Question

Given $x-\frac{1}{x}=3$, find the value of $x^4+\frac{1}{x^4}$

• A. 119
• B. 125
• C. 130
• D. 145

Answer 1

Answer:

A. 119

Given,

$x-\frac{1}{x}=3$

We need to find:

$x^4+\frac{1}{x^4}$

Step 1: Square the given equation

$\left(x-\frac{1}{x}\right)^2=3^2$
$x^2+\frac{1}{x^2}-2=9$

$x^2+\frac{1}{x^2}=11$

Step 2: Square again

$\left(x^2+\frac{1}{x^2}\right)^2$
$= x^4+\frac{1}{x^4}+2$

Substitute $(x^2+\frac{1}{x^2}=11):$

$11^2=x^4+\frac{1}{x^4}+2$
$121=x^4+\frac{1}{x^4}+2$
$x^4+\frac{1}{x^4}=119$

Answer:

$\boxed{119}$