Q. If the roots of the equation $ax^{2} + bx + c = 0$ of the form $\frac{k + 1}{k}\ and\ \frac{k + 2}{k + 1},\ then\ (a + b + c)^{2}$ is equal to
A) $2b^{2} - ac$
B) $\sum_{}^{}a^{2}$
C) $b^{2} - 4ac$
D) $b^{2} - 2ac$
Explanation
(c)
$\because\ \frac{k + 1}{k} + \frac{k + 2}{k + 1} = - \frac{b}{a}$ ...(i)
and $\frac{k + 1}{k}.\frac{k + 2}{k + 1} = \frac{c}{a}$
$\Rightarrow \ \frac{k + 2}{k} = \frac{c}{a}$
$\Rightarrow \frac{2}{k} = \frac{c}{a} - 1 = \frac{c - a}{a}$
$\Rightarrow k = \frac{2a}{c - a}$
On putting the value of $k$ in the Eq. (i), we get
$\frac{c + a}{2a} + \frac{2c}{c + a} = - \frac{b}{a}$
$\Rightarrow (c + a)^{2} + 4ac = - 2b(a + c)$
$\Rightarrow (a + b + c)^{2} = b^{2} - 4ac$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.