# Quadratic Equation: Sum and Product of the Roots

How to find the quadratic equation from the sum and product of the roots (and vice versa): 2 formulas, 4 examples, and their solutions.

## From Roots to Quadratic Equation

### Formula

If the roots of a quadratic equation are
r1 and r2,

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

So, if you know the roots,
you can write the quadratic equation.

## Example 1

### Solution

The roots are 3 and 4.

x2 - (3 + 4)x + 3⋅4 = 0.

-(3 + 4) = -7
+3⋅4 = +12

So
x2 - 7x + 12 = 0

## Example 2

### Solution

If one root is 2 + i,
then the other root is,
the conjugate of 2 + i,
2 - i.

This is true because
the imaginary number part comes from
±√b2 - 4ac.
And this ± makes the conjugate roots.

The roots are 2 + i and 2 - i.

x2 - [(2 + i) + (2 - i)]x + (2 + i)(2 - i) = 0.

-[(2 + i) + (2 - i)]
= -[2 + 2]
= -4

+(2 + i)(2 - i)
= +22 + 12

Divide Complex Numbers: Formula

+22 + 12 = +4 + 1

+4 + 1 = +5

So
x2 - 4x + 5 = 0

## From Quadratic Equation to Sum and Product of the Roots

### Formula

ax2 + bx + c = 0
(a ≠ 0),

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

you can find
the sum and the product of the roots.

## Example 3

### Solution

It says
one root is 2.

Then set the other root r.

So the roots are
2 and r.

1x2 + 6x + c = 0.

The x2 term and the x term are known.

The roots are 2 and r.

Then
r + 2 = -6/1.

-6/1 = -6

Move +2 to the right side.

Then r = -8.

So the other root is -8.

## Example 4

### Solution

It says
one root is 5.

Then set the other root r.

So the roots are
5 and r.

3x2 + bx - 15 = 0.

The x2 term and the constant term are known.

The roots are 5 and r.

Then
5⋅r = -15/3.

-15/3 = -5

Divide both sides by 5

Then r = -1.

So the other root is -1.