Q. The value of $a$ for which the sum of the squares of the roots of the equation $x^{2} - (a - 2)x - a - 1 = 0$ assumes the least value, is
A) 0
B) 1
C) 2
D) 3
Explanation
(b)
Let $\alpha\$and $\beta$ be the roots, then
$\alpha + \beta = (a - 2)$ and $\alpha\beta = - (a + 1)$
Now, $\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha\beta$
$= (a - 2)^{2} + 2(a + 1)$
$= (a - 1)^{2} + 5$
$\Rightarrow \ \alpha^{2} + \beta^{2} \geq 5$
Thus, the minimum value of $\alpha^{2} + \beta^{2}$ is 5 at $a = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.