Question

If $P(x)=a x^{2}+b x+c$ and $Q(x)=-a x^{2}+b x+c$, where $a c \neq 0$, then $P(x) Q(x)$ has atleast two real roots.

Answer 1

$P(x) \cdot Q(x)=\left(a x^{2}+b x+c\right)\left(-a x^{2}+b x+c\right)$
Now, $ D_{1}=b^{2}-4 a c$ and $D_{2}=b^{2}+4 a c$
Clearly, $ D_{1}+D_{2}=2 b^{2} \geq 0$
$\therefore$ Atleast one of $D_{1}$ and $D_{2}$ is (+ ve). Hence, atleast two real roots.
Hence, statement is true.