Quadratic Equations and the Fundamental Theorem of Algebra
With complex numbers, every quadratic equation has a solution. When the discriminant $b^2 - 4ac$ is negative, the quadratic formula produces two complex solutions that are conjugates of each other. The Fundamental Theorem of Algebra generalizes this: every polynomial equation of degree $n \geq 1$ has exactly $n$ solutions in $\mathbb{C}$ (counting multiplicity).
Quick Reference
| Discriminant | Nature of solutions | Formula |
|---|---|---|
| $b^2 - 4ac > 0$ | Two distinct real roots | $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ |
| $b^2 - 4ac = 0$ | One repeated real root | $x = \dfrac{-b}{2a}$ |
| $b^2 - 4ac < 0$ | Two complex conjugate roots | $x = \dfrac{-b}{2a} \pm i\dfrac{\sqrt{4ac-b^2}}{2a}$ |
The Quadratic Formula for Complex Solutions
Consider the general quadratic equation:
$ ax^2 + bx + c = 0, \quad a \neq 0. $Derivation via completing the square (click to expand)
Divide by $a$:
$x^2 + \frac{b}{a}x = -\frac{c}{a}$Add $\left(\dfrac{b}{2a}\right)^2$ to both sides to complete the square:
$x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}$The left side is a perfect square:
$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2 - 4ac}{4a^2}$Taking the square root of both sides:
$x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}$Solving for $x$ gives the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$When $b^2 - 4ac \geq 0$, this produces real solutions. But when $b^2 - 4ac < 0$, write $b^2 - 4ac = -(4ac - b^2)$ where $4ac - b^2 > 0$. Then:
$ \sqrt{b^2 - 4ac} = \sqrt{-(4ac - b^2)} = i\sqrt{4ac - b^2} $The two solutions are:
$ x_1 = \frac{-b}{2a} + i\frac{\sqrt{4ac - b^2}}{2a}, \qquad x_2 = \frac{-b}{2a} - i\frac{\sqrt{4ac - b^2}}{2a} $Notice that $x_2 = \bar{x}_1$: the two complex solutions are always complex conjugates of each other when the coefficients $a$, $b$, $c$ are real.
Example 1. Solve $x^2 + x + 1 = 0$.
Solution. Here $a = 1$, $b = 1$, $c = 1$. The discriminant is:
$b^2 - 4ac = 1 - 4 = -3 < 0$Applying the quadratic formula:
$x = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$The two solutions are:
$x_1 = -\frac{1}{2} + \frac{\sqrt{3}}{2}\,i, \qquad x_2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}\,i$Note that $x_2 = \bar{x}_1$.
Example 2. Solve $2x^2 - 4x + 5 = 0$.
Solution. Here $a = 2$, $b = -4$, $c = 5$. The discriminant is:
$b^2 - 4ac = 16 - 40 = -24 < 0$Applying the quadratic formula:
$ \begin{aligned} x &= \frac{4 \pm \sqrt{-24}}{4} = \frac{4 \pm 2i\sqrt{6}}{4} = 1 \pm \frac{\sqrt{6}}{2}\,i \end{aligned} $The two solutions are $x_1 = 1 + \dfrac{\sqrt{6}}{2}\,i$ and $x_2 = 1 - \dfrac{\sqrt{6}}{2}\,i$.
The Fundamental Theorem of Algebra
Fundamental Theorem of Algebra (Gauss, 1799). Every polynomial equation of the form
$a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0$where $a_n, \ldots, a_0$ are real or complex numbers and $n \geq 1$, has at least one solution in $\mathbb{C}$. Equivalently, every such polynomial has exactly $n$ roots in $\mathbb{C}$, counting multiplicity.
This theorem has profound consequences:
The complex numbers are algebraically closed. This means no further extension of the number system is needed to solve polynomial equations. Unlike the real numbers, where $x^2 + 1 = 0$ has no solution, the complex numbers contain solutions to every polynomial equation.
Degree equals number of roots. A polynomial of degree $n$ factors completely over $\mathbb{C}$ as:
$ a_n(x - r_1)(x - r_2) \cdots (x - r_n) $where $r_1, r_2, \ldots, r_n \in \mathbb{C}$ are the $n$ roots (not necessarily distinct).
Complex roots of real polynomials come in conjugate pairs. If the polynomial has real coefficients and $z = a + bi$ (with $b \neq 0$) is a root, then $\bar{z} = a - bi$ is also a root.
Example 3. A degree-4 polynomial with real coefficients has roots $1$, $-2$, and $3 + i$. What are all four roots?
Solution. Since the polynomial has real coefficients, complex roots come in conjugate pairs. The root $3 + i$ forces the conjugate $3 - i$ to also be a root. The four roots are:
$1, \quad -2, \quad 3 + i, \quad 3 - i$Frequently Asked Questions
What does the discriminant tell us about complex solutions?
For a quadratic $ax^2 + bx + c = 0$ with real coefficients:
- $b^2 - 4ac > 0$: two distinct real solutions
- $b^2 - 4ac = 0$: one repeated real solution
- $b^2 - 4ac < 0$: two complex conjugate solutions (no real solutions)
