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.
Quadratic 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.
Example2x2 - 3x + 5 = 0
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.
Discriminant
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.
ExampleRoots of x2 + 3x + 4 = 0
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.