If P(x) = ax^2 + bx + c and Q(x) = -ax^2 + dx | vedclass

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(B) The equation $P(x) \cdot Q(x) = 0$ implies that either $P(x) = 0$ or $Q(x) = 0$. Let $D_1$ be the discriminant of $P(x) = ax^2 + bx + c$, so $D_1

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at least two real roots

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(B) The equation $P(x) \cdot Q(x) = 0$ implies that either $P(x) = 0$ or $Q(x) = 0$. Let $D_1$ be the discriminant of $P(x) = ax^2 + bx + c$, so $D_1 = b^2 - 4ac$. Let $D_2$ be the discriminant of $Q(x) = -ax^2 + dx + c$, so $D_2 = d^2 - 4(-a)(c) = d^2 + 4ac$. Adding the two discriminants, we get $D_1 + D_2 = (b^2 - 4ac) + (d^2 + 4ac) = b^2 + d^2$. Since $b^2 + d^2 \geq 0$, at least one of $D_1$ or $D_2$ must be non-negative. If $D_1 \geq 0$, $P(x) = 0$ has at least two real roots (or one repeated root). If $D_2 \geq 0$, $Q(x) = 0$ has at least two real roots. Since $ac 
eq 0$, the quadratic terms exist. Thus, the equation $P(x) \cdot Q(x) = 0$ has at least two real roots.

If $P(x) = ax^2 + bx + c$ and $Q(x) = -ax^2 + dx + c$ where $ac \neq 0$, then $P(x) \cdot Q(x) = 0$ has $(a, b, c, d \in \mathbb{R})$:

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