Binomial Equations

What are binomial equations

1.Binomial equations are a particular class of algebraic equations that involve only two terms. Their general form is the following:

$$ax^n + b = 0$$

2.In this expression $a$ and $b$ are constants, while $x$ is the unknown raised to the power $n$, where $n$ is a positive integer.

3.The strategy used to solve a binomial equation depends on the value of $n$. When $n$ is odd, the equation has a unique real solution. When $n$ is even, the existence of real solutions depends on the sign of the radicand: the equation has no real solution when solving it requires an even root of a negative number. In more advanced settings the solutions may also be complex, a case treated in the study of the roots of complex numbers.

Equations of degree one or two

4.When $n = 1$, the equation reduces to a linear equation of the form $ax + b = 0$, solved by isolating the variable:

$$ax + b = 0 \quad \rightarrow \quad x = -\frac{b}{a}$$

5.When $n = 2$, the equation reduces to a quadratic equation of the form $ax^2 + b = 0$. This can be solved with the quadratic formula or, more directly, by extracting the square root of $-\frac{b}{a}$:

$$x = \pm\sqrt{-\frac{b}{a}}$$

6.

This expression yields real solutions only when $-\frac{b}{a} \geq 0$. Otherwise the equation has complex solutions.

Equations with degree greater than two

7.When $n$ is greater than $2$, the equation can be solved by extracting the $n$-th root of $-\frac{b}{a}$, distinguishing two cases according to whether $n$ is even or odd.

8.When $n$ is even and $-\frac{b}{a}$ is positive, the equation has two distinct solutions, which reduce to a single solution when $-\frac{b}{a}$ equals $0$:

$$x = \pm\sqrt[n]{-\frac{b}{a}}$$

9.When $n$ is even and $-\frac{b}{a}$ is negative, the equation has no solution in the field of real numbers.

10.When $n$ is odd, the equation always has a single real solution:

$$x = \sqrt[n]{-\frac{b}{a}}$$

11.The $n$-th root of a negative number exists in the reals only when the index is odd. The extraction follows the same method used for positive numbers, and the result of an odd root of a negative number is itself negative. When the index is even, no real result exists.

Example 1

12.Consider the binomial equation $x^3 - 27 = 0$. Isolating the power and extracting the cube root gives:

$$ \begin{align} x^3 &= 27 \\[6pt] x &= \sqrt[3]{27} \\[6pt] &= 3 \end{align} $$

13.Here $n$ is odd and the number under the root is positive, so the equation admits a single real solution:

$$x = 3$$

Example 2

14.Consider the binomial equation $x^3 + 8 = 0$. Isolating the power and extracting the cube root gives:

$$ \begin{align} x^3 &= -8 \\[6pt] x &= \sqrt[3]{-8} \\[6pt] &= -2 \end{align} $$

15.Here $n$ is odd and the number under the root is negative, so the equation admits a single real solution:

$$x = -2$$

Example 3

16.Consider the binomial equation $x^4 - 16 = 0$. Isolating the power and extracting the fourth root gives:

$$ \begin{align} x^4 &= 16 \\[6pt] x &= \pm\sqrt[4]{16} \\[6pt] &= \pm 2 \end{align} $$

17.Here $n$ is even and the number under the root is positive, so the equation admits two distinct real solutions, symmetric with respect to the origin:

$$x = -2 \quad x = 2$$

Example 4

18.Consider the binomial equation $x^4 + 5 = 0$. Isolating the power gives:

$$ \begin{align} x^4 &= -5 \\[6pt] x &= \sqrt[4]{-5} \end{align} $$

19.Here $n$ is even and the number under the root is negative, so the equation admits no real solution. The set of real solutions is therefore empty:

$$S = \varnothing$$

20.These examples illustrate the most direct cases. In practice, more complex polynomial equations can sometimes be reduced to binomial equations, which are then resolved by the straightforward methods shown here.