# Sum and Product of the Roots of a Quadratic Equation

How to solve the sum and product of the roots of a quadratic equation problems: formulas, proofs, examples, and their solutions.

## Formula 1

If the roots of a quadratic equation are r1 and r2,
then the quadratic equation is x2 - (r1 + r2)x + r1r2 = 0.

So if the roots of a quadratic equation are given,
then you can directly write the quadratic equation.

## Proof: Formula 1

If x = r1 and x = r2,
then x - r1 = 0 and x - r2 = 0.

that satisfy these two conditions is
(x - r1)(x - r2) = 0.

Solving a quadratic equation by factoring

Use the FOIL method
and arrange the trinomial.

Then x2 - (r1 + r2)x + r1r2 = 0.

## Example 1

3 + 4 = 7
3⋅4 = 12

x2 - 7x + 12 = 0.

## Example 2

(2 + i) + (2 - i) = 4
(2 + i)(2 - i) = 5

x2 - 4x + 5 = 0.

## Formula 2

For the quadratic equation ax2 + bx + c = 0:

r1 + r2 = -b/a
r1r2 = c/a

So if the coefficients or the constant term are given,
you can find the sum or the product of the roots.

## Proof: Formula 2

Divide both sides by a.

Change the middle term +(b/a)x to -(-b/a)x.

Compare the terms of
this equation and the previous formula:
x2 - (b/a))x + c/a = 0
x2 - (r1 + r2)x + r1r2 = 0

Then r1 + r2 = -b/a
and r1r2 = c/a.

## Example 3

The coefficients of x2 and x are given:
1, -6.

And one of the roots is 2.

Then, if the other root is r,
2 + r = -6/1.

Then r = -8.

## Example 4

The coefficients of x2 and x are given:
2, -1.

And one of the roots is -3.

Then, if the other root is r,
-3 + r = -(-1)/2.

Then r = 7/2.

## Example 5

The coefficient of x2 and the constant term are given:
3, -12.

And one of the roots is 5.

Then, if the other root is r,
5r = (-12)/3.

Then r = -4/5.