Quadratic Equations and the Fundamental Theorem of Algebra

With complex numbers, every quadratic equation has a solution. When the discriminant $b^{2} - 4 a c$ 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 $ℂ$ (counting multiplicity).

Quick Reference

Discriminant	Nature of solutions	Formula
$b^{2} - 4 a c > 0$	Two distinct real roots	$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$b^{2} - 4 a c = 0$	One repeated real root	$x = \frac{- b}{2 a}$
$b^{2} - 4 a c < 0$	Two complex conjugate roots	$x = \frac{- b}{2 a} \pm i \frac{\sqrt{4 a c - b^{2}}}{2 a}$

The Quadratic Formula for Complex Solutions

Consider the general quadratic equation:

a x^{2} + b x + c = 0, a \neq 0.

Derivation via completing the square (click to expand)

Divide by $a$ :

x^{2} + \frac{b}{a} x = - \frac{c}{a}

Add ${(\frac{b}{2 a})}^{2}$ to both sides to complete the square:

x^{2} + \frac{b}{a} x + {(\frac{b}{2 a})}^{2} = {(\frac{b}{2 a})}^{2} - \frac{c}{a}

The left side is a perfect square:

{(x + \frac{b}{2 a})}^{2} = \frac{b^{2}}{4 a^{2}} - \frac{c}{a} = \frac{b^{2} - 4 a c}{4 a^{2}}

Taking the square root of both sides:

x + \frac{b}{2 a} = \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a}

Solving for $x$ gives the quadratic formula:

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

When $b^{2} - 4 a c \geq 0$ , this produces real solutions. But when $b^{2} - 4 a c < 0$ , write $b^{2} - 4 a c = - (4 a c - b^{2})$ where $4 a c - b^{2} > 0$ . Then:

\sqrt{b^{2} - 4 a c} = \sqrt{- (4 a c - b^{2})} = i \sqrt{4 a c - b^{2}}

The two solutions are:

x_{1} = \frac{- b}{2 a} + i \frac{\sqrt{4 a c - b^{2}}}{2 a}, x_{2} = \frac{- b}{2 a} - i \frac{\sqrt{4 a c - b^{2}}}{2 a}

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} - 4 a c = 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, x_{2} = - \frac{1}{2} - \frac{\sqrt{3}}{2} i

Note that $x_{2} = {\bar{x}}_{1}$ .

Example 2. Solve $2 x^{2} - 4 x + 5 = 0$ .

Solution. Here $a = 2$ , $b = - 4$ , $c = 5$ . The discriminant is:

b^{2} - 4 a c = 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 $x_{1} = 1 + \frac{\sqrt{6}}{2} i$ and $x_{2} = 1 - \frac{\sqrt{6}}{2} i$ .

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} + \dots + a_{1} x + a_{0} = 0

where $a_{n}, \dots, a_{0}$ are real or complex numbers and $n \geq 1$ , has at least one solution in $ℂ$ . Equivalently, every such polynomial has exactly $n$ roots in $ℂ$ , 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 $ℂ$ as:

a_{n} (x - r_{1}) (x - r_{2}) \dots (x - r_{n})

where $r_{1}, r_{2}, \dots, r_{n} \in ℂ$ are the $n$ roots (not necessarily distinct).

Complex roots of real polynomials come in conjugate pairs. If the polynomial has real coefficients and $z = a + b i$ (with $b \neq 0$ ) is a root, then $\bar{z} = a - b i$ 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, - 2, 3 + i, 3 - i

Frequently Asked Questions

What does the discriminant tell us about complex solutions?

For a quadratic $a x^{2} + b x + c = 0$ with real coefficients:

$b^{2} - 4 a c > 0$ : two distinct real solutions
$b^{2} - 4 a c = 0$ : one repeated real solution
$b^{2} - 4 a c < 0$ : two complex conjugate solutions (no real solutions)

A negative discriminant is not a problem once we work in

ℂ

Why do complex solutions always come in conjugate pairs for real-coefficient polynomials?

p (x)

has real coefficients and

z = a + b i

is a root, then

p (z) = 0

. Taking the complex conjugate of both sides and using the fact that conjugation distributes over addition and multiplication (and that the conjugate of any real coefficient is itself), one obtains

p (\bar{z}) = \overset{―}{p (z)} = \bar{0} = 0

. So

\bar{z}

is also a root.

Does every polynomial have exactly

n

distinct roots?

No. A polynomial of degree

n

has exactly

n

roots counting multiplicity. Some roots may be repeated. For example,

x^{2} - 2 x + 1 = (x - 1)^{2}

has the root

x = 1

with multiplicity 2, but only one distinct root.

Can the Fundamental Theorem of Algebra be applied to equations with complex coefficients?

Yes. The theorem applies to any polynomial with complex coefficients, not just real ones. The only difference is that when the coefficients are complex, there is no guarantee that complex roots come in conjugate pairs.

Quadratic Equations and the Fundamental Theorem of Algebra

Quadratic Equations and the Fundamental Theorem of Algebra

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The Quadratic Formula for Complex Solutions

The Fundamental Theorem of Algebra

Frequently Asked Questions

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