Question

If p and q are 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. 0
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is D 5

$x^2-(\alpha -2)-\alpha -1=0$

Then sum of roots=$\alpha -2$

product of roots = $-\alpha -1$

$(p+q{)}^2=p^2+q^2+2pq$

Thus $(\alpha -2{)}^2=p^2+q^2+2(-\alpha -1)$

$p^2+q^2={\alpha }^2-4\alpha +4+2\alpha +2={\alpha }^{{}^2}-2\alpha +6={\alpha }^2-2\alpha +1+5=(\alpha -1{)}^2+5$

Thus minimum velue of $p^2+q^2$ is 5