# The Sum and Product of Roots Theorem

From the quadratic formula, we know that the numbers $r_1$ and $r_2$ are the roots of the quadratic equation $ax^2 + bx +c =0$ where $a \neq 0$ if and only if

$r_1 + r_2 = -\frac{b}{a}$

and

$r_1r_2 = \frac{c}{a}$

In this proof, we are going to show that they are indeed the root of the quadratic equation. We call this the Sum and Product of Roots Theorem.

The equations above come from the fact that the roots of the quadratic equation are

$r_1 = \displaystyle \frac{-b + \sqrt{b^2 -4ac}}{2a}$ and $r_2 \displaystyle \frac{- b - \sqrt{b^2 - 4ac}}{2a}$.

For the sum we have,

$r_1 + r_2 = \displaystyle \frac{-b + \sqrt{b^2 -4ac}}{2a} + \displaystyle \frac{- b - \sqrt{b^2 - 4ac}}{2a} = \displaystyle \frac{-b + \sqrt{b^2 -4ac}}{2a} - \displaystyle \frac{ b - \sqrt{b^2 - 4ac}}{2a}$

$= \displaystyle \frac{-2b}{2a}$
$= - \displaystyle \frac{b}{a}$.

Now, for the product, we have

$\displaystyle r_1r_2 = \left (\frac{-b + \sqrt{b^2 -4ac}}{2a} \right ) \left ( \frac{ b - \sqrt{b^2 - 4ac}}{2a} \right ) = \frac{b^2 - (b^2 - 4ac)}{4a^2}$

$= \displaystyle \frac{4ac}{4a^2}$

$= \displaystyle \frac{c}{a}$.

In the next section, we prove that if

$r_1 + r_2 = -\frac {b}{a}$ and $r_1r_2 = \frac{c}{a}$

then, $r_1$ and $r_2$ are the roots of $ax^2 + bx + c = 0$.

To do this, we multiply both sides of the preceding equations by $a$ giving us

(1) $ar_1 + ar_2 = -b$ and

(2) $ar_1r_2 = c$.

From (1) , $ar_1 + ar_2 = -b$ implies that $ar_1 = -b - ar_2$.

Substituting the value of $ar_1$ in equation (2), we have

$(-b - ar_2)(r_2) = c$.

which is equivalent to

$-br_2 - ar^2_2 = c$.

Subtracting $c$ from both sides and rearranging the terms, we have

$- ar^2_2 - br_2 - c = 0$.

Multiplying both sides by $- 1$, we have

$ar^2_2 + br_2 + c = 0$

This shows that $r_2$ is a root of $ax^2 + bx + c = 0$.

Proving that $r_1$ is also a root is left as an exercise.

Reference

Senk, Sharon L., et al. “University of Chicago School Mathematics Project Advanced Algebra.” Glenview, IL: Scott, Foresman and Company (1990).