Question

if x is equal to 9 + 4 under root 5 and X Y is equal to one, then 1 / X square + 1 /y square is

Answer 1

$<I >Hey here is your answer$

GIVEN :-
x = 9 + 4√5
xy = 1

$\it \: to \: find \: = > \frac{1}{ {x}^{2} } + \frac{1}{ {y}^{2} }$

Let, solve the sum :-

here x = 9 + 4√5 (given). -(1)

xy = 1
putting x = 9 + 4√5 in equation:-
(9 + 4√5) y = 1

$y \: = \frac{1}{9 + 4 \sqrt{5} }\: \: \: \: \: \: \: - (2)$
by rationalising method :-

$y \: = \frac{1}{9 + 4 \sqrt{5} } \\ \\ \frac{1}{9 + 4 \sqrt{5} } \times \frac{9 - 4 \sqrt{5} }{9 - 4 \sqrt{5} } \\ \\ \frac{9 - 4 \sqrt{5} }{ {(9)}^{2} - {(4 \sqrt{5}) }^{2} } \\ \\ \frac{9 - 4 \sqrt{5} }{81 - 80} \\ \\ 9 - 4 \sqrt{5}$
so, y = 9 - 4√5

now :-

to solve
$\it \frac{1}{ {x}^{2} } + \frac{1}{ {y}^{2} }$
we have to find the values of 1/x and 1/y first to make question easy :-

x = 9 + 4√5
1/x = 1 / 9+4√5

by rationalising :-

$\frac{1}{x} \: = \frac{1}{9 + 4 \sqrt{5} } \\ \\ \frac{1}{9 + 4 \sqrt{5} } \times \frac{9 - 4 \sqrt{5} }{9 - 4 \sqrt{5} } \\ \\ \frac{9 - 4 \sqrt{5} }{ {(9)}^{2} - {(4 \sqrt{5}) }^{2} } \\ \\ \frac{9 - 4 \sqrt{5} }{81 - 80} \\ \\ 9 - 4 \sqrt{5}$
so, 1/x = 9 - 4√5

now y = 1 / 9+4√5 (see 2)
1/y = 9 + 4√5

now ;
Let's find the value of
$\it \frac{1}{ {x}^{2} } + \frac{1}{ {y}^{2} } \\ \\ {( \frac{1}{x} )}^{2} + {( \frac{1}{y}) }^{2} \\ \\ (9 - 4 \sqrt{5} {)}^{2} + {(9 + 4 \sqrt{5} )}^{2} \\ \\ {(9)}^{2} + {(4 \sqrt{5} )}^{2} - 2 \times 9 \times 4 \sqrt{5} + {(9)}^{2} + {(4 \sqrt{5} )}^{2} + 2 \times 9 \times 4 \sqrt{5} \\ \\ 81 + 80 - 72 \sqrt{5 } + 81 + 80 + 70 \sqrt{5} \\ \\ 81 + 80 + 81 + 80 \\ \\ \bf 322 \: answer$
hope this helps you

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