Question

if
${x}^{2} + \frac{1}{ {x}^{2} } = 62$
then find the value of
$x + \frac{1}{x}$
​

Answer 1

Answer:

$\large\bold\red{\pm8}$

Step-by-step explanation:

Given,

${x}^{2} + \frac{1}{ {x}^{2} } = 62 \\ \\ = > {(x)}^{2} + {( \frac{1}{x} )}^{2} + 2 \times x \times \frac{1}{x} = 62 + 2 \\ \\ = > {(x + \frac{1}{x} )}^{2} = 64 \\ \\ = > {(x + \frac{1}{x} )}^{2} = {( \pm8)}^{2} \\ \\ = > \large\bold{ x + \frac{1}{x} = \pm8}$

Answer 2

${\large\bf{\mid{\overline{\underline{Given:-}}}\mid}}$

• ${x}^{2} + \frac{1}{ {x}^{2} } = 62$

$\large\boxed{\underline{\mathcal{\red{Q}\green{u}\pink{e}\orange{s}\blue{ti}\red{on.??}}}}$

• $x + \frac{1}{x}$

$\Large\bold\star\underline{\underline\textbf{Formula\:used}}$

• (a²+b²+2ab) = (a+b)²

$\Large\underline{\underline{\sf{Solution}:}}$

${x}^{2} + \frac{1}{ {x}^{2} } = 62 \\ \\ \textbf{adding \: 2 \: both \: sides \: } \\ \\ \red\leadsto \: {x}^{2} + \frac{1}{ {x}^{2} } \: + 2 = 62 + 2 \\ \\ \red\leadsto \: {x}^{2} + \frac{1}{ {x}^{2} } \: + 2 \times x \times \frac{1}{x} = 64 \\ \\ \red\leadsto \: (x + \frac{1}{x} )^{2} = 64 \\ \\ \red\leadsto \: (x + \frac{1}{x} ) = \sqrt{64} \\ \\ \red\leadsto \: \red{\large\boxed{\bold{x + \frac{1}{x} = + 8 \: \: \: or \: - 8}}}$

$\large\underline\textbf{Hope it Helps You.}$