Question

if x=√3+√2/√3-√2
find ,
(i) x^2+1/x^2
(ii)x^4+1/x^4
​

Answer 1

Answer:

x+

x

1

=3−2

2

+

3−2

2

1

x

1

=

3−2

2

1

×

3+2

2

3+2

2

=

1

3+2

2

x+

x

1

=3−2

2

+3+2

2

=6

Squaring on 60th sides.

(x+

x

1

)

2

=36

x

2

+

x

2

1

=36−2⇒x

2

+

x

2

1

=34

Squaring on 60th sides.

(x

2

+

x

2

1

)

2

=(34)

2

x

4

+

x

4

1

=1156−2⇒x

4

+

x

4

1

=1154.

Answer 2

Question :

If $\sf{x = \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}}$ , find :

• $\sf{x^{2} + \dfrac{1}{x^{2}}}$
• $\sf{x^{4} + \dfrac{1}{x^{4}}}$

Given :

The value of x :

• $\sf{x = \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}}$

To find :

The value of :-

• $\sf{x^{2} + \dfrac{1}{x^{2}} = ?}$
• $\sf{x^{4} + \dfrac{1}{x^{4}} ?}$

Knowledge required :

• $\sf{(a + b)^{2} = a^{2} + b^{2} + 2ab}$

• $\sf{(a - b)^{2} = a^{2} + b^{2} - 2ab}$

• $\sf{a^{2} + b^{2} = (a + b)^{2} - 2ab}$

• $\sf{a^{2} - b^{2} = (a + b)(a - b)}$

• $\sf{\bigg(x + \dfrac{1}{x}\bigg)^{2} = x^{2} + \dfrac{1}{x^{2}} + 2}$

Solution :

First let us find the value of x (Simplified form) :

By Rationaling the value of x we get :

$:\implies \sf{x = \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}}$

By multiplying (√3 + 2) in both the Numerator, we get :

$:\implies \sf{x = \dfrac{(\sqrt{3} + \sqrt{2}) \times (\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2}) \times (\sqrt{3} + \sqrt{2})}}$

$:\implies \sf{x = \dfrac{(\sqrt{3} + \sqrt{2})^{2}}{(\sqrt{3} - \sqrt{2}) \times (\sqrt{3} + \sqrt{2})}}$

$:\implies \sf{x = \dfrac{3 + 2 \times \sqrt{3} \times \sqrt{2} + 2}{3 - 2}}$

$:\implies \sf{x = 3 + 2 \times \sqrt{3} \times \sqrt{2} + 2}$

$:\implies \sf{x = 5 + 2\sqrt{6}}$

$:\implies \sf{x = 5 + 2\sqrt{6}}$

$\boxed{\therefore \sf{x = 5 + 2\sqrt{6}}}$⠀⠀⠀⠀⠀⠀⠀...Eq.(i)

Now let's find the value of 1/x.

$:\implies \sf{\dfrac{1}{x} = \dfrac{1}{5 + 2\sqrt{6}}} \quad \sf{[\because x = 5 + 2\sqrt{6}]}$

By multiplying (5 - 2√6) in both the Numerator and Denominator, we get :

$:\implies \sf{\dfrac{1}{x} = \dfrac{1 \times (5 - 2\sqrt{6})}{(5 + 2\sqrt{6}) \times (5 - 2\sqrt{6})}}$

$:\implies \sf{\dfrac{1}{x} = \dfrac{5 - 2\sqrt{6}}{5^{2} - 2\sqrt{6}^{2}}}$

$:\implies \sf{\dfrac{1}{x} = \dfrac{5 - 2\sqrt{6}}{25 - 24}}$

$:\implies \sf{\dfrac{1}{x} = 5 - 2\sqrt{6}}$

$\boxed{\therefore \sf{\dfrac{1}{x} = 5 - 2\sqrt{6}}}$⠀⠀⠀⠀⠀⠀⠀...Eq.(ii)

By adding Eq.(i) and Eq.(ii), we get :

$:\implies \sf{x + \dfrac{1}{x} = 5 + 2\sqrt{6} + 5 - 2\sqrt{6}}$

$:\implies \sf{x + \dfrac{1}{x} = 5 + 5}$

$:\implies \sf{x + \dfrac{1}{x} = 10}$

$\boxed{\therefore \sf{x + \dfrac{1}{x} = 10}}$⠀⠀⠀⠀⠀⠀⠀...Eq.(iii)

By squaring Eq.(iii), we get :

$:\implies \sf{\bigg(x + \dfrac{1}{x}\bigg)^{2} = 10^{2}}$

$:\implies \sf{x^{2} + \dfrac{1}{x^{2}} + 2= 100}$

$:\implies \sf{x^{2} + \dfrac{1}{x^{2}} = 100 - 2}$

$:\implies \sf{x^{2} + \dfrac{1}{x^{2}} = 98}$

$\boxed{\therefore \sf{x^{2} + \dfrac{1}{x^{2}} = 98}}$⠀⠀⠀⠀⠀⠀⠀...Eq.(iv)

By squaring Eq.(iv), we get :

$:\implies \sf{\bigg(x^{2} + \dfrac{1}{x^{2}}\bigg)^{2} = 98^{2}}$

$:\implies \sf{x^{4} + \dfrac{1}{x^{4}} + 2 = 9604}$

$:\implies \sf{x^{4} + \dfrac{1}{x^{4}} = 9604 - 2}$

$:\implies \sf{x^{4} + \dfrac{1}{x^{4}} = 9602}$

$\boxed{\therefore \sf{x^{4} + \dfrac{1}{x^{4}} = 9602}}$