Quadratic Equation: Complex Roots

How to find the complex roots of a quadratic equation and see if the quadratic equation has two complex roots: 2 formulas, 2 examples, and their solutions.

Formula

Formula

Recall that
for a quadratic equation
ax2 + bx + c = 0
(a ≠ 0),

x = [-b ± √b2 - 4ac] / 2a.

This is the quadratic formula.

Let's use this
to find the complex roots.

Example 1

Example

Solution

The given quadratic equation is
2x2 - 3x + 5 = 0.

a = 2
b = -3
c = +5

Then, by the quadratic formula,
x = [+3 ± √(-3)2 - 4⋅2⋅5] / [2⋅2].

+3 = 3
(-3)2 = 9
-4⋅2⋅5 = -40

2⋅2 = 4

9 - 40 = -31

-31 = √31i

Imaginary Number i

So x = [3 ± √31i]/4.

Formula

Formula

For a quadratic equation
ax2 + bx + c = 0
(a ≠ 0),

x = [-b ± √b2 - 4ac] / 2a.

The discriminant, D, is the number
inside the radical sign:
b2 - 4ac.

Nature of the Roots

Recall that
the discriminant D determines
the nature of the roots.

If you know what a complex number is,
you can describe the last case differently.

If D is minus (< 0),
then the quadratic equation has
two complex roots.

Two complex roots are not real roots.
So the previous word,
no real roots,
is true.

Example 2

Example

Solution

The given quadratic equation is
1x2 + 3x + 4 = 0.

a = 1
b = +3
c = +4

Then D = 32 - 4⋅1⋅4.

32 = 9
-4⋅1⋅4 = -16

9 - 16 = -7

D = -7

D is minus.

Then the quadratic equation has
two complex roots.

So
two complex roots
is the answer.