0 votes 0 votes The roots of the equation $x^{2/3}+x^{1/3}-2=0$ are : $1, -8$ $-1, -2$ $\frac{2}{3}, \frac{1}{3}$ $-2, -7$ Quantitative Aptitude nielit2019feb-scientistd quantitative-aptitude polynomials + – Lakshman Bhaiya asked Apr 3, 2020 • recategorized Nov 8, 2020 by Krithiga2101 Lakshman Bhaiya 13.7k points 709 views answer comment Share See all 0 reply Please log in or register to add a comment.
1 votes 1 votes (A) $1, -8$ $x^{\frac{2}{3}} + x^{\frac{1}{3}} – 2 = 0\\ \> \\ \text{put }x=1 \\ \implies 1 + 1-2 = 0 \\ \> \\ \text{put } x= -8 \\ \implies (-8)^{\frac{2}{3}} + (-8)^{\frac{1}{3}} – 2 = 4-2 -2 = 0$ Nikhil_dhama answered May 25, 2021 Nikhil_dhama 906 points comment Share See 1 comment See all 1 1 comment reply neethu_seb 3.4k points commented Dec 11, 2022 reply Share 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 0 votes 0 votes Please log in or register to add a comment.