Question

if X square + one upon x square equal to 47 find the value of x plus one upon x
$x ^{2} + \frac{1}{x {}^{2} } = 47 find x + \frac{1}{x}$

Answer 1

$\underline{ \large{\text{Answer}}} : \\ \\ \text{Given that,} \\ \\ \it{{x}^{2} + \frac{1}{ {x}^{2}} = 47 } \\ \\ \implies \it{ {(x + \frac{1}{x} )}^{2} - 2(x)( \frac{1}{x}) = 47 } \\ \\ \implies \it{ {(x + \frac{1}{x} )}^{2} - 2 = 47} \\ \\ \implies \it{ {(x + \frac{1}{x}) }^{2} = 47 + 2 = 49 } \\ \\ \implies \it{{ {(x + \frac{1}{x}) }^{2} = {7}^{2} }} \\ \\ \therefore \boxed{ \it{x + \frac{1}{x} = \pm 7 }} \\ \\ \bigstar \: \underline{ \large{ \text{MarkAsBrainliest}}} \: \bigstar$