If $P(x) = ax^2 + bx + c$ and $Q(x) = -ax^2 + dx + c$, then prove that $P(x) \cdot Q(x) = 0$ has at least two real roots? How should i solve the same? I assumed the roots be $ \alpha, \beta $ for $ P(x) $ and $ \gamma, \delta $ for $ Q(x) $. Product of roots turn out to be of the opposite signs, being $$ \alpha \cdot \beta = \dfrac{c}{a} \\ \gamma \cdot \delta = - \dfrac{c}{a} $$
Thus if I multiply the equation, I should get 4 solutions to the equation.
So I equated the quadratic formula giving solutions for $ P(x) $ with $ Q(x) $ and cancelled out the common term $ 2a $
Now if I multiply the 4 terms left over together, I should get $ c^2 $ shouldn't I? Since I am multiplying the roots and I already cancelled out $ 2a $ when I multiplied it?
I do not understand how to proceed, or if any of my assumptions above are correct. Kindly comment on the same and possibly guide me through it?

EDIT It seems that this answer may have been posted before and is outlined in a comment to the selected answer.
Another way to approach this problem is to consider each of the ways in which we may get zero or one real root for the product $P(x)\cdot Q(x)$ and show that they are not possible.
$P(x)\cdot Q(x)$ will have zero real roots if both $b^2-4ac<0$ and $d^2+4ac<0$. Equivalently, there will be zero real roots if $b^2<4ac$ and $d^2<-4ac$. One of $4ac$ and $-4ac$ must be negative and both $b^2$ and $d^2$ are at least zero, so both statements cannot be simultaneously true.
There will be one real root if either of the following holds.
\begin{align}
b^2-4ac<0&\textrm{ and }d^2+4ac=0\\
b^2-4ac=0&\text{ and }d^2+4ac<0
\end{align}
In the first case, $d^2=-4ac$ and that implies that $b^2+d^2<0$, which is not possible. In the second case, $b^2=4ac$ and that implies that $d^2+b^2<0$, which is not possible.
If $P(x)\cdot Q(x)$ may not have zero or one real root, it must have at least two.
