Question

(4x^2+1/x^2)=21 find the value of x

Answer 1

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Answer 2

$4 {x}^{2} + \frac{1}{ {x}^{2} } = 21 \\ ( {2x})^{2} + \frac{1}{ {x}^{2} } = 21 \\ {(2x + \frac{1}{x})}^{2} - 4 = 21 \\ 2x + \frac{1}{x} = +5 \: or \: - 5$

similarly

${(2x - \frac{1}{x} )}^{2} + 4 = 21 \\ 2x - \frac{1}{x} = + \sqrt{17} \: or - \sqrt{17}$

adding both equations we get

4x =

$5 + \sqrt{17} \\ 5 - \sqrt{17} \\ - 5 + \sqrt{17} \\ - 5 - \sqrt{17}$

to get x divide these by 4