Question : If $\sqrt{x}+\frac{1}{\sqrt{x}}=3, x>0$, then $x^2\left(x^2-47\right)=?$
Option 1: $0$
Option 2: $2$
Option 3: $-2$
Option 4: $-1$
Correct Answer: $-1$
Solution : As we know, $(x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2$ ⇒ $(\sqrt{x} + \frac{1}{\sqrt{x}})^2 = x+\frac{1}{x} +2$ ⇒ $x + \frac{1}{x} + 2 = 3^2 = 9$ ⇒ $x + \frac{1}{x} = 9 - 2 = 7$ ⇒ $(x + \frac{1}{x})^2= 7^2$ ⇒ $x^2 + \frac{1}{x^2} + 2 = 49$ ⇒ $\frac{(x^4 + 1)}{x^2} = 49 - 2 = 47$ ⇒ $x^4 + 1 = 47x^2$ ⇒ $x^4 - 47x^2 = – 1$ ⇒ $x^2(x^2 - 47) = – 1$ Hence, the correct answer is $-1$.
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