One of the roots of the quadratic equation is positive and the other is negative

Question

One of the roots of the quadratic $ax^2 + bx + c = 0$ is positive and the other is negative. Tell me the sign of a,b,c so that this happens.

Answer 1

We need distinct real roots, so necessarily $\Delta=b^2-4ac>0$. Furthermore, the product of the roots, $\frac ca$ must be negative. As the sign of $\frac ca$ is the same as the sign of $ac$, we have $ac<0$.

We may note that $ac<0$ implies $\Delta>0$, so the first condition is redundant, and finally, the equation $ax^2+bx+c=0$ has a positive root and a negative root if and only if $ac<0$.

Answer 2

Hint:

Viete equalitites: the sum of the roots is $\;-\cfrac ba\;$ and their product is $\;\cfrac ca\;$ .

Answer 3

Let $a$ be positive. Then by Vieta $c$ must be negative; here the sign of $b$ doesn't matter. If conversely $c$ is negative in that case we know that there must be two different solution one of different signs; again the sign of $b$ doesn't matter. Similarly if $a$ is negative. Hence $ac$ must be negative and the sign of $b$ doesn't matter.

Answer 4

We need $\sqrt{b^2-4ac}\gt b$. This happens iff $b^2-4ac\gt b^2$. Which happens whenever $ac\lt0$. So $a$ and $c$ must have opposite signs.