Binomial equations

A binomial equation is an algebraic equation.

Definition

The general form of a binomial equation is

$$a{x}^n+b=0,$$

where a, b are any real or complex numbers. We also assume that a, b ≠ 0 and that n is a positive integer.

Special cases of a binomial equation:

• n = 1: the equation has the form ax + b = 0 and is a linear equation.
• n = 2: the equation has the form ax² + b = 0 and is a quadratic equation.

We can rewrite any binomial equation so that its leading coefficient is 1 by dividing the equation by a. The equation then has the form:

$$x^n+\frac{b}{a}=0$$

We can also make a substitution and write c in place of the fraction b/a. So if c = b/a, the equation has the form:

$$x^n+c=0$$

How to solve a binomial equation

First we make the leading coefficient 1, then move c to the right-hand side to get the equation

$$x^n=-c$$

Next we take the nth root of both sides.

$$x=\sqrt[n]{-c}$$

We then solve this using the formula for roots of complex numbers, which looks like this:

$$\sqrt[n]{z}=\sqrt[n]{|z|}[\cos (\frac{\alpha }{n}+\frac{2k\pi }{n})+i\sin (\frac{\alpha }{n}+\frac{2k\pi }{n})]$$

where

$$z=|z|(\cos \alpha +i\sin \alpha )$$

and the whole formula holds for k = 0, 1, …, n − 1.

Example: find the roots of the equation x³ − 4 = 0. After making the leading coefficient 1 and taking the root, the equation has the form:

$$x=\sqrt[3]{4}$$

Because the number −c = 4 has no imaginary part, the angle α is zero and we can leave it out of the formula. The kth root of the equation has the form:

$$x_k=\sqrt[3]{4}[\cos \left(\frac{2k\pi }{3}\right)+i\sin \left(\frac{2k\pi }{3}\right)]$$

Now we must apply this formula for k = 0, 1, 2. For k = 0 we get:

$$x_0=\sqrt[3]{4}(\cos 0+i\sin 0)=\sqrt[3]{4}$$

For k = 1:

$$\begin{array}{rcl}{x}_1& =& \sqrt[3]{4}(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3})\\ & =& \sqrt[3]{4}(-\frac{1}{2}+i\frac{\sqrt{3}}{2})\\ & =& \frac{1}{2}\sqrt[3]{4}(-1+i\sqrt{3})\end{array}$$

For k = 2:

$$\begin{array}{rcl}{x}_2& =& \sqrt[3]{4}(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3})\\ & =& \sqrt[3]{4}(-\frac{1}{2}-i\frac{\sqrt{3}}{2})\\ & =& -\frac{1}{2}\sqrt[3]{4}(1+i\sqrt{3})\end{array}$$

And that is everything: we have three roots, x₀, x₁ and x₂.