Quadratic equations with complex solutions

If you get a negative discriminant when solving a quadratic equation, the equation has no solution in the real numbers. However, the equation always has a solution in the complex numbers.

Motivation

Let’s try to solve the quadratic equation x² + 2x + 5 = 0. First we work out the discriminant: D = 4 − 20 = −16. As we already know, this equation has no solution in the real numbers. The advantage of complex numbers, though, is that they let us take the square root of a negative number too. Since i² = −1, we can rewrite our discriminant as −16 = 16i². Now we can take its square root and get 4i. We work out the solutions of the quadratic equation:

$$x_1=\frac{-2+\sqrt{16i^2}}{2}=\frac{-2+4i}{2}=\frac{2(-1+2i)}{2}=2i-1.$$

$$x_2=\frac{-2-\sqrt{16i^2}}{2}=\frac{-2-4i}{2}=\frac{2(-1-2i)}{2}=-2i-1.$$

Formulae and relationships

From the previous example we can derive a formula for the roots of a quadratic equation when the discriminant is negative:

$$x_{1,2}=\frac{-b\pm i\sqrt{|D|}}{2a}.$$

It follows that every quadratic equation has a solution. If the discriminant is not negative, the real numbers are enough; if it is negative, we must solve the equation in the complex numbers. In general, we can say that every quadratic equation has a solution in the complex numbers.

The formulae for the sum and product of roots also hold for these complex roots. First the derivation for the sum:

$$\begin{array}{rcl}{x}_1+x_2& =& \frac{-b+i\sqrt{|D|}}{2a}+\frac{-b-i\sqrt{|D|}}{2a}\\ & =& \frac{-b+i\sqrt{|D|}-b-i\sqrt{|D|}}{2a}\\ & =& \frac{-2b}{2a}=-\frac{b}{a}\end{array}$$

And now for the product. We use the fact that the discriminant is negative, and because we need to take its square root, we work with minus D, which gives a positive number. So instead of |D| we write −D; the result is the same.

$$\begin{array}{rcl}{x}_1\cdot x_2& =& \frac{-b+i\sqrt{-D}}{2a}\cdot \frac{-b-i\sqrt{-D}}{2a}\\ & =& \frac{b^2+b\cdot i\sqrt{-D}-b\cdot i\sqrt{-D}-(-D)i^2}{4a^2}\\ & =& \frac{b^2+D{i}^2}{4a^2}\\ & =& \frac{b^2-D}{4a^2}\\ & =& \frac{b^2-(b^2-4ac)}{4a^2}\\ & =& \frac{4ac}{4a^2}=\frac{c}{a}\end{array}$$

Example

Solve the quadratic equation 2x² + 6x + 9 = 0. We work out the discriminant: D = 36 − 72 = −36. It is negative, so we use the formula for complex roots and just substitute:

$$x_1=\frac{-b+i\sqrt{|D|}}{2a}=\frac{-6+i\sqrt{36}}{4}=\frac{-6+6i}{4}=\frac{3(i-1)}{2}=-\frac{3}{2}+\frac{3i}{2}$$

$$x_2=\frac{-b-i\sqrt{|D|}}{2a}=\dots =-\frac{3}{2}-\frac{3i}{2}$$