Question

if x is equal to 3 + under root 8 and Y is equal to 3 minus under root 8 then 1 divided by X square + 1 / square is equal to

Answer 1

Heya !!!

Identities used :

$(x + y)(x - y) = {x}^{2} - {y}^{2} \\ \\ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \\ \\ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$



$x = 3 + \sqrt{8} \\ \\ \frac{1}{x} = \frac{1}{3 + \sqrt{8} }$

On rationalizing the denominator we get,

$\frac{1}{x} = \frac{1}{3 + \sqrt{8} } \times \frac{3 - \sqrt{8} }{3 - \sqrt{8} } \\ \\ \frac{1}{x} = \frac{3 - \sqrt{8} }{(3 + \sqrt{8} )(3 - \sqrt{8} )} \\ \\ \frac{1}{x} = \frac{3 - \sqrt{8} }{ {(3)}^{2} - {( \sqrt{8}) }^{2} } \\ \\ \frac{1}{x} = \frac{3 - \sqrt{8} }{9 - 8} \\ \\ \frac{1}{x} = 3 - \sqrt{8}$

On squaring both the sides we get

${( \frac{1}{x} )}^{2} = (3 - \sqrt{8} )^{2} \\ \\ \frac{1}{ {x}^{2} } = ( {(3)}^{2} + {( \sqrt{8} )}^{2} - 2(3)( \sqrt{8} )) \\ \\ \frac{1}{ {x}^{2} } = 9 + 8 - 6 \sqrt{8} \\ \\ \frac{1}{ {x}^{2} } = 17 - 6 \sqrt{8}$

Now,

y = 3 - √8

$\frac{1}{y} = \frac{1}{3 - \sqrt{8} }$

On rationalizing the denominator we get,

$\frac{1}{y} = \frac{1}{3 - \sqrt{8} } \times \frac{3 + \sqrt{8} }{3 + \sqrt{8} } \\ \\ \frac{1}{y} = \frac{3 + \sqrt{8} }{(3 + \sqrt{8} )(3 - \sqrt{8}) } \\ \\ \frac{1}{y} = \frac{3 + \sqrt{8} }{ {(3)}^{2} - {( \sqrt{8} )}^{2} } \\ \\ \frac{1}{y} = \frac{3 + \sqrt{8} }{9 - 8} \\ \\ \frac{1}{y} = 3 + \sqrt{8}$

On squaring both the sides we get,

${( \frac{1}{y} )}^{2} = {(3 + \sqrt{8} )}^{2} \\ \\ \frac{1}{ {y}^{2} } = ( {(3)}^{2} + {( \sqrt{8}) }^{2} + 2(3)( \sqrt{8} ) \\ \\ \frac{1}{ {y}^{2} } = 9 + 8 + 6 \sqrt{8} \\ \\ \frac{1}{ {y}^{2} } = 17 + 6 \sqrt{8}$

So,
Putting the values in

$\frac{1}{ {x}^{2} } + \frac{1}{ {y}^{2} } = (17 - 6 \sqrt{8} ) + (17 + 6 \sqrt{8} ) \\ \\ \frac{1}{ {x}^{2} } + \frac{1}{ {y}^{2} } = 17 - 6 \sqrt{8} + 17 + 6 \sqrt{8} \\ \\ \frac{1}{ {x}^{2} } + \frac{1}{ {y}^{?} } = 17 + 17 \\ \\ \frac{1}{ {x}^{2} } + \frac{1}{ {y}^{2} } = 34$

Hope this helps ☺