Question

Let p and q be the roots of the quadratic equation $x^2$ - ($\alpha$ - 2) x - $\alpha$ - 1 = 0. What is the minimum possible value of $p^2+q^2$?

• A. 3
• B. 4
• C. 5

Answer 1

The correct option is C 5

Sum of the root = p + q $=\frac{\{\alpha -2\}}{1}$
$=\alpha -2$
Product of the roots = pq $=\frac{-(\alpha +1)}{1}=-(\alpha +1)$
Now, $p^2+q^2=(p+q{)}^2-2pq$
$=(\alpha -2{)}^2-2(-1\left)\right(\alpha +1)={\alpha }^2-4\alpha +4+2\alpha +2={\alpha }^2-2\alpha +6$
We have to find the minimum possible value of ${\alpha }^2-2\alpha +6$
$D=(-2{)}^2-4\times 1\times 6=-20$
and coefficient of ${\alpha }^2$ is +ve.
Rough diagram of ${\alpha }^2-2\alpha +6$ is

minimum value $=\frac{-D}{4a}=\frac{-(-20)}{\frac{4}{1}}=\frac{\left(20\right)}{4}=5$