Question

If $(x+\frac{1}{x})=4$, find the value of $(x^4+\frac{1}{x^4})$

• A. $120$
• B. $194$
• C. $160$
• D. $142$

Answer 1

The correct option is B $194$

$x+\frac{1}{x}=4$
Squaring both sides,
$=(x+\frac{1}{x}{)}^2=4^2$
$=x^2+(\frac{1}{x}{)}^2+2(x\left)\right(\frac{1}{x})=16$
$=x^2+\frac{1}{x^2}+2=16$
$=x^2+\frac{1}{x^2}=14$
Again Squaring both sides,we get,
$=(x^2+\frac{1}{x^2}{)}^2={14}^2$
$=(x^2{)}^2+(\frac{1}{x^2}{)}^2+2\left(x^2\right)\left(\frac{1}{x^2}\right)=196$
$=x^4+\frac{1}{x^4}=196-2$
$=x^4+\frac{1}{x^4}=194$