Question

if
${x}^{2} + \frac{1}{ {x}^{2} } = 18$
find the value of
$x - \frac{1}{x}$
​

Answer 1

$\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ANSWER}}}$

Given,

${x}^{2} + \frac{1}{ {x}^{2} } = 18$

we know that,

${x}^{2} + \frac{1}{ {x}^{2} } = {(x - \frac{1}{ x} )}^{2} + 2 \\ \\ \implies \: 18 = {(x - \frac{1}{x} )}^{2} + 2 \\ \\ \implies \: {(x - \frac{1}{x} )}^{2} = 18 - 2 = 16 \\ \\ \implies \: x - \frac{1}{x} = \sqrt{16} = \:\pm\:4$