\frac{9}{2}-\frac{1}{2}\sqrt{77}

On equating rational and irrational terms, we get

a-b\sqrt{77}=\frac{9}{2}-\frac{1}{2}\sqrt{77}

Hence, we get a =\frac{9}{2} , b =\frac{1}{2}

(vi) We know that rationalisation factor for4-3\sqrt5\ in\ 4+3\sqrt5 . We will multiply numerator and denominator of the given expression\frac{4+3\sqrt5}{4-3\sqrt5}\ by\ 4+3\sqrt5 , to get

\frac{4+3\sqrt5}{4-3\sqrt5}\times\frac{4+3\sqrt5}{4+3\sqrt5}=\frac{4^2+(3\sqrt5)^2+2\times4\times3\sqrt5}{4^2-(3\sqrt5)^2}\\ =\frac{16+45+24\sqrt5}{16-45}\\ =\frac{61+24\sqrt5}{-29}\\ =-\frac{61}{29}-\frac{24}{29}\sqrt5

On equating rational and irrational terms, we get

a+b\sqrt5=-\frac{61}{29}-\frac{24}{29}\sqrt5

Hence, we get a =-\frac{61}{29} , b =-\frac{24}{29}

Question 7. If x=2+\sqrt3, find the value ofx^3+\frac{1}{x^3}

Solution:

We know thatx^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)\left(x^2-1+\frac{1}{x^2}\right) . We have to find the value ofx^3+\frac{1}{x^3}

Asx=2+\sqrt3

Therefore,

\frac{1}{x}=\frac{1}{2+\sqrt3}

We know that rationalization factor for2+\sqrt3\ is\ 2-\sqrt3 . We will multiply numerator and denominator of the given expression\frac{1}{2+\sqrt3}\ by\ 2-\sqrt3

to get,

\frac{1}{x}=\frac{1}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}\\ =\frac{2-\sqrt3}{2^2-\sqrt3^2}\\ =\frac{2-\sqrt3}{4-3}\\ =2-\sqrt3

Putting the value of x\ and\ \frac{1}{x} , we get

x^3+\frac{1}{x^3}=(2+\sqrt3+2-\sqrt3)((2+\sqrt3)^2-1+(2-\sqrt3)^2)\\ =4(2^2+(\sqrt3)^2+2\times2\times\sqrt3-1+2^2+(\sqrt3)^2-2\times2\times\sqrt3)\\ =4(4+3+4\sqrt3-1+4+3-4\sqrt3)\\ =52

Hence the value of the given expression is 52.

Question 8. If x=3+\sqrt8. Find the value ofx^3+\frac{1}{x^3}

Solution:

We know that x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2 . We have to find the value of x^2+\frac{1}{x^2}\ as\ x=3+\sqrt8

Therefore,

\frac{1}{x}=\frac{1}{3+\sqrt8}

We know that rationalization factor for 3+\sqrt8\ is\ 3-\sqrt8 .

We will multiply numerator and denominator of the given expression\frac{1}{3+\sqrt8}\ by\ 3-\sqrt8

to get,

\frac{1}{x}=\frac{1}{3+\sqrt8}\times\frac{3-\sqrt8}{3-\sqrt8}\\ =\frac{3-\sqrt8}{3^2-\sqrt8^2}\\ =\frac{3-\sqrt8}{9-8}\\ =3-\sqrt8

Putting the value ofx\ and\ \frac{1}{x}

We get,

x^2+\frac{1}{x^2}=(3+\sqrt8+3-\sqrt8)^2-2\\ =(6)^2-2\\ =36-2\\ =34

Hence the given expression is simplified to 34.

Question 9. Find the value of \frac{6}{\sqrt5-\sqrt3} , it being given that\sqrt3=1.732\ and\ \sqrt5=2.236

Solution:

We know that for rationalization factor we will multiply denominator and numerator of the given expression\frac{6}{\sqrt5-\sqrt3}\ by\ \sqrt5+\sqrt3

to get,

\frac{6}{\sqrt5-\sqrt3}\times\frac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}=\frac{6\sqrt5+6\sqrt3}{\sqrt5^2-\sqrt3^2}\\ =\frac{6\sqrt5+6\sqrt3}{5-3}\\ =\frac{6\sqrt5+6\sqrt3}{2}\\ =3\sqrt5+3\sqrt3

Putting the values of \sqrt5\ and\ \sqrt3

we get,

3\sqrt5+3\sqrt3=3(2.236)+3(1.732)\\ =6.708+5.196\\ =11.904

Hence, value of the given expression is 11.904.

Question 10. Find the value of each of the following correct to three place of decimals, it being given that\sqrt2=1.4142,\ \sqrt3=1.732,\ \sqrt5=2.2360,\ \sqrt6=2.4495\ and\ \sqrt{10}=3.162

(i)\frac{3-\sqrt5}{3+2\sqrt5}

(ii)\frac{1+\sqrt2}{3-2\sqrt2}

Solution:

(i) We know that rationalization factor for3+2\sqrt5\ is\ 3-2\sqrt5

We will multiply numerator and denominator of the given expression\frac{3-\sqrt5}{3+2\sqrt5}\ by\ 3-2\sqrt5

to get

\frac{3-\sqrt5}{3+2\sqrt5}\times\frac{3-2\sqrt5}{3-2\sqrt5}=\frac{3^2-3\times2\times\sqrt5-3\times\sqrt5+2\times\sqrt5^2}{3^2-(2\sqrt5)^2}\\ =\frac{9-9\sqrt5+10}{9-20}\\ =\frac{19-9\sqrt5}{-11}\\ =\frac{9\sqrt5-19}{11}

Putting the values of\sqrt5

We get

\frac{9\sqrt5-19}{11}=\frac{9(2.236)-19}{11}\\ =\frac{20.124-19}{11}\\ =\frac{1.124}{11}\\ =0.102

Hence, the given expression is simplified to 0.102.

(ii) We know that rationalization factor for3-2\sqrt2\ is\ 3+2\sqrt2

We will multiply numerator and denominator of the given expression\frac{1+\sqrt2}{3-2\sqrt2}\ by\ 3+2\sqrt2

to get

\frac{1+\sqrt2}{3-2\sqrt2}\times\frac{3+2\sqrt2}{3+2\sqrt2}=\frac{3+2\times\sqrt2+3\times\sqrt2+2\times\sqrt2^2}{3^2-(2\sqrt2)^2}\\ =\frac{3+2\sqrt2+3\sqrt2+4}{9-8}\\ =\frac{7+5\sqrt2}{1}\\ =7+5\sqrt2

Putting the values of\sqrt2

We get

7+5\sqrt2=7+5(1.4142)\\ =7+7.071\\ =14.071

Hence, the given expression is simplified to 14.071.

Question 11. Ifx=\frac{\sqrt3+1}{2} , find the value of4x^3+2x^2-8x+7

Solution:

We have,x=\frac{\sqrt3+1}{2}

It can be simplified as

2x-1=\sqrt3

On squaring both sides, we get

(2x-1)^2=\sqrt3^2\\ (2x)^2+1-2\times2x=3\\ 4x^2+1-4x=3\\ 4x^2-4x-2=0

The given equation can be rewritten as4x^2+2x^2-8x+7=x(4x^2-4x-2)+\frac{6}{4}(4x^2-4x-2)+3+7

Therefore, we have

4x^3+2x^2-8x+7=x(0)+\frac{6}{4}(0)+3+7\\ =3+7\\ =10

Hence, the value of given expression is 10.

Summary

The practice questions in this set cover a range of rationalisation scenarios, including simple surds, compound surds, and expressions with multiple terms in the denominator. Students are challenged to rationalize denominators containing square roots, cube roots, and higher-order roots. Some problems involve algebraic manipulation alongside rationalisation, requiring students to apply multiple mathematical skills simultaneously. The questions progress in difficulty, starting with straightforward examples and advancing to more complex expressions that demand a deeper understanding of algebraic principles and rationalisation techniques.