Question

If
$x { }^{2} + 1 \div {x}^{2} = 51$
then evaluate the value of
$x + 1 \div x$
​

Answer 1

${x}^{2} + \frac{1}{ {x}^{2} } = 51 \\ \ ( x + \frac{1}{x} )^{2} + 2(x) \frac{1}{ x} = 51 \\( x + \frac{x}{1} ) ^{2} = 51 - 2 \\ ( x + \frac{x}{1} ) ^{2} = 49 \\ ( x + \frac{x}{1} ) = \sqrt{49} \\ ( x + \frac{x}{1} ) = \frac{ + }{ - } 7$

Answer 2

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Given,

${x}^{2} + \frac{1}{ {x}^{2} } = 51 \\ \\ x + \frac{1}{x} = \: to \: find \: \\ \\ now \\ \: {(x + \frac{1}{x}) }^{2} = {x}^{2} + 2 \times \cancel x \times \frac{1}{ \cancel x} + \frac{1}{ {x}^{2} } \\ \implies \: {(x + \frac{1}{x} )}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2 \\ \implies \: {(x + \frac{1}{x} )}^{2} =51 + 2 = 53 \\ \implies \: (x + \frac{1}{x} ) = \pm \sqrt{53}$