Question

if x=9-4√5 find the value of x²+1/x²​

Answer 1

Answer:

\begin{gathered}x = 9 - 4 \sqrt{5} \\ \\ \frac{1}{x} = \frac{1}{9 - 4 \sqrt{5} } \\ \\ \frac{1}{x} = \frac{1}{9 - 4 \sqrt{5} } \times \frac{9 + 4 \sqrt{5} }{9 + 4 \sqrt{5} } \\ \\ \frac{1}{x} = \frac{9 + 4 \sqrt{5} }{ {(9)}^{2} - {(4 \sqrt{5} )}^{2} } \\ \\ \frac{1}{x} = \frac{9 + 4 \sqrt{5} }{81 - 80} \\ \\ \frac{1}{x} = 9 + 4 \sqrt{5} \end{gathered}

x=9−4

5

x

1

=

9−4

5

1

x

1

=

9−4

5

1

×

9+4

5

9+4

5

x

1

=

(9)

2

−(4

5

)

2

9+4

5

x

1

=

81−80

9+4

5

x

1

=9+4

5

Now,

\begin{gathered} {x}^{2} + \frac{1}{ { x}^{2} } = {(9 - 4 \sqrt{5}) }^{2} + {(9 + 4 \sqrt{5} )}^{2} \\ \\ = 81 + 80 - 72 \sqrt{5} + 81 + 80 + 72 \sqrt{5} \\ \\ = 81 + 80 + 81 + 80 \\ \\ = 161 + 161 \\ \\ = 322\end{gathered}

x

2

+

x

2

1

=(9−4

5

)

2

+(9+4

5

)

2

=81+80−72

5

+81+80+72

5

=81+80+81+80

=161+161

=322

Step-by-step explanation:

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Answer 2

GIVEN :-

• $\sf \: x = 9 - 4 \sqrt{5}$

TO FIND :-

• $\sf \: {x}^{2} + \frac{1}{ {x}^{2} } =$

SOLUTION :-

First we need to calculate 1/x

$\sf\frac{1}{x} = \frac{1}{9 - 4 \sqrt{5} }$

$\sf \: \: \: \: \: = \bigg( \frac{1}{9 - 4 \sqrt{5} } \bigg). \bigg( \frac{9 + 4 \sqrt{5} }{9 + 4 \sqrt{5} } \bigg)$

$\sf \: \: \: \: \: = \bigg( \frac{9 + 4 \sqrt{5} }{ {9}^{2} - { (4 \sqrt{5})}^{2} } \bigg)$

$\sf \: \: \: \: \: = \bigg( \frac{9 + 4 \sqrt{5} }{ 81 - 80 } \bigg)$

$\sf \: \: \: \: \: = 9 + 4 \sqrt{5}$

Now Appling

$\sf \: x + \frac{1}{x} = (9 - 4 \sqrt{5} + 9 + 4 \sqrt{5} )$

$\sf \: x + \frac{1}{x} = 18$

Now squaring both sides

$\sf \: \bigg(x + \frac{1}{x} \bigg)^{2} = 18^{2}$

$\sf \: {x}^{2} + \frac{1}{ {x}^{2} } + 2.(x). \frac{1}{(x)} = 324$

$\sf \: {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 324$

$\sf \: {x}^{2} + \frac{1}{ {x}^{2} } = 324 - 2$

$\green{ \boxed{ \sf \: {x}^{2} + \frac{1}{ {x}^{2} } = 322 }}$