consider the quadratics, $ P(x) = ax^2 + bx +c$ ,and, $Q(x)= -ax^2 + dx+c$ , given that ac $\neq$ 0. What can we say about the roots of PQ? More particularly, does PQ have at least two real roots?
my work:
I don't really know where to start even, I suppose it has something to do with discriminant but what does one do after writing discriminant of each quadratic?
Taking complex numbers into consideration, P(x) and Q(x) both have two roots, and PQ, as a quartic polynomial, has four roots. But I think you are actually refering to Real roots rather than roots. We could easily find that any root of P(x) and Q(x) is the root of PQ, since PQ is the product of P(x) and Q(x). And we could also find that any root for PQ is a root for either P(x) or Q(x) or both. We could prove that by assuming this is not true, then there would be
$PQ=0=P(x)*Q(x)$
Where
$P(x)\ne0,Q(x)\ne0$
And that is impossible because the product of two non-zero numbers cannot be zero. Therefore we could deduce that the root for PQ, as a set, is the union of the roots of the P(x) and Q(x). So, as a result, We could find the roots for PQ by simply finding the roots for P(x) and Q(x), which is done by considering the signs of$\Delta_p=b^2-4ac, \Delta_q=d^2+4ac$. Noticing that $ac\ne0 ,b^2\ge0,d^2\ge0$, we could infer that there is at least one positive $\Delta$, since it is impossible for both $4ac$ and $-4ac$ to be negative. Thus, at least one polynomial has two real roots, implying that PQ has at least two real roots.
By Victor, Huang