Question

If
${a}^{2} + {b}^{2} + \frac{1}{ {a}^{2} } + \frac{1}{ {b}^{2} } = 4$
Then find the value of
$\sqrt{ {a}^{2} + {b}^{2} }$
Any irrelevant answers WILL BE REPORTED​

Answer 1

Answer:

Step-by-step explanation:

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→   $a^2+b^2+\frac{1}{a^2}+\frac{1}{b^2}$

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

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

→   $ab(a+b)+a^2+b^2=4$

→   $a^2+b^2=4-ab(a+b)$

→   $\sqrt{a^2+b^2}=2-\sqrt{ab(a+b)}$

       ∴$hence\:proved$

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