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

  • Last Updated : 16 May, 2021

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}

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(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.




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