Question

If
${a}^{2} + {b}^{2} + \frac{1} {a}^{2} + \frac{1} {b}^{2} = 4$
then find the value of
$\sqrt{ {a}^{2} + {b}^{2} }$

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Answer 1

${a}^{2} + {b}^{2} + \frac{1}{ {a}^{2} } + \frac{1}{ {b}^{2} } = 4$

$\frac{ {a}^{4} {b}^{2} + {b}^{4} {a}^{2} + {a}^{2} + {b}^{2} }{ {a}^{2} {b}^{2} } = 4$

$4 {a}^{2} {b}^{2} = {a}^{2} {b}^{2} ( {a}^{2} + {b}^{2} + 1 + 1)$

${a}^{2} + {b}^{2} + 1 + 1 = 4$

or

${a}^{2} + {b}^{2} = 2$

Therefore

$\sqrt{ {a}^{2} + {b}^{2} } = \sqrt{2} = 1.414$