Question

if
$x + \frac{1}{x} = 7$
then
${x}^{2} + \frac{1}{ {x}^{2} } =$

Answer 1

Hey Friend ☺

x + 1/x = 7.

( x + 1/x )^2 = ( x )^2 + ( 1/x )^2 + 2 ( x )( 1/x )

》( x + 1/x )^2 = x^2 + 1/x^2 + 2

Since ( a + b )^2 = a^2 + b^2 + 2ab

》( 7 ) ^2 = x^2 + 1/x^2 + 2

》49 = x^2 + 1/x^2 + 2

》x^2 + 1/x^2 = 49 - 2

》x^2 + 1/x^2 = 47

Hope it helps you ..!!

✌

Answer 2

Hello Mate!

$if \: x + \frac{1}{x} = 7 \\ then \: {(x + \frac{1}{x}) }^{2} = {7}^{2} \\ {x}^{2} + {( \frac{1}{x} )}^{2} + 2 \times x \times \frac{1}{x} = 49 \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 49 \\ {x}^{2} + \frac{1}{ {x}^{2} } = 49 - 2 \\ {x}^{2} + \frac{1}{ {x}^{2} } = 47$

Hope it helps☺!✌