Question

If $x+\frac{1}{x}=-2$, find the value of $(x^2+\frac{1}{x^2})\times$$(x^4+\frac{1}{x^4})$

• A. $2$
• B. $1$
• C. $4$
• D. $-1$

Answer 1

The correct option is C $4$

Given $x+\frac{1}{x}=-2$

squaring on both sides

$(x+\frac{1}{x}{)}^2=(-2{)}^2$

$⟹x^2+\frac{1}{x^2}+2=4$

$⟹x^2+\frac{1}{x^2}=2$ ------(1)

squaring on both sides

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

$⟹x^4+\frac{1}{x^4}+2=4$

$⟹x^4+\frac{1}{x^4}=2$ ------(2)

product of (1) and (2) is

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