Nonreal Roots of Quadratic Equations

Extending Beyond the Real Numbers

Start with the equation:

x^2+1=0

To satisfy this equation, $x^2$ would have to equal $-1$ . No real number has a negative square, so the equation has no real solution. We extend the number system with the imaginary unit $i$ , which satisfies the following definition:

i^2=-1

A complex number has the form

z=a+bi\qquad(a,b\in\mathbb R)

The number $2+5i$ is complex. It is not purely imaginary because its real part is not zero. A number such as $3i$ is purely imaginary.

A Negative Discriminant

For a quadratic equation $ax^2+bx+c=0$ with real coefficients, the discriminant is

D=b^2-4ac

If $D<0$ , the equation has no real roots. It has two distinct non-real complex roots, and those roots form a conjugate pair.

We still begin with the quadratic formula:

x=\frac{-b\pm\sqrt{D}}{2a}

Because $D<0$ , write $D=-|D|$ . Then

\sqrt D=\sqrt{-|D|}=i\sqrt{|D|}

Finding and Classifying Nonreal Roots

A negative discriminant determines the form of the two nonreal roots and shows that they form a conjugate pair.

Finding Two Nonreal Roots

Solve

x^2+4x+5=0

The coefficients are $a=1$ , $b=4$ , and $c=5$ . First calculate the discriminant:

D=4^2-4\cdot1\cdot5=16-20=-4

Since $D=-4<0$ , the roots are non-real. Substitute the values into the quadratic formula:

x=\frac{-4\pm\sqrt{-4}}{2\cdot1}

The calculation gives:

\begin{aligned} &x_1=-2+i \\ &\text{and} \\ &x_2=-2-i \end{aligned}

We can check one root directly:

(-2+i)^2+4(-2+i)+5

The other root follows by changing the sign of the imaginary part.

Classifying Roots Before Solving

Consider

2x^2+x+3=0

Here $a=2$ , $b=1$ , and $c=3$ . Its discriminant is

D=1^2-4\cdot2\cdot3=1-24=-23

The negative discriminant tells us that the equation has two non-real complex roots. Their exact values are

x=\frac{-1\pm\sqrt{-23}}{2\cdot2}

The calculation gives:

\begin{aligned} &x_1=\frac{-1+i\sqrt{23}}4 \\ &\text{and} \\ &x_2=\frac{-1-i\sqrt{23}}4 \end{aligned}

What the Real Graph Shows

The first example can be rewritten as

f(x)=x^2+4x+5=(x+2)^2+1

For every real $x$ , the value of $(x+2)^2$ is nonnegative, so $f(x)\ge1$ . The graph stays above the $x$ -axis and has no real intercepts. The complex roots are not extra points on this real graph. They belong to the extended number system.

Why the Roots Come in Conjugate Pairs

For real coefficients, the quadratic formula contains

-b\pm i\sqrt{|D|}

The two choices of the sign change only the imaginary part. This produces $u+vi$ and $u-vi$ , called complex conjugates. If one is a root of a polynomial with real coefficients, its conjugate is also a root.

Practice Problem

Solve $2x^2+2x+5=0$ over the complex numbers.

Worked Solution

Start with the discriminant:

D=2^2-4\cdot2\cdot5=4-40=-36

Because $D<0$ , use $\sqrt{-36}=6i$ in the quadratic formula:

x=\frac{-2\pm\sqrt{-36}}{2\cdot2}

So the conjugate roots are

\begin{aligned} &x_1=-\frac12+\frac32i \\ &\text{and} \\ &x_2=-\frac12-\frac32i \end{aligned}

Their sum is $-1=-\frac ba$ and their product is $\frac52=\frac ca$ , which checks the result against Vieta's formulas.