Quadratic Equations with Complex Solutions

When the discriminant is negative

The quadratic formula solves $ax^2 + bx + c = 0$:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

The discriminant $b^2 - 4ac$ decides what kind of solutions appear.

Discriminant	Solutions
positive	two real
zero	one real, repeated
negative	two complex, a conjugate pair

In Algebra 1 a negative discriminant meant “no real solutions” and the problem stopped. With complex numbers, it keeps going.

Solve $x^2 + 2x + 5 = 0$.

$$b^2 - 4ac = 4 - 20 = -16$$ $$x = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i$$

Check $-1 + 2i$ by substituting.

$$(-1 + 2i)^2 + 2(-1 + 2i) + 5 = (1 - 4i + 4i^2) + (-2 + 4i) + 5$$ $$= (1 - 4i - 4) - 2 + 4i + 5 = 0 \quad\checkmark$$

What complex solutions mean on the graph

• y = x^2 + 2x + 5

The parabola’s lowest point is $(-1, 4)$, above the axis, so it has no $x$-intercepts. The solutions $-1 \pm 2i$ exist all the same. They are not real numbers, so they have no place on the $x$-axis.

Notice that the real part, $-1$, is the $x$-coordinate of the vertex. The real part of a conjugate pair is the axis of symmetry, $-\tfrac{b}{2a}$, for the same reason as in the real case: the two solutions are $-\tfrac{b}{2a}$ plus and minus the same amount.

Why complex solutions come in conjugate pairs

Write the formula as

$$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$

When the discriminant is negative, the second term is purely imaginary. The first term is real. So the two solutions share the real part $-\tfrac{b}{2a}$ and have opposite imaginary parts:

$$p + qi \quad\text{and}\quad p - qi$$

The $\pm$ in the formula is what makes them conjugates. A quadratic with real coefficients can never have a single non-real solution on its own.

Factoring over the complex numbers

Polynomial identities keep working when complex numbers are allowed. $x^2 + 4$ has no real factors, but the difference of squares applies once $4$ is written as $-(2i)^2$:

$$x^2 + 4 = x^2 - (2i)^2 = (x + 2i)(x - 2i)$$

Check: $(x + 2i)(x - 2i) = x^2 - 4i^2 = x^2 + 4$ ✓. The two roots, $\pm 2i$, are again a conjugate pair.

The Fundamental Theorem of Algebra

Fundamental Theorem of Algebra. Every polynomial of degree $n \ge 1$ has exactly $n$ roots in the complex numbers, counted with multiplicity.

For quadratics it can be seen directly. The formula always produces two values, $-\tfrac{b}{2a} \pm \tfrac{\sqrt{D}}{2a}$, and some complex number squares to $D$ whatever its sign. When $D = 0$, the two coincide as a repeated root, which counts twice.

$D$	Roots	Count with multiplicity
$D > 0$	two different real roots	$2$
$D = 0$	one real root, repeated	$2$
$D < 0$	two complex conjugates	$2$

Every row has two. Without complex numbers the count would drop to zero in the last row — which is exactly the gap $i$ was invented to close. The lesson on the Fundamental Theorem of Algebra carries the count to every degree.

Worked examples

Common mistakes

Practice problems