Q. If $\alpha,\beta$ are the roots of the equation $ax^{2} + bx + c = 0$, then the equation whose roots are $2\alpha + 3\beta$ and $3\alpha + 2\beta$, is
A) $acx^{2} + (a + c)bx - (a + c)^{2} = 0$
B) $acx^{2} - (a + c)bx + (a + c)^{2} = 0$
C) $abx^{2} - (a + b)cx + (a + b)^{2} = 0$
D) None of the above
Explanation
(d)
Here, $\alpha + \beta = - \frac{b}{a},\ \ \ \alpha\beta = \frac{c}{a}$
The required equation is
$x^{2} - 5x(\alpha + \beta) + (2\alpha + 3\beta)(3\alpha + 2\beta) = 0$
$\Rightarrow \ \ \ \ \ x^{2} + \frac{5b}{a}x\left\lbrack 6(\alpha + \beta)^{2} \right\rbrack + \lbrack 3\alpha\beta\rbrack = 0$
$\Rightarrow \ \ x^{2} + \frac{5b}{a}x + \left\lbrack 6\frac{b^{2}}{a^{2}} + \frac{c}{a} \right\rbrack = 0$
$\Rightarrow \ \ a^{2}x^{2} + 5abx + 6b^{2} + ac = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.