Q. If $\alpha$ and $\beta$ are the solutions of the quadratic equation $ax^{2} + bx + c = 0$ such that $\beta = \alpha^{1/3}$, then
A) $(ac)^{1/3} + (ab)^{1/3} + c = 0$
B) $\left( a^{3}b \right)^{1/4} + \left( ab^{3} \right)^{1/4} + c = 0$
C) $\left( a^{3}c \right)^{1/4} + \left( ac^{3} \right)^{1/4} + b = 0$
D) $\left( a^{4}c \right)^{1/3} + \left( ac^{4} \right)^{1/3} + b = 0$
Explanation
(c)
Here, $\alpha + \beta = - \frac{b}{a},\ \ \alpha\beta = \frac{c}{a}$
But given that $\beta = \alpha^{1/3}$
$\therefore\ \alpha + \alpha^{1/3} = - \frac{b}{a}$ and $\alpha.\alpha^{1/3} = \frac{c}{a}$
$\Rightarrow \ \ \ \alpha^{4/3} = \frac{c}{a}\ \ \Rightarrow \ \ \ \ \ \alpha = \left( \frac{c}{a} \right)^{3/4}$
$\therefore\ \ \alpha + \alpha^{1/3} = - \frac{b}{a}$
$\Rightarrow \ \left( \frac{c}{a} \right)^{3/4} + \left( \frac{c}{a} \right)^{1/4} = - \frac{b}{a}$
$\Rightarrow \left( ac^{3} \right)^{1/4} + \left( a^{3}c \right)^{1/4} + b = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.