Proof for showing $1 + \frac{c}{a} + \frac{b}{a} > 0$ for any quadratic equation,where its roots are non real.

Question

The question is as follows :

If the equation $ax^2 + bx + c$ has non real roots, prove that $1 + c/a + b/a > 0 $.

Looking at the question,the first thing that came to my mind was to use the relationship that the discriminant is less than 0 for non - real roots. But that wasn't of any help. How do I proceed?

Answer 1

If $a>0$, we have a parabola above the $x$-axis, hence $f(1)=a+b+c>0$.

On the other hand, $a<0$ implies $f(1)<0$. The given inequality follows.

Answer 2

For the case a>0:
wlog assume $a=1$. We can write the polynomial equation not touching x-axis as $(x-x_0)^2+\alpha^2$ for some $\alpha>0$. Expansion gives $x^2-2x_0x+x_0^2+\alpha^2$. Therefore $$a+b+c=1-2x_0+x_0^2+\alpha^2=(1-x_0)^2+\alpha^2 > 0$$

a<0 case can be done similarly.