Quadratic equations

Description of how to calculate quadratic equations with complex numbers

Quadratic Equations and Complex Numbers


This article describes an important application of complex numbers, the solution of quadratic equations. An example of the solvability of a quadratic equation has already been described in the article Complex Numbers, namely \(x^2 = -1\).

Here we describe, with complex numbers all quadratic equations are solvable.

We come back to the equation \(x^2= -1\). This quadratic equation is not solvable in the domain of real numbers; but in the world of complex numbers.

The equation \(z^2= -1\) has the complex solution \(z = i\). The unknown \(z\) instead of \(x\) indicates that a complex solution is allowed.


Obviously, the following solution is also correct

\((-i)^2=(-1)^2·i^2=1·(-1)=-1\)

So \(-i\) is a solution of the equation \(z^2 = -1\).

The quadratic equation \(z^2 = -1\) has two solutions, namely the two conjugate complex numbers \(z_1 = i\) and \(z_2 = -i\).



As next example, we solve the equation \(z^2 = -2\). We write

\(z^2=2·(-1)\)

The two conjugate complex solutions are

\(z_1=\sqrt{2i}\)   and   \(z_2=-\sqrt{2i}\)


The equation \(z^2=-2=2·(-1)\) is equivalent to the equation

\(\displaystyle\frac{z^2}{2}=\left(\frac{z}{\sqrt{2}}\right)^2=-1\) with the two solutions   \(\displaystyle\frac{z_1}{\sqrt{2}}=i\)   und   \(\displaystyle\frac{z_2}{\sqrt{2}}=-1\)
That is \(z_1=\sqrt{2i}\)   and   \(z_2=-\sqrt{2i}\).


Completing the quadratic equations


The quadratic equations discussed in the previous section, in which \(z\) occurred only as \(z^2\), were easily solved.In this section we will treat a quadratic equation of the form   \(az^2+bz^2 +c = 0\)   with   \(a,b,c\)   real and   \(a\not =0\).

As an example we calculate the following quadratic equation

\(z^2-6z+13=0\)

We try to apply the solution method from the previous section to this equation. To write the left side of the equation as a square, we use the method of quadratic addition. In our example, this looks like this.

\(0=z^2-6z+13=z^2-2·3z+3^2-3^2+13=(z-3)^2-9+13=(z-3)^2+4)\)

The equation   \(z^2-6z+13=0\)   is equivalent to the equation \((z-3)^2+4=0\). The reshaped equation looks no less complicated than before, but it can be solved like the equations discussed in the previous section

The equation \((z-3)^2+4=0\) is equivalent to \((z-3)^2 = -4=4·(-1)\)

If we take the root of it we get \(z-3=2i\) or \(z-3 = -2i\)

The solutions are \(z_1 = 3+2i\) and \(z_2=3-2i\)


General solution formula for quadratic equations


We again use the general quadratic equation of the form

\(az^2+bz+c=0\)

A division by \(a\) results

\(\displaystyle z^2+\frac{b}{a}·z+\frac{c}{a}=0\)

The summand

\(\displaystyle\frac{b}{a}·z=2·\frac{b}{2a}·z\)

is taken as the product of a binomial formula and the quadratic addition

\(\displaystyle +\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2\)


This results in

\(\displaystyle 0=z^2+\frac{b}{a}·z+\frac{c}{a}=z^2+2·\frac{b}{2a}·z+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2+\frac{c}{a}=\left(z+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}{a}\)
\(\displaystyle = \left(z+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{4ac}{4a·a}=\left(z+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\)

The equation \(az^2+bz+c=0\) is thus equivalent to the equation

\(\displaystyle \left(z+\frac{b}{2a}\right)^2 = \frac{b^2-4ac}{(2a)^2}\)

Whether the number on the right is positive or negative depends on whether it is \(b^2-4ac\) positive or negative. The denominator \((2a)^2\) is always positive. The term \(D=b^2-4ac\) is the discriminant of the quadratic equation. It therefore depends on the sign of the discriminant whether the quadratic equation has a real or not a real solution.

To solve the equation, we set \(D\) for \(b^2-4ac\) and write

\(\displaystyle \left(z+\frac{b}{2a}\right)^2=\frac{D}{(2a)^2}\)

For the equation we get the two solutions

    \(\displaystyle z_1+\frac{b}{2a}= \frac{ω}{2a}\)   and   \(\displaystyle z_2+\frac{b}{2a}= \frac{-ω}{2a}\)

also   \(\displaystyle z_1+\frac{-b}{2a}+ \frac{ω}{2a}\)   and   \(\displaystyle z_2+\frac{-b}{2a}- \frac{ω}{2a}\)

We write the short solution

\(\displaystyle z_{1,2}\frac{-b±ω}{2a}\)


Summary


The solution of the quadratic equation \(az^2+bz+c=0\) with \(a,b,c\) real and \(a ≠0\) is

\(\displaystyle z_{1,2}\frac{-b±ω}{2a}\)

The complex number ω is \(ω^2=D=b^2-4ac\)

Depending on the sign of the discriminant \(D\), the solutions are as follows

Ist \(D>0\), then are \(z_1\) und \(z_2\) real.

Ist \(D=0\), then are \(z_1=z_2\) real.

Ist \(D < 0\), then are \(z_1\) and \(z_2\) conjugated complex to each other.