<s>[INST] 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
 [/INST] Set $PQ=\left( ax^2+bx+c\right) \left( -ax^2+dx+c\right)$ to zero, to get $$\left( ax^2+bx+c\right) \left( -ax^2+dx+c\right)=0.$$
Check the discriminant for both $P$ and $Q$.
For $P$, you get $D_P=b^2-4ac$. For $Q$, you get $D_Q=d^2+4ac$.
Suppose $P$ has no real roots, such that $D_P=b^2-4ac < 0 \implies4ac>b^2$. In that case, $D_Q=d^2+4ac>0$ and $Q$ thus has two roots, so your polynomial $PQ$ has at least two roots.
Next, suppose $P$ has 1 real root, in which case $4ac=b^2$ and so $D_1 = d^2+b^2$ which has 2 roots, so $PQ$ has 3.
Lastly, suppose $P$ has 2 real roots, and thus $b^2>4ac$, in which case $Q$ may have zero, one, or two real roots.
Therefore, $PQ$ has at least two real roots.
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