Q. The solution of the equation $2x^{3} - x^{2} - 22x - 24 = 0$ when two of the roots in the ration $3:4$, is
A) $3,\ 4,\ \frac{1}{2}$
B) $- \frac{3}{2},\ - 2,\ 4$
C) $- \frac{1}{2},\ \frac{3}{2},\ 2$
D) $\frac{3}{2},\ 2,\ \frac{5}{2}$
Explanation
(b)
Given equation is $2x^{3} - x^{2} - 22x - 24 = 0$
On putting $x = 2,\ - 2$ only $x = - 2$ satisfies this equation
So, $x = - 2$ is a root of this equation and from the given options only (b) has this root
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.