Let $f(x)$ be a quadratic with nonreal roots. Show that if the coefficient of $x^2$ in $f(x)$ is positive, then $f(x)>0$ for all $x$

Question

I do not understand the solution the book gives:

"Completing the square gives $f(x)=a(x-h)^2+k$ for some constants $h$ and $k$. The roots of $f(x)$ are the solutions to the equation $a(x-h)^2+k=0$. Rearranging this equation gives $(x-h)^2=-k/a$. We are given that $a>0$ and that the roots of $f(x)$ are nonreal. So, we must have $k>0$, as well. Since $k>0$ and $a(x-h)^2\geq0$ for all $x$, we have $f(x)=a(x-h)^2+k\gt0$ for all $x$."

Can someone please explain this solution in more detail, especially the deductions made?

Answer 1

First complete the square. The roots $r_0, r_1$ are those complex numbers such that $a(r_0 - h)^2 + k = 0$ and $a(r_1 - h)^2 + k = 0$. Rearranging, we get $(r - h)^2 = -k/a$, which implies $r = h \pm \sqrt{-k/a}$.

We know the coefficient of $x^2$ is positive, i.e. $a$ is positive. Now notice that we are given that $r$ is non-real, hence we must have $-k/a$ being negative. Hence, $k$ must be positive.

Now, we have $a > 0$, and we know $(x-h)^2 \geq 0$ since it is the square of a real number (we assume $x$ is real), and $k > 0$. Hence, $a(x-h)^2 + k$ is greater than zero.

Answer 2

We can do it slightly differently using the old form: $f(x) = ax^2+bx +c$ with $a > 0$. The condition that $f$ has non-real roots implies $b^2-4ac < 0$. By completing square, we have: $f(x) = a\left(x+\dfrac{b}{2a}\right)^2+ \dfrac{4ac-b^2}{4a}\geq \dfrac{4ac-b^2}{4a} > 0$.