Question

if x =√5+√3/√5-√3 find x²+1/x²​

Answer 1

$\large \underline \bold{Given :-}$

$\: \: \: \: \: \:$$\sf{x =\dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}}$

$\large \underline \bold{To \: Find :-}$

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

$\large \underline \bold{Usable \: Identity -}$

$\sf{1) \: (a + b)^{2} + (a - b)^{2} = 2(a^{2} + b^{2})}$

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

$\sf{3) \: (a + b)^{2} = a^{2} + 2ab + b^{2}}$

$\large \underline \bold{Solution :-}$

$\: \: \:$$\sf{x =\dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}}$

$\: \: \:$$\sf{\dfrac{1}{x} =\dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}}$

Now ,

$\: \: \:$$\sf{x + \dfrac{1}{x} =\dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}}$

$\: \: \:$$\sf{x + \dfrac{1}{x} =\dfrac{(\sqrt{5} + \sqrt{3})^{2} + (\sqrt{5} - \sqrt{3})^{2}}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})}}$

$\: \: \:$$\sf{x + \dfrac{1}{x} =\dfrac{2[(\sqrt{5})^{2} + (\sqrt{3})^{2}]}{(\sqrt{5})^{2} - (\sqrt{3})^{2}}}$

$\: \: \:$$\sf{x + \dfrac{1}{x} =\dfrac{2(5 + 3)}{5 - 3}}$

$\: \: \:$$\sf{x + \dfrac{1}{x} = \dfrac{\cancel{2} (8)}{\cancel{2}}}$

$\: \: \:$$\sf{x + \dfrac{1}{x} = 8}$

$\:$Doing square both side -

$\: \:$$\sf{(x + \dfrac{1}{x})^{2} = (8)^{2}}$

$\: \:$$\sf{x^{2} + \dfrac{1}{x^{2}} + 2(\cancel{x})(\dfrac{1}{\cancel{x}}) = 64}$

$\: \:$$\sf{x^{2} + \dfrac{1}{x^{2}} + 2 = 64}$

$\: \:$$\small \bold{x^{2} + \dfrac{1}{x^{2}} = 62}$

Answer 2

Step-by-step explanation:

\large \underline \bold{Given :-}

Given:−

\: \: \: \: \: \: \sf{x =\dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}}x=

5

−

3

5

+

3

\large \underline \bold{To \: Find :-}

ToFind:−

\: \: \: \: \: \: \sf{x^{2} + \dfrac{1}{x^{2}} = \: ?}x

2

+

x

2

1

=?

\large \underline \bold{Usable \: Identity -}

UsableIdentity−

\sf{1) \: (a + b)^{2} + (a - b)^{2} = 2(a^{2} + b^{2})}1)(a+b)

2

+(a−b)

2

=2(a

2

+b

2

)

\sf{2) \: (a + b)(a - b) = a^{2} - b^{2}}2)(a+b)(a−b)=a

2

−b

2

\sf{3) \: (a + b)^{2} = a^{2} + 2ab + b^{2}}3)(a+b)

2

=a

2

+2ab+b

2

\large \underline \bold{Solution :-}

Solution:−

\: \: \: \sf{x =\dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}}x=

5

−

3

5

+

3

\: \: \: \sf{\dfrac{1}{x} =\dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}}

x

1

=

5

+

3

5

−

3

Now ,

\: \: \: \sf{x + \dfrac{1}{x} =\dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}}x+

x

1

=

5

−

3

5

+

3

+

5

+

3

5

−

3

\: \: \: \sf{x + \dfrac{1}{x} =\dfrac{(\sqrt{5} + \sqrt{3})^{2} + (\sqrt{5} - \sqrt{3})^{2}}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})}}x+

x

1

=

(

5

−

3

)(

5

+

3

)

(

5

+

3

)

2

+(

5

−

3

)

2

\: \: \: \sf{x + \dfrac{1}{x} =\dfrac{2[(\sqrt{5})^{2} + (\sqrt{3})^{2}]}{(\sqrt{5})^{2} - (\sqrt{3})^{2}}}x+

x

1

=

(

5

)

2

−(

3

)

2

2[(

5

)

2

+(

3

)

2

]

\: \: \: \sf{x + \dfrac{1}{x} =\dfrac{2(5 + 3)}{5 - 3}}x+

x

1

=

5−3

2(5+3)

\: \: \: \sf{x + \dfrac{1}{x} = \dfrac{\cancel{2} (8)}{\cancel{2}}}x+

x

1

=

2

2

(8)

\: \: \: \sf{x + \dfrac{1}{x} = 8}x+

x

1

=8

\: Doing square both side -

\: \: \sf{(x + \dfrac{1}{x})^{2} = (8)^{2}}(x+

x

1

)

2

=(8)

2

\: \: \sf{x^{2} + \dfrac{1}{x^{2}} + 2(\cancel{x})(\dfrac{1}{\cancel{x}}) = 64}x

2

+

x

2

1

+2(

x

)(

x

1

)=64

\: \: \sf{x^{2} + \dfrac{1}{x^{2}} + 2 = 64}x

2

+

x

2

1

+2=64

\: \: \small \bold{x^{2} + \dfrac{1}{x^{2}} = 62}x

2

+

x

2

1

=62