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).

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