Suppose that $p$ and $q$ are constants such that the smallest possible value of $x^2+px+q$ is $0$. Express $q$ in terms of $p$.
I am unsure what it is asking. I feel it is asking something very simple. However I am unsure of what it is. Any solutions or walk-throughs would be appreciated.
Let $f(x)=x^2+px+q$. Then $f^\prime (x)=2x+p$ and when $f$ takes it's minimum value $f^\prime(x)=0$, i.e. $x=-p/2$. Thus $0=f(-p/2)=p^2/4-p^2/2+q$ or $q=p^2/4.$
Hint: Apply completing the square
$$x^2+px+q=\left(x+\frac{p}2 \right)^2-\frac{p^2}{4}+q$$
If the minimum value of the quadratic is $0$, then it has a double root and opens upwards, so can be factored as $(x-r)^2=0$. Expanding this and equating coefficients, we get $p=-2r$ and $q=r^2$, therefore $q=p^2/4$.
