Question

If $f(x) = ax^2+bx+c$ and $g(x) = -ax^2+bx+c$, then $f(x) \cdot g(x) = 0$ has

• A. Two real roots
• B. at least two real roots
• C. four real roots
• D. at most two real roots

Answer 1

Answer: (B)

at least two real roots

Answer 2

The equation $f(x) \cdot g(x) = 0$ implies $f(x) = 0$ or $g(x) = 0$. The discriminant of $f(x) = ax^2+bx+c=0$ is $\Delta_1 = b^2-4ac$. The discriminant of $g(x) = -ax^2+bx+c=0$ is $\Delta_2 = b^2-4(-a)(c) = b^2+4ac$. For real roots, we need $\Delta_1 \ge 0$ or $\Delta_2 \ge 0$. If $ac > 0$, then $b^2+4ac > b^2 \ge 0$, so $\Delta_2 > 0$. If $ac < 0$, then $b^2-4ac > b^2 \ge 0$, so $\Delta_1 > 0$. If $ac = 0$, then either $a=0$ or $c=0$. If $a=0$, $f(x)=bx+c$ and $g(x)=bx+c$, so $f(x)g(x)=(bx+c)^2=0$ has one root if $b \ne 0$, or no roots if $b=0, c \ne 0$. If $c=0$, $f(x)=ax^2+bx$ and $g(x)=-ax^2+bx$. $f(x)g(x) = (ax^2+bx)(-ax^2+bx) = b^2x^2 - a^2x^4 = x^2(b^2-a^2x^2)=0$. This has roots $x=0$ (twice) and $x = \pm b/a$ (if $a \ne 0$). So at least two real roots. In all cases, at least one of the discriminants is non-negative, guaranteeing at least two real roots.