Sign of Quadratic Function Between Roots

Theorem

Let $a \in \R_{>0}$ be a (strictly) positive real number.

Let $\alpha$ and $\beta$, where $\alpha < \beta$, be the roots of the quadratic function:

$\map Q x = a x^2 + b x + c$

whose discriminant $b^2 - 4 a c$ is (strictly) positive.

Then:

$\begin {cases} \map Q x < 0 & : \text {when $\alpha < x < \beta$} \\ \map Q x > 0 & : \text {when $x < \alpha$ or $x > \beta$} \end {cases}$

Proof

Because $b^2 - 4 a c > 0$, we have from Solution to Quadratic Equation with Real Coefficients that the roots of $\map Q x$ are real and unequal.

This demonstrates the existence of $\alpha$ and $\beta$, where by hypothesis we state that $\alpha < \beta$.

We can express $\map Q x$ as:

$\map Q x = a \paren {x - \alpha} \paren {x - \beta}$

When $\alpha < x < \beta$ we have that:

$x - \alpha > 0$

$x - \beta < 0$

and so:

$\map Q x = a \paren {x - \alpha} \paren {x - \beta} < 0$

$\Box$

When $x < \alpha$ we have that:

$x - \alpha < 0$

$x - \beta < 0$

and so:

$\map Q x = a \paren {x - \alpha} \paren {x - \beta} > 0$

$\Box$

When $x > \beta$ we have that:

$x - \alpha > 0$

$x - \beta > 0$

and so:

$\map Q x = a \paren {x - \alpha} \paren {x - \beta} > 0$

$\Box$

Hence the result.

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: Exercise $\S 1.12 \ (4)$