SELINA Solution Class 9 Chapter 1 Rational and Irrational Numbers Exercise 1C

Question 1.1

State, with reason, of the following is surd or not : √180

Sol:

√180 = ${\displaystyle \sqrt{2\times 2\times 5\times 3\times 3}}$ = 6√5 which is irrational.
∴ √180 is a surds.

Question 1.2

State, with reason, of the following is surd or not :
${\displaystyle \sqrt[4]{27}}$

Sol:

${\displaystyle \sqrt[4]{27}=\sqrt[4]{3\times 3\times 3}}$ which is irrational.

∴${\displaystyle \sqrt[4]{24}}$ is a surds.

Question 1.3

State, with reason, of the following is surd or not : 
${\displaystyle \sqrt[5]{128}}$

Sol:

${\displaystyle \sqrt[5]{128}=\sqrt[5]{2\times 2\times 2\times 2\times 2\times 2\times 2}=2\sqrt[5]{4}}$

∴ ${\displaystyle \sqrt[5]{128}}$ is a surds. 

Question 1.4

State, with reason, of the following is surd or not : 
${\displaystyle \sqrt[3]{64}}$

Sol:

${\displaystyle \sqrt[3]{64}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2}=4}$ which is rational.

∴ ${\displaystyle \sqrt[3]{64}}$ is not a surds.

Question 1.5

State, with reason, of the following is surd or not : 
${\displaystyle \sqrt[3]{25}.\sqrt[3]{40}}$

Sol:

${\displaystyle \sqrt[3]{25}.\sqrt[3]{40}=\sqrt[3]{5\times 5\times 2\times 2\times 2\times 5}=2\times 5=10}$

∴ ${\displaystyle \sqrt[3]{25}.\sqrt[3]{40}}$ is not a surds.

Question 1.6

State, with reason, of the following is surd or not : 
${\displaystyle \sqrt[3]{-125}}$

Sol:

${\displaystyle \sqrt[3]{-125}=\sqrt[3]{-5\times -5\times -5}=-5}$

∴ ${\displaystyle \sqrt[3]{-125}}$ is not a surds. 

Question 1.7

State, with reason, of the following is surd or not : √π

Sol:

We observe that ${\displaystyle \sqrt{\pi }}$​ is an irrational number but π is not a rational number.

∴π​ is a surd.

Question 1.8

State, with reason, of the following is surd or not : 
${\displaystyle \sqrt{3+\sqrt{2}}}$

Sol:

${\displaystyle \sqrt{3+\sqrt{2}}}$ is not a surds because 3 + √2 is irrational.

Question 2.1

Write the lowest rationalising factor of : 5√2

Sol:

5√2 x 5√2 = 5 x 2 = 10 which is rational.

∴ lowest rationalizing factor is √2 

Question 2.2

Write the lowest rationalising factor of : √24

Sol:

√24 = ${\displaystyle \sqrt{2\times 2\times 2\times 3}=2\surd 6}$

∴ lowest rationalizing factor is √6.

Question 2.3

Write the lowest rationalising factor of : √5 - 3

Sol:

( √5 - 3 )( √5 + 3 ) = ( √5 )² - (3)² = 5 - 9 = -4

∴ lowest rationalizing factor is ( √5 + 3 )

Question 2.4

Write the lowest rationalising factor of : 7 - √7

Sol:

7 - √7
( 7 - √7 )( 7 + √7 ) = 49 - 7 = 42  
Therefore, lowest rationalizing factor is ( 7 + √7 ).

Question 2.5

Write the lowest rationalising factor of : √18 - √50

Sol:

√18 - √50
√18 - √50 = ${\displaystyle \sqrt{2\times 3\times 3}-\sqrt{5\times 5\times 2}}$
                 = 3√2 - 5√2 = -2√2
∴ lowest rationalizing factor is √2

Question 2.6

Rationalise the denominators of : ${\displaystyle \frac{\surd 3+1}{\surd 3-1}}$

Sol:

= ${\displaystyle \frac{\surd 3+1}{\surd 3-1}\times \frac{\surd 3+1}{\surd 3+1}}$

= ${\displaystyle \frac{{(\surd 3+1)}^2}{{(\surd 3)}^2-{\left(1\right)}^2}}$

= ${\displaystyle \frac{3+1+2\surd 3}{3-1}}$

= ${\displaystyle \frac{4+2\surd 3}{2}}$

= ${\displaystyle \frac{2(2+\surd 3)}{2}}$

= 2 + √3

Question 2.7

Write the lowest rationalising factor of : √13 + 3

Sol:

( √13 + 3 )( √13 - 3 ) = ( √13 )² - 3² = 13 - 9 = 4

Its lowest rationalizing factor is √13 - 3.

Question 2.8

Write the lowest rationalising factor of : 15 - 3√2

Sol:

15 - 3√2
15 - 3√2 = 3( 5 - √2 )
               = 3( 5 - √2 )( 5 + √2 )
               = 3 x ${\displaystyle [5^2-{\left(\sqrt{2}\right)}^2]}$
               = 3 x [ 25 - 2 ]
               = 3 x 23
               = 69
Its lowest rationalizing factor is 5 + √2.

Question 2.9

Write the lowest rationalising factor of : 3√2 + 2√3

Sol:

3√2 + 2√3
= ( 3√2 + 2√3 )( 3√2 - 2√3 )
= ( 3√2)² - (2√3)²
= 9 x 2 - 4 x 3
= 18 - 12
= 6
its lowest rationalizing factor is 3√2 - 2√3.

Question 3.1

Rationalise the denominators of : ${\displaystyle \frac{3}{\sqrt{5}}}$

Sol:

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

Question 3.2

Rationalise the denominators of : ${\displaystyle \frac{2\sqrt{3}}{\sqrt{5}}}$

Sol:

${\displaystyle \frac{2\sqrt{3}}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{2}{2}\sqrt{15}}$

Question 3.3

Rationalise the denominators of : ${\displaystyle \frac{1}{\sqrt{3}-\sqrt{2}}}$

Sol:

${\displaystyle \frac{1}{\sqrt{3}-\sqrt{2}}\times \left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\right)=\frac{\sqrt{3+\sqrt{2}}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}=\frac{\sqrt{3}+\sqrt{2}}{3-2}}$

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

Question 3.4

Rationalise the denominators of : ${\displaystyle \frac{3}{\sqrt{5}+\sqrt{2}}}$

Sol:

${\displaystyle \frac{3}{\sqrt{5}+\sqrt{2}}\times \left(\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}\right)}$

= ${\displaystyle \frac{3(\sqrt{5}-\sqrt{2})}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{2}\right)}^2}}$

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

= ${\displaystyle \sqrt{5}-\sqrt{2}}$

Question 3.5

Rationalise the denominators of : ${\displaystyle \frac{2-\surd 3}{2+\surd 3}}$

Sol:

${\displaystyle \frac{2-\surd 3}{2+\surd 3}\times \frac{2-\surd 3}{2-\surd 3}}$

= ${\displaystyle \frac{{(2-\surd 3)}^2}{{\left(2\right)}^2-{(\surd 3)}^2}=\frac{4+3-4\surd 3}{4-3}}$

= ${\displaystyle \frac{7-4\surd 3}{1}}$ 
= 7 -  4√3

Question 3.6

Rationalise the denominators of : ${\displaystyle \frac{\sqrt{3}+1}{\sqrt{3}-1}}$

Sol:

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

= ${\displaystyle \frac{{(\sqrt{3}+1)}^2}{{\left(\sqrt{3}\right)}^3-{\left(1\right)}^2}}$

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

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

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

= 2 + √3

Question 3.7

Rationalise the denominators of : ${\displaystyle \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}$

Sol:

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

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

= ${\displaystyle \frac{3+2-2\sqrt{6}}{3-2}}$

= 5 - 2√6

Question 3.8

Rationalise the denominators of : ${\displaystyle \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}}$

Sol:

${\displaystyle \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}\times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}}$

= ${\displaystyle \frac{6+5-2\surd 30}{{(\surd 6)}^2-{(\surd 5)}^2}}$

= ${\displaystyle \frac{11-2\surd 30}{6-5}}$

= 11 - 2√30

Question 3.9

Rationalise the denominators of : ${\displaystyle \frac{2\surd 5+3\surd 2}{2\surd 5-3\surd 2}}$

Sol:

${\displaystyle \frac{2\surd 5+3\surd 2}{2\surd 5-3\surd 2}\times \frac{2\surd 5+3\surd 2}{2\surd 5+3\surd 2}}$

= ${\displaystyle \frac{{(2\sqrt{5}+3\sqrt{2})}^2}{{\left(2\sqrt{5}\right)}^2-{\left(3\sqrt{2}\right)}^2}}$

= ${\displaystyle \frac{4\times 5+9\times 2+12\sqrt{10}}{20-18}}$

= ${\displaystyle \frac{20+18+12\sqrt{10}}{2}}$

= ${\displaystyle \frac{38+12\sqrt{10}}{2}}$

= ${\displaystyle \frac{2(19+6\sqrt{10})}{2}}$

= 19 + 6√10

Question 4.1

Find the values of 'a' and 'b' in each of the following : 
${\displaystyle \frac{2+\sqrt{3}}{2-\sqrt{3}}=a+b\sqrt{3}}$

Sol:

${\displaystyle \frac{2+\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}=a+b\sqrt{3}}$

= ${\displaystyle \frac{{(2+\sqrt{3})}^2}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2}=a+b\sqrt{3}}$

= ${\displaystyle \frac{4+3+4\sqrt{3}}{4-3}=a+b\sqrt{3}}$

7 + 4√3 = a + b√3

a = 7, b = 4

Question 4.2

Find the values of 'a' and 'b' in each of the following:
${\displaystyle \frac{\sqrt{7}-2}{\sqrt{7}+2}=a\sqrt{7}+b}$ 

Sol:

${\displaystyle \frac{\sqrt{7}-2}{\sqrt{7}+2}=a\sqrt{7}+b}$ 

${\displaystyle \frac{{(\sqrt{7}-2)}^2}{{\left(\sqrt{7}\right)}^2-{\left(2\right)}^2}=a\sqrt{7}+b}$

${\displaystyle \frac{7+4-4\sqrt{7}}{7-4}=a\sqrt{7}+b}$

${\displaystyle \frac{11-4\sqrt{7}}{3}=a\sqrt{7}+b}$

${\displaystyle a=-\frac{4}{3},b=\frac{11}{3}}$

Question 4.3

Find the values of 'a' and 'b' in each of the following: 
${\displaystyle \frac{3}{\sqrt{3}-\sqrt{2}}=a\sqrt{3}-b\sqrt{2}}$

Sol:

${\displaystyle \frac{3}{\sqrt{3}-\sqrt{2}}=a\sqrt{3}-b\sqrt{2}}$

${\displaystyle \frac{3}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=a\sqrt{3}-b\sqrt{2}}$

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

${\displaystyle (3\sqrt{3}+3\sqrt{2})=a\sqrt{3}-b\sqrt{2}}$

⇒ a = 3, b = -3 

Question 4.4

Find the values of 'a' and 'b' in each of the following:
${\displaystyle \frac{5+3\sqrt{2}}{5-3\sqrt{2}}=a+b\sqrt{2}}$

Sol:

${\displaystyle \frac{5+3\sqrt{2}}{5-3\sqrt{2}}=a+b\sqrt{2}}$

${\displaystyle \frac{5+3\sqrt{2}}{5-3\sqrt{2}}\times \frac{5+3\sqrt{2}}{5+3\sqrt{2}}=a+b\sqrt{2}}$

${\displaystyle \frac{{(5+3\sqrt{2})}^2}{{\left(5\right)}^2-{\left(3\sqrt{2}\right)}^2}=a+b\sqrt{2}}$

${\displaystyle \frac{25+18+30\sqrt{2}}{25-18}=a+b\sqrt{2}}$

${\displaystyle \frac{43+30\sqrt{2}}{7}=a+b\sqrt{2}}$

${\displaystyle a=\frac{43}{7}, b=\frac{30}{7}}$

Question 5.1

Simplify :
${\displaystyle \frac{22}{2\sqrt{3}+1}+\frac{17}{2\sqrt{3}-1}}$

Sol:

${\displaystyle \frac{22}{2\sqrt{3}+1}+\frac{17}{2\sqrt{3}-1}}$

= ${\displaystyle \frac{22(2\sqrt{3}-1)+17(2\sqrt{3}+1)}{(2\sqrt{3}+1)(2\sqrt{3}-1)}}$

= ${\displaystyle \frac{44\sqrt{3}-22+34\sqrt{3}+17}{{\left(2\sqrt{3}\right)}^2-1}}$

=${\displaystyle \frac{78\sqrt{3}-5}{12-1}}$

= ${\displaystyle \frac{78\sqrt{3}-5}{11}}$

Question 5.2

Simplify :
${\displaystyle \frac{\sqrt{2}}{\sqrt{6}-\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{6}+\sqrt{2}}}$

Sol:

${\displaystyle \frac{\sqrt{2}}{\sqrt{6}-\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{6}+\sqrt{2}}}$

= ${\displaystyle \frac{\sqrt{2}}{\sqrt{6}-2}-\frac{\sqrt{3}}{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{2}(\sqrt{6}+\sqrt{2})-\sqrt{3}(\sqrt{6}-\sqrt{2})}{{\left(\sqrt{6}\right)}^2-{\left(\sqrt{2}\right)}^2}}$

= ${\displaystyle \frac{\sqrt{12}+2-\sqrt{18}+\sqrt{6}}{6-2}}$

= ${\displaystyle \frac{2\sqrt{3}+2-3\sqrt{2}+\sqrt{6}}{4}}$

Question 6.1

If x =${\displaystyle \frac{\sqrt{5}-2}{\sqrt{5}+2}}$ and y = ${\displaystyle \frac{\sqrt{5}+2}{\sqrt{5}-2}}$; find :
x²

Sol:

x² = ${\displaystyle {\left[\frac{\sqrt{5}-2}{\sqrt{5}+2}\right]}^2=\frac{5+4-4\sqrt{5}}{5+4+4\sqrt{5}}=\frac{9-4\sqrt{5}}{9+4\sqrt{5}}}$

= ${\displaystyle \frac{9-4\sqrt{5}}{9+4\sqrt{5}}\times \left[\frac{9-4\sqrt{5}}{9-4\sqrt{5}}\right]=\frac{{(9-4\sqrt{5})}^2}{{\left(9\right)}^2-{\left(4\sqrt{5}\right)}^2}}$ 

= ${\displaystyle \frac{81+80-72\sqrt{5}}{81-80}=161-72\sqrt{5}}$

Question 6.2

If x =${\displaystyle \frac{\sqrt{5}-2}{\sqrt{5}+2}}$ and y = ${\displaystyle \frac{\sqrt{5}+2}{\sqrt{5}-2}}$; find : y²

Sol:

 y² = ${\displaystyle {\left[\frac{\sqrt{5}+2}{\sqrt{5}-2}\right]}^2=\frac{5+4+4\sqrt{5}}{5+4-4\sqrt{5} }=\frac{9+4\sqrt{5}}{9-4\sqrt{5}}}$

= ${\displaystyle \frac{9+4\sqrt{5}}{9-4\sqrt{5}}\times \frac{9+4\sqrt{5}}{9+4\sqrt{5}}=\frac{{(9+4\sqrt{5})}^2}{{\left(9\right)}^2-{\left(4\sqrt{5}\right)}^2}=\frac{81+80+72\sqrt{5}}{81-80}}$

= ${\displaystyle 161+72\sqrt{5}}$

Question 6.4

If x =${\displaystyle \frac{\sqrt{5}-2}{\sqrt{5}+2}}$ and y = ${\displaystyle \frac{\sqrt{5}+2}{\sqrt{5}-2}}$; find :  xy

Sol:

xy = ${\displaystyle \frac{(\sqrt{5}-2)(\sqrt{5}+2)}{(\sqrt{5}+2)(\sqrt{5}-2)}=1}$

Question 6.4

If x =${\displaystyle \frac{\sqrt{5}-2}{\sqrt{5}+2}}$ and y = ${\displaystyle \frac{\sqrt{5}+2}{\sqrt{5}-2}}$; find :
x² + y² + xy.

Sol:

x² + y² + xy 
= 161 - 72√5 + 161 +72√5 + 1

= 322 + 1 = 323

Question 7.1

If m = ${\displaystyle \frac{1}{3-2\sqrt{2}}\text{and}n=\frac{1}{3+2\sqrt{2}},}$ find m²

Sol:

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

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

m = ${\displaystyle \frac{3+2\sqrt{2}}{{\left(3\right)}^2-{\left(2\sqrt{2}\right)}^2}}$

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

m = 3 +2√2

⇒ m² = ( 3 + 2√2)²

           = (3)² + 2 x 3 x 2√2 + (2√2)²

           = 9 + 12√2  + 8

           = 17 + 12√2

Question 7.2

If m = ${\displaystyle \frac{1}{3-2\sqrt{2}}\text{and}n=\frac{1}{3+2\sqrt{2}},}$ find mn

Sol:

mn = ( 3 + 2√2 )( 3 - 2√2 ) = (3)² - (2√2)² = 9 - 8 = 1

Question 7.3

If m = ${\displaystyle \frac{1}{3-2\sqrt{2}}\text{and}n=\frac{1}{3+2\sqrt{2}},}$ find n²

Sol:

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

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

n = ${\displaystyle \frac{3-2\sqrt{2}}{{\left(3\right)}^2-{\left(2\sqrt{2}\right)}^2}}$

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

n = 3 - 2√2

⇒ n²  = ( 3 - 2√2)²

           = (3)² - 2 x 3 x 2√2 + (2√2)²

           = 9 - 12√2  + 8

           = 17 - 12√2

Question 8.1

If x = 2√3 + 2√2 , find : ${\displaystyle \frac{1}{x}}$

Sol:

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

= ${\displaystyle \frac{2\sqrt{3}-2\sqrt{2}}{12-8}}$

= ${\displaystyle \frac{2(\sqrt{3}-\sqrt{2})}{42}}$

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

Question 8.2

If x = 2√3 + 2√2 , find : ${\displaystyle (x+\frac{1}{x})}$

Sol:

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

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

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

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

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

Question 8.3

If x = 2√3 + 2√2 , find : ${\displaystyle {(x+\frac{1}{x})}^2}$

Sol:

${\displaystyle {(x+\frac{1}{x})}^2={[5\sqrt{3}+3\frac{\sqrt{2}}{2}]}^2}$

= ${\displaystyle \frac{75+18+30\sqrt{6}}{4}}$

= ${\displaystyle \frac{93+30\sqrt{6}}{4}}$

Question 9

If x = 1 - √2, find the value of ${\displaystyle {(x-\frac{1}{x})}^3}$

Sol:

Given that x = 1 - √2
We need to find the value of ${\displaystyle {(x-\frac{1}{x})}^3}$

Since x = 1 - √2, we have 
${\displaystyle \frac{1}{x}=\frac{1}{1-\sqrt{2}}\times \frac{1+\sqrt{2}}{1+\sqrt{2}}}$

⇒ ${\displaystyle \frac{1}{x}=\frac{1+\sqrt{2}}{{\left(1\right)}^2-{\left(\sqrt{2}\right)}^2}}$      [ Since ( a - b )( a + b ) = a² - b² ]

⇒ ${\displaystyle \frac{1}{x}=\frac{1+\sqrt{2}}{1-2}}$

⇒ ${\displaystyle \frac{1}{x}=\frac{1+\sqrt{2}}{-1}}$

⇒ ${\displaystyle \frac{1}{x}=-(1+\sqrt{2})}$              .....(1)

Thus, ${\displaystyle (x-\frac{1}{x})=(1-\surd 2)-(-(1+\sqrt{2}))}$

⇒ ${\displaystyle (x-\frac{1}{x})=1-\surd 2+1+\surd 2}$

⇒ ${\displaystyle (x-\frac{1}{x})=2}$

⇒ ${\displaystyle {(x-\frac{1}{x})}^3=2^3}$

⇒ ${\displaystyle {(x-\frac{1}{x})}^3=8}$

Question 10

If x = 5 - 2√6, find ${\displaystyle x^2+\frac{1}{x^2}}$

Sol:

Given x = 5 - 2√6

We need to find ${\displaystyle x^2+\frac{1}{x^2}}$

Since x = 5 - 2√6 , we have

${\displaystyle \frac{1}{x}=\frac{1}{5-2\surd 6}}$

⇒ ${\displaystyle \frac{1}{x}=\frac{1}{5-2\surd 6}\times \frac{5+2\surd 6}{5+2\surd 6}}$

⇒ ${\displaystyle \frac{1}{x}=\frac{5-2\surd 6}{(5-2\surd 6)(5+2\surd 6)}}$

⇒ ${\displaystyle \frac{1}{x}=\frac{5+2\surd 6}{5^2-{(2\surd 6)}^2}}$

⇒ ${\displaystyle \frac{1}{x}=\frac{5+2\surd 6}{25-24}}$

⇒ ${\displaystyle \frac{1}{x}=\frac{5+2\surd 6}{1}}$

⇒ ${\displaystyle \frac{1}{x}=(5+2\surd 6)}$                          .....(1)

Thus, ${\displaystyle (x-\frac{1}{x})}$ = ( 5 - 2√6 ) - ( 5 + 2√6)

⇒ ${\displaystyle (x-\frac{1}{x})}$ = 5 - 2√6 - 5 - 2√6

⇒ ${\displaystyle (x-\frac{1}{x})}$ = - 4√6                       ....(2)

Now consider ${\displaystyle {(x-\frac{1}{x})}^2}$ :

Thus
${\displaystyle (x-\frac{1}{x})=x^2+\frac{1}{x^2}-2x\times \frac{1}{x}}$    [ since ( a - b )² = a² - 2ab + b² ]

⇒ ${\displaystyle {(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2}$

⇒ ${\displaystyle {(x-\frac{1}{x})}^2+2=x^2+\frac{1}{x^2}}$     .....(3)

Thus, from equations (2) and (3), we have

${\displaystyle \left(x^2\right)+\left(\frac{1}{x^2}\right)={(-4\surd 6)}^2+2}$

⇒ ${\displaystyle \left(x^2\right)+\left(\frac{1}{x^2}\right)=96+2}$

⇒ ${\displaystyle \left(x^2\right)+\left(\frac{1}{x^2}\right)=98}$

Question 11

Show that :
${\displaystyle \frac{1}{3-2\surd 2}-\frac{1}{2\surd 2-\surd 7}+\frac{1}{\surd 7-\surd 6}-\frac{1}{\surd 6-\surd 5}+\frac{1}{\surd 5-2}=5}$

Sol:

L.H.S = ${\displaystyle \frac{1}{3-2\surd 2}-\frac{1}{2\surd 2-\surd 7}+\frac{1}{\surd 7-\surd 6}-\frac{1}{\surd 6-\surd 5}+\frac{1}{\surd 5-2}}$
= ${\displaystyle \frac{1}{3-\surd 8}-\frac{1}{\surd 8-\surd 7}+\frac{1}{\surd 7-\surd 6}-\frac{1}{\surd 6-\surd 5}+\frac{1}{\surd 5-2}}$

= ${\displaystyle \frac{1}{3-\surd 8}\times \frac{3+\surd 8}{3+\surd 8}-\frac{1}{\surd 8-\surd 7}\times \frac{\surd 8+\surd 7}{\surd 8+\surd 7}+\frac{1}{\surd 7-\surd 6}\times \frac{\surd 7+\surd 6}{\surd 7+\surd 6}-\frac{1}{\surd 6-\surd 5}\times \frac{\surd 6+\surd 5}{\surd 6+\surd 5}+\frac{1}{\surd 5-2}\times \frac{\surd 5+2}{\surd 5+2}}$

= ${\displaystyle \frac{3+\surd 8}{{\left(3\right)}^2-{(\surd 8)}^2} -\frac{\surd 8+\surd 7}{{(\surd 8)}^2-{(\surd 7)}^2}+\frac{\surd 7+\surd 6}{{(\surd 7)}^2-{(\surd 6)}^2}-\frac{\surd 6-\surd 5}{{(\surd 6)}^2-{(\surd 5)}^2}+\frac{\surd 5-2}{{(\surd 5)}^2-{\left(2\right)}^2}}$

= ${\displaystyle \frac{3+\surd 8}{9-8} -\frac{\surd 8+\surd 7}{8-7} +\frac{\surd 7+\surd 6}{7-6}-\frac{\surd 6-\surd 5}{6-5}+\frac{\surd 5-2}{5-4}}$

= 3 + √8 - √8 - √7 + √7 + √6 - √6 - √5 + √5 + 2

= 3 + 2

= 5

= R.H.S.

Question 12

Rationalise the denominator of : ${\displaystyle \frac{1}{\surd 3-\surd 2+1}}$

Sol:

${\displaystyle \frac{1}{\surd 3-\surd 2+1}}$

= ${\displaystyle \frac{1}{(\surd 3-\surd 2)+1}\times \frac{(\surd 3-\surd 2)-1}{(\surd 3-\surd 2)-1}}$

= ${\displaystyle \frac{\surd 3-\surd 2-1}{{(\surd 3-\surd 2)}^2-{\left(1\right)}^2}}$

= ${\displaystyle \frac{\surd 3-\surd 2-1}{{(\surd 3)}^2-2\surd 6+{(\surd 2)}^2-1}}$

= ${\displaystyle \frac{\surd 3-\surd 2-1}{3-2\surd 6+2-1}}$

= ${\displaystyle \frac{\surd 3-\surd 2-1}{4-2\surd 6}}$

= ${\displaystyle \frac{(\surd 3-\surd 2)-1}{2(2-\surd 6)}}$

= ${\displaystyle \frac{\surd 3-\surd 2-1}{2(2-\surd 6)}\times \frac{2+\surd 6}{2+\surd 6}}$

= ${\displaystyle \frac{2\surd 3-2\surd 2-2+\surd 18-\surd 12-\surd 6}{2[{\left(2\right)}^2-{(\surd 6)}^2]}}$

= ${\displaystyle \frac{2\surd 3-2\surd 2-2+3\surd 2-2\surd 3-\surd 6}{2[4-6]}}$

= ${\displaystyle \frac{\surd 2-2-\surd 6}{2(-2)}}$

= ${\displaystyle \frac{\surd 2-2-\surd 6}{-4}}$

= ${\displaystyle \frac{1}{4}(2+\surd 6-\surd 2)}$

Question 13.1

If √2 = 1.4 and √3 = 1.7, find the value of : ${\displaystyle \frac{1}{\surd 3-\surd 2}}$

Sol:

√2 = 1.4 and √3 = 1.7

${\displaystyle \frac{1}{\surd 3-\surd 2}}$

= ${\displaystyle \frac{1}{\surd 3-\surd 2}\times \frac{\surd 3+\surd 2}{\surd 3+\surd 2}}$

= ${\displaystyle \frac{\surd 3+\surd 2}{{(\surd 3)}^2-{(\surd 2)}^2}}$

= ${\displaystyle \frac{\surd 3+\surd 2}{3-2}}$

= √3 + √2

= 1.7 + 1.4

= 3.1

Question 13.2

If √2 = 1.4 and √3 = 1.7, find the value of : ${\displaystyle \frac{1}{3+2\surd 2}}$

Sol:

√2 = 1.4 and √3 = 1.7

${\displaystyle \frac{1}{3+2\surd 2}}$

= ${\displaystyle \frac{1}{3+2\surd 2}\times \frac{3-2\surd 2}{3-2\surd 2}}$

= ${\displaystyle \frac{3-2\surd 2}{{\left(3\right)}^2-{(2\surd 2)}^2}}$

= ${\displaystyle \frac{3-2\surd 2}{9-8}}$

= 3 - 2√2

= 3 - 2( 1.4 )

= 3 - 2.8

= 0.2

Question 13.3

Simplify : ${\displaystyle \frac{2-\surd 3}{\surd 3}}$

Sol:

${\displaystyle \frac{2-\surd 3}{\surd 3}}$

= ${\displaystyle \frac{2}{\surd 3}-\frac{\surd 3}{\surd 3}}$

= ${\displaystyle \frac{2\surd 3}{\surd 3.\surd 3}-1}$

= ${\displaystyle \frac{2\surd 3}{3}-1}$

Question 14

Evaluate : ${\displaystyle \frac{4-\surd 5}{4+\surd 5}+\frac{4+\surd 5}{4-\surd 5}}$

Sol:

${\displaystyle \frac{4-\surd 5}{4+\surd 5}+\frac{4+\surd 5}{4-\surd 5}}$

= ${\displaystyle \frac{4-\surd 5}{4+\surd 5}\times \frac{4-\surd 5}{4-\surd 5}+\frac{4+\surd 5}{4-\surd 5}\times \frac{4+\surd 5}{4+\surd 5}}$

= ${\displaystyle \frac{{(4-\surd 5)}^2}{{\left(4\right)}^2-{(\surd 5)}^2}+\frac{{(4+\surd 5)}^2}{{\left(4\right)}^2-{(\surd 5)}^2}}$

= ${\displaystyle \frac{16+5-8\surd 5}{16-5}+\frac{16+5+8\surd 5}{16-5}}$

= ${\displaystyle \frac{21-8\surd 5}{11}+\frac{21+8\surd 5}{11}}$

= ${\displaystyle \frac{21-8\surd 5+21+8\surd 5}{11}}$

= ${\displaystyle \frac{42}{11}=3\frac{9}{11}}$

Question 15

If ${\displaystyle \frac{2+\surd 5}{2-\surd 5}=x\text{and} \frac{2-\surd 5}{2+\surd 5}=y}$; find the value of x² - y².

Sol:

x =${\displaystyle \frac{2+\surd 5}{2-\surd 5}}$                           

   =${\displaystyle \frac{2+\surd 5}{2-\surd 5}\times \frac{2+\surd 5}{2+\surd 5}}$      

   = ${\displaystyle \frac{{(2+\surd 5)}^2}{2^2-{(\surd 5)}^2}}$

   = ${\displaystyle \frac{4+4\surd 5+5}{4}-5}$

    = ${\displaystyle \frac{9+4\surd 5}{-1}}$

    =  - ( 9 - 4√5 )

y = ${\displaystyle \frac{2-\surd 5}{2+\surd 5}}$

   =${\displaystyle \frac{2-\surd 5}{2+\surd 5}\times \frac{2-\surd 5}{2-\surd 5}}$

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

   = ${\displaystyle \frac{4-4\surd 5+5}{4}-5}$

    = ${\displaystyle \frac{9-4\surd 5}{-1}}$

    =  - ( 9 + 4√5 )

∴ x² - y² = ( - 9 - 4√5 )² - ( - 9 + 4√5 )²

               = 81 + 72√5 + 80 - ( 81 - 72√5 + 80 )

               = 81 + 72√5 + 80 - 81 + 72√5 - 80
               = 144√5