Question

The function $x^2+px+q$ with $p$ and $q$ greater than the zero has its minimum value when:

• A. $x=-p$
• B. $x=\frac{-p}{2}$
• C. $x=-2p$
• D. $x=\frac{p^2}{4q}$

Answer 1

Answer: (B)

$x=\frac{-p}{2}$

$Letf\left(x\right)=x^2+px+qp,q>0$

${ax}^2+bx+c$ has its minimum value at $x=-\frac{b}{2a}$

Here, a=1, b=p, c=q.

$\therefore f\left(x\right)$ has its minimum value at $x=-\frac{p}{2\left(1\right)}=-\frac{p}{2}$