System of two quadratics equation, $P(x)$ and $Q(x)$ If $P(x) = ax^2 + bx + c$ and $Q(x) = – ax^2 + dx + c$, $ac \ne 0$, then the equation $P(x) . Q(x) = 0$ has
(A) Exactly two real roots
(B) At least two real roots
(C) Exactly four real roots
(D) No real roots
My approach is as follow Let $T(x)=P(x).Q(x)$
$T\left( x \right) =  - a^2{x^4} + a\left( {d - b} \right){x^3} + \left( {bd} \right){x^2} + c\left( {d + b} \right)x + {c^2}$
Not able to approach from here

If the coefficient of $x^2$ and the constant term of a quadratic equation are of opposite sign $($both non-zero$)$ then discriminant of the quadratic equation is positive, which means roots are real.
So either $P(x)$ or $Q(x)$ has coefficient of $x^2$ and the constant term with opposite sign $($both are non-zero$)$. So one of them has $2$ real roots.
Therefore $P(x)Q(x)=0$ has at least $2$ real roots.
We can also solve using your approach
$T\left( x \right) =  - a^2{x^4} + a\left( {d - b} \right){x^3} + \left( {bd} \right){x^2} + c\left( {d + b} \right)x + {c^2}$
$T(0)>0$ and for sufficiently large $k$( $k$ is positive) we have $T(k)<0$ and $T(-k)<0$.
So there should be a real root in  the interval $(-k,0)$ and $(0,k)$ so at least $2$ real roots exist.
