Binomial equations are a specific type of algebraic equation of the form:

\[a^n x + b = 0\]

  • \(a\) and \(b\) are constants.
  • \(x\) is a variable raised to a power of \(n\), where \(n\) is a positive integer.

Different strategies exist to solve these equations based on the value of \(n\) and the following factors.


Equations with a degree less than two

If the value of \(n\) equals \(1\), the given equation reduces to a simple linear equation in the form of \(ax + b = 0\) that can be solved by isolating the variable \(x\) and finding its corresponding value.

\[ ax+b= 0 \quad \implies \quad x = \frac{-b}{a} \]

If the value of \(n\) equals \(2\), the given equation reduces to a quadratic equation in the form of \(ax^2 + b = 0\) that can be solved using the quadratic formula or in a more straightforward calculating the square root of the term \(\large{\frac{-b}{a}}\):

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


Equations with a degree greater the than two

If \(n\) is greater than 2, we are dealing with a relatively simple case of an equation with a degree higher than two. Generally, such equations can be solved by calculating the nth root of the value \(\large{\frac{-b}{a}}\) while considering two two distinct cases, depending on whether \(n\) os even or \(n\) odd:

When \(n\) is even and \(\large{\frac{-b}{a}}\) is positive, we have two distinct solutions, or a single solution if \(\large{\frac{-b}{a}}\) equals \(0\), in the form:

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

When \(n\) is even and \(\large{\frac{-b}{a}}\) is negative, the equation has no solutions in the real number field.


When \(n\) is odd, there is always a single solution in the form:

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

In fact, the nth root of a negative number is feasible only when the root’s index is odd. This arithmetic operation is mathematically valid and can be executed using the same methodology as real positive numbers. The result of extracting the nth root of a negative number is also negative, except when the index is an even number. In such cases, there is no real solution.


Solve the binomial equation \(x^3-27 = 0\).

We have:

\begin{align*} 3x^3 &= 27 \\[0.6em] x &= \sqrt[\Large{3}]{27} \end{align*}

In this case, \(n\) is odd, and the number under the root is positive, so the equation admits a single real solution.

The solution to the equation is:

\[ x = 3 \]

Solve the binomial equation \(x^3 + 8 = 0\).

We have:

\begin{align*} x^3 &= -8 \\[0.6em] x &= \sqrt[\large{3}]{-8} \end{align*}

In this case, \(n\) is odd, and the number under the root is negative, so the equation admits a single real solution.

The solution to the equation is:

\[ x = -2 \]


Solve the binomial equation \(x^4 +5 = 0\).

We have:

\begin{align*} x^4 &= -5 \\[0.6em] x &= \sqrt[\large{4}]{-5} \end{align*}

In this case, \(n\) is even, and the number under the root is negative, so the equation does not admit real solutions.

The solution to the equation is:

\[\nexists x \in \mathbb{R}\]


Indeed, the examples presented are merely generic illustrations of easily solvable equations. In practice, it is possible to encounter instances where complex polynomial equations may be reduced to binomial equations, which, as demonstrated, can be resolved using straightforward methods