# Quadratic expression with $D \leq 0$

If the equation

$$ax^2 +bx +c = 0$$

doesn't have $2$ distinct real roots and

$$a+c>b,$$

then how could I prove that

$$f(x) = ax^2 + bx +c \geq 0.$$

I tried to prove $a>0$, since discriminant is already less than or equal to zero therefore if I prove $a>0$ then $f(x)$ will satisfy the above condition. But I'm not able to get $a>0$.

• Your first expression isn’t an equation. – amd Sep 14 '18 at 5:24

The given condition $a+c>b$ means we are given that $f(-1)>0$. So if there exists $x_0 \in \mathbb{R}$ such that $f(x_0)<0$, then $f(x)=0$ will have two distinct real roots. This means $f(x) \geq 0$ for all $x$.