Question : If $(x+\frac{1}{x})=6$ and $x>1$, find the value of $(x^2–\frac{1}{x^2})$.
Option 1: $18 \sqrt{2}$
Option 2: $30 \sqrt{2}$
Option 3: $24 \sqrt{2}$
Option 4: $12 \sqrt{10}$
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Correct Answer: $24 \sqrt{2}$
Solution : Given: $(x+\frac{1}{x})=6$, and $x>1$. We know, $(x–\frac{1}{x})^2=(x+\frac{1}{x})^2–4×x×\frac{1}{x}$ ⇒ $(x–\frac{1}{x})^2=(6)^2–4$ ⇒ $(x–\frac{1}{x})=\sqrt{32}=4\sqrt2$ ⇒ $x^2–\frac{1}{x^2}=(x+\frac{1}{x})(x–\frac{1}{x})$ ⇒ $x^2–\frac{1}{x^2}=(6)(4\sqrt2)=24\sqrt2$ Hence, the correct answer is $24\sqrt2$.
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