Question

State true or false:

If $x\ne 0$ and $x+\frac{1}{x}=2$, then:
$x^2+\frac{1}{x^2}=x^3+\frac{1}{x^3}=x^4+\frac{1}{x^4}$

• A. True
• B. False

Answer 1

The correct option is A True

$x+\frac{1}{x}=2$
Squaring both sides we get,
$=>(x+\frac{1}{x}{)}^2=2^2$
$=>x^2+(\frac{1}{x}{)}^2+2(x\left)\right(\frac{1}{x})=4$
$=>x^2+\frac{1}{x^2}+2=4$
$=>x^2+\frac{1}{x^2}=4-2$
$=>x^2+\frac{1}{x^2}=2$
Now,
$=>x^3+\frac{1}{x^3}=(x+\frac{1}{x})\left[\right(x{)}^2+(\frac{1}{x}{)}^2-(x\left)\right(\frac{1}{x}\left)\right]$
$=\left(2\right)(2-1)$
$=2$
Thus, $x^2+\frac{1}{x^2}$=$x^3+\frac{1}{x^3}$
Now,
$x^2+\frac{1}{x^2}=2$
Squaring both sides we get,
$=>(x^2{)}^2+(\frac{1}{x^2}{)}^2+2\left(x^2\right)\left(\frac{1}{x^2}\right)=4$
$=>x^4+\frac{1}{x^4}+2=4$
$=>x^4+\frac{1}{x^4}=4-2$
$=>x^4+\frac{1}{x^4}=2$
Thus,
$x^2+\frac{1}{x^2}=x^3+\frac{1}{x^3}=x^4+\frac{1}{x^4}$