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?

Assume $a, b, c, d$ real. (I am taking the original form of the question.)
Then the discriminant of $a x^{2} + b x + c$ is $b^{2} - 4 a c$, while the discriminant of $-a x^{2} + d x + c$ is $d^{2} + 4 ac$.
Since the sum of the two discriminants is non-negative:
$$
(b^{2} - 4 a c) + (d^{2} + 4 ac) = b^{2} + d^{2} \ge 0,
$$
at least one of the two discriminants is non-negative, so at least one of the two equations has real solutions.
