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

The parabolas represented by the quadratic functions $ \ P(x) \ = \ ax^2 + bx + c \ $ and $ \ Q(x) \ = \ – ax^2 + dx + c \ \  , \ \ ac \ne 0 \ \ , $ "open" in opposite "vertical" directions and have a common $ \ y-$intercept.  The absolutely minimal number of $ \ x-$intercepts that could be arranged is one, by placing the vertices of both parabolas at the origin.  But these are $  \ y \ = \ ax^2 \ $ and $  \ y \ = \ -ax^2 \ \ , $  which are not permitted by the condition $ \ ac \ne 0 \ \ ; \ $ a "vertical shift" in either direction necessarily produces two $ \ x-$intercepts, and so two real zeroes for $ \ (ax^2 + c)·(-ax^2 + c) \ \ . $
Thus, the unavoidable minimum number of real zeroes for $ \ P(x)·Q(x) \ $ is two.  We can arrange for more than two by placing the two parabolas so that they are tangent at the origin at a point other than their vertices, with  $  \ y \ = \ ax^2 + bx \ $ and $  \ y \ = \ -ax^2 + bx \   \     ( d    =   b ) \ \ .  $  With their vertices then at $ \ x \ = \ \pm \frac{b}{2a} \ \ , $ there are three $ \ x-$intercepts, including one at the origin.  Again, we may not have $ \ c = 0 \ \ , $ but a "vertical shift" in either direction, that is kept small enough that the vertices remain on opposite sides of the $ \ x-$axis, then leads to four $ \ x-$intercepts (the one at the origin "bifurcating" into two).
We demonstrate geometrically then that we must have at least two zeroes for  $ \ P(x)·Q(x) \ $ and may easily obtain up to four [choice $ \ \mathbf{(B)}  $ ] .  If we generally "break" the symmetry of the vertices about the origin described in the previous paragraph, we can easily obtain more than two $  \ x-$intercepts for the parabolas.
We can also see this by putting the polynomials into "vertex form":
$$ a·\left(x \ + \ \frac{b}{2a} \right)^2 \ + \ \left(c \ - \ \frac{b^2}{4a} \right) \ \ , \ \ -a·\left(x \ - \ \frac{d}{2a} \right)^2 \ + \ \left(c \ + \ \frac{d^2}{4a} \right) $$
and finding their zeroes from
$$  \left(x \ + \ \frac{b}{2a} \right)^2 \ \ = \ \ -\frac{1}{a}·\left(c \ - \ \frac{b^2}{4a} \right) \ \ , \ \  \left(x \ - \ \frac{d}{2a} \right)^2 \ \ = \ \  \frac{1}{a}·\left(c \ + \ \frac{d^2}{4a} \right) \ \ . $$
Often, it will be the case that the right  side of one of these equations is positive while the right side of the other is negative, giving us only two real zeroes.  But it is also possible for both right sides to be positive when $ \ \frac{b^2}{4a^2} \ > \ \frac{c}{a} \ > \ -\frac{d^2}{4a^2} \ \ , $ producing four real zeroes.  [This is equivalent to the condition for the polynomial discriminants.]
