Toolkit 31

Signs and Roots of Quadratics

f(x)=ax^2+bx+c,\qquad \Delta=b^2-4ac

Proof

Let

f(x)=ax^2+bx+c, \qquad a\ne 0,

and let

\Delta = b^2 - 4ac.

We consider three cases.

The quadratic has two distinct real roots

r_1 = \frac{-b-\sqrt{\Delta}}{2a}, \qquad r_2 = \frac{-b+\sqrt{\Delta}}{2a},

where $r_1 < r_2$ . Thus,

f(x) = a(x-r_1)(x-r_2).

From the sign chart above, we immediately obtain

\operatorname{sgn}(f(x)) = \begin{cases} \operatorname{sgn}(a), & x<r_1, \\ 0, & x=r_1, \\ -\operatorname{sgn}(a), & r_1<x<r_2, \\ 0, & x=r_2, \\ \operatorname{sgn}(a), & x>r_2. \end{cases}

The quadratic has one repeated real root

r = -\frac{b}{2a}.

Hence,

f(x) = a(x-r)^2.

From the sign chart above, we immediately obtain

\operatorname{sgn}(f(x)) = \operatorname{sgn}(a)

for $x \ne r$ , while

f(r) = 0.

Completing the square,

\begin{aligned} f(x) &= a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right) \\ &= a\left(\left(x+\frac{b}{2a}\right)^2 + \frac{4ac-b^2}{4a^2}\right). \end{aligned}

Since $\Delta = b^2-4ac < 0$ , we have $4ac-b^2 > 0$ . Therefore,

\left(x+\frac{b}{2a}\right)^2 + \frac{4ac-b^2}{4a^2} > 0

for every real $x$ . Thus,

\operatorname{sgn}(f(x)) = \operatorname{sgn}(a)

for every real $x$ . These three cases give the complete root and sign classification of a quadratic polynomial.

\square