Question : If $\left(y^2+\frac{1}{y^2}\right)=74$ and $y>1$, then find the value of $\left(y-\frac{1}{y}\right)$.
Option 1: $6 \sqrt{2}$
Option 2: $-2 \sqrt{19}$
Option 3: $2 \sqrt{19}$
Option 4: $-6 \sqrt{2}$
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Correct Answer: $6 \sqrt{2}$
Solution :
$y^2+\frac{1}{y^2}=74$
Subtracting 2 from both sides,
$⇒y^2+\frac{1}{y^2}-2=74-2$
⇒ $(y-\frac{1}{y})^2=72$
⇒ $y-\frac{1}{y}=\pm\sqrt{72}$
As $y>1$,
$\therefore y-\frac{1}{y}=\sqrt{72}=\sqrt{36×2}=6\sqrt2$
Hence, the correct answer is $6\sqrt{2}$.
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