Question

If p and $q(\ne 0)$ are the roots of the equation $x^2+px+q=0,$ then the least value of $x^2+px+q\left(xϵR\right)$ is:

• A. $-\frac{1}{4}$
• B. $\frac{1}{4}$
• C. $\frac{-9}{4}$
• D. $\frac{9}{4}$

Answer 1

The correct option is C $\frac{-9}{4}$

$y_{min}=\frac{-b^2+4ac}{4a}atx=-\frac{b}{2a}\therefore y_{min}=\frac{-p^2+4\times 1\times q}{4\times 1}=\frac{4q-p^2}{4}$
Now, since p and q are the roots of the equation $x^2+px+q=0.$
$\therefore p+q=-p\Rightarrow q=-2p$
and $pq=q\Rightarrow p=1$ $(\because q\ne 0)$
$q=-2$
$\therefore y_{min}=\frac{4q-p^2}{4}=\frac{4\times (-2)-(1{)}^2}{4}=-\frac{9}{4}.$