Question : If $x^{4}+\frac{1}{x^{4}}=16$, then what is the value of $x^{2}+\frac{1}{x^{2}}$?
Option 1: $3 \sqrt{2}$
Option 2: $2 \sqrt{2}$
Option 3: $5 \sqrt{2 }$
Option 4: $4 \sqrt{2}$
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Correct Answer: $3 \sqrt{2}$
Solution :
Given, $x^{4}+\frac{1}{x^{4}}=16$
Adding 2 on both sides,
$x^{4}+\frac{1}{x^{4}}+2=16+2$
⇒ $x^{4}+\frac{1}{x^{4}}+2(x^2× \frac{1}{x^2})=18$
⇒ $(x^{2}+\frac{1}{x^{2}})^2=18$
⇒ $x^{2}+\frac{1}{x^{2}}=3\sqrt{2}$
Hence, the correct answer is $3\sqrt{2}$.
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