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 $-1$.
Related Questions
Know More about
Staff Selection Commission Sub Inspector ...
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Get Updates Brochure
Your Staff Selection Commission Sub Inspector Exam brochure has been successfully mailed to your registered email id “”.
