Question

What values of x satisfy $x^{\frac{2}{3}}+x^{\frac{1}{3}}-2\le 0$?

• A. $-8\le x\le 1$
• B. $-1\le x\le 8$
• C. $1<x<8$
• D. $1\ge x\ge 8$

Answer 1

The correct option is A $-8\le x\le 1$

$x^{\frac{2}{3}}+x^{\frac{1}{3}}-2\le 0$
$Put{x}^{\frac{1}{3}}=yorx=y^3\Rightarrow y^2+y-2\le 0Or{y}^2+2y-y-2\le 0Or(y-1)(y+2)\le 0\Rightarrow -2\le y\le 1Or(-2{)}^3\le x\le 1^{3}or-8\le x\le 1$