Question

The real roots of the equation $x^{\frac{2}{3}}+x^{\frac{1}{3}}-2=0$ are

• A. $1,8$
• B. $-1,-8$
• C. $-1,8$
• D. $1,-8$

Answer 1

Answer: (D)

$1,-8$

The given equation is $x^{\frac{2}{3}}+x^{\frac{1}{3}}-2=0$
Put $x^{\frac{1}{3}}=y,$ then equation reduces to $y^2+y-2=0$
$\Rightarrow (y-1)(y+2)=0$
$\Rightarrow y=1$ or $y=-2$
$\Rightarrow x^{\frac{1}{3}}=1$ or $x^{\frac{1}{3}}=-2$
$\therefore x=(1{)}^3$ or $x=(-2{)}^3=-8$
Hence, the real roots of the given equations are $1,-8.$