# The Sum and Product of Roots Theorem

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

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

and

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

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

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

For the sum we have,

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

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

Now, for the product, we have

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

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

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

In the next section, we prove that if

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

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

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

(1) $latex ar_1 + ar_2 = -b$ and

(2) $latex ar_1r_2 = c$.

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

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

$latex (-b – ar_2)(r_2) = c$.

which is equivalent to

$latex -br_2 – ar^2_2 = c$.

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

$latex – ar^2_2 – br_2 – c = 0$.

Multiplying both sides by $latex – 1$, we have

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

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

Proving that $latex r_1$ 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).