Imaginary roots of a quadratic equation

Imaginary roots of a quadratic equationWe know that solving a quadratic equation ax2 + bx + c = 0, where a, b and c, with a ≠ 0, is:

x1 = [- b + √ (b2 - 4ac)] / 2a

x2 = [- b - √ (b2 - 4ac)] / 2a

Furthermore, we know that if the discriminant "b2 - 4ac ≥ 0 ”the solution is made up of real numbers. But when "b2 - 4ac <0 ", there is no solution in real numbers, but two solutions that include imaginary numbers and that satisfy the given equation.

Example 1: Solve the quadratic equation 2x2 - 2x + 5 = 0.

The coefficients are:

a = 2; b = - 2 and c = 5

We substitute the coefficients in the quadratic formula:

x = [- b ± √ (b2 - 4ac)] / 2a

x = {- (- 2) ± √ [(-2)2 - 4 (2) (5)]} / 2 (2)

We solve

x = {2 ± √ [(-2)2 - 4 (2) (5)]} / 2 (2)

x = [2 ± √ - 36] / 4

Remembering that i = √ -1, we have:

x = [2 ± (√ 36 · √ -1)] / 4 = (2 ± 6i) / 4

x1 = (2 + 6i) / 4

x2 = (2 - 6i) / 4

Example 1: Solve the quadratic equation x2 - 6x + 25 = 0.

The coefficients are:

a = 1; b = - 6 and c = 25

We substitute the coefficients in the quadratic formula:

x = [- b ± √ (b2 - 4ac)] / 2a

x = {- (- 6) ± √ [(-6)2 - 4 (1) (25)]} / 2 (1)

We solve

x = [6 ± √ - 64] / 2

Remembering that i = √ -1, we have:

x = [6 ± (√ 64 · √ -1)] / 2 = (6 ± 8i) / 2

x1 = (6 + 8i) / 2 = 3 + 4i

x2 = (6 - 8i) / 2 = 3 - 4i