NIELIT 2019 Feb Scientist D - Section D: 15

Question

The roots of the equation $x^{2/3}+x^{1/3}-2=0$ are :

• A. $1, -8$
• B. $-1, -2$
• C. $\frac{2}{3}, \frac{1}{3}$
• D. $-2, -7$

Answer 1

(A) $1, -8$

$x^{\frac{2}{3}} + x^{\frac{1}{3}} – 2 = 0\\  \> \\  \text{put }x=1 \\  \implies 1 + 1-2 = 0 \\ \> \\ \text{put } x= -8 \\ \implies (-8)^{\frac{2}{3}} + (-8)^{\frac{1}{3}} – 2 = 4-2 -2 = 0$

Comments

Another method, putting $x^{\frac{1}{3}} = y$, so equation will become $y^{2} +y-2 =0$

On solving, we will get values as y = -2 , 1 

So, x = $y^{3}$ = $-2^{3}$, $1^{3}$ = -8, 1