Question

if x=9+4√5and xy=1,then find 1/x^2+1/y^2​

Answer 1

Answer:

Please find the attachment.

Answer 2

GIVEN :–

• x = 9+4√5 and xy = 1

TO FIND :–

• Value of 1/x² +1/y² = ?

SOLUTION :–

$\\ \implies\bf x = 9+4 \sqrt{5}$

• And –

$\\\implies\bf xy=1$

$\\\implies\bf(9+4 \sqrt{5})y=1$

$\\\implies\bf y= \dfrac{1}{9+4 \sqrt{5}}$

$\\\implies\bf y= \dfrac{1}{9+4 \sqrt{5}} \times\dfrac{9 - 4 \sqrt{5} }{9 - 4 \sqrt{5} }$

$\\\implies\bf y= \dfrac{9- 4 \sqrt{5} }{(9)^{2} - (4 \sqrt{5})^{2} }$

$\\\implies\bf y= \dfrac{9- 4 \sqrt{5} }{81-80}$

$\\\implies\bf y= \dfrac{9- 4 \sqrt{5} }{1}$

$\\\implies\bf y=9- 4 \sqrt{5}$

• Now Let's find –

$\\\implies\bf P = \dfrac{1}{ {x}^{2}} + \dfrac{1}{ {y}^{2} }$

$\\\implies\bf P = \dfrac{ {x}^{2} + {y}^{2} }{({x}^{2})( {y}^{2} )}$

$\\\implies\bf P = \dfrac{ {x}^{2} + {y}^{2} }{{(xy)}^{2}}$

$\\\implies\bf P = \dfrac{ {(x + y)}^{2} - 2xy}{{(xy)}^{2}}$

• Now put the values –

$\\\implies\bf P = \dfrac{ {(9+4 \sqrt{5}+9 - 4 \sqrt{5})}^{2} - 2(1)}{{(1)}^{2}}$

$\\\implies\bf P = \dfrac{ {(9+9)}^{2} - 2(1)}{{(1)}^{2}}$

$\\\implies\bf P = \dfrac{ {(18)}^{2} - 2}{1}$

$\\\implies\bf P =324 - 2$

$\\\implies\bf P =322$

• Hence –

$\\\implies \large \green{ \boxed{\bf\dfrac{1}{ {x}^{2}} + \dfrac{1}{{y}^{2}}=322}}$