Question

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

Answer 1

Given : $P\left(x\right)=a{x}^2+bx+c$

$Q\left(x\right)=-a{x}^2++dx+c$

Also $ac\ne 0$ then $P\left(x\right)Q\left(x\right)=0$

Discriminant of $P\left(x\right)=b^2-4ac$

and discriminant of $Q\left(x\right)=d^2+4ac$

And it is given that $ac\ne 0$

$\Rightarrow$ ac can be negative or positive

If ac is -ve

Discriminant of $P\left(x\right)=(+)ve$

and Discriminant of Q(x) can be -ve or +ve

if ac is +ve

Discriminant of P(x) can be +ve or -ve

Discriminant of Q(x) is +ve

T or both cases P(x) Q(x)=0 has at least two real roots.