Question

If $P\left(x\right)={ax}^2+bx+c$ and $Q\left(x\right)=-{ax}^2+bx+c$ , where $ac\ne 0$ , then show that $P\left(x\right)Q\left(x\right)$=0 has at least two real roots.

Answer 1

$P\left(x\right).Q\left(x\right)=(a{x}^2+bx+c)(-a{x}^2+bx+c)$
Now, $D_1=b^2-4ac$ and $D_2=b^2+4ac$
Clearly, $D_1+D_2=2b^2\ge 0$
Therefore at least one of $D_1$ and $D_2$ is (+ve).
Hence, at least two real roots.
Hence statement is true