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: $12 \sqrt{10}$
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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