# Rational Equations

Rational equations feature at least one fraction in which the numerator and denominator are polynomials. Such equations are categorized as rational because they can be expressed as the ratio of two polynomials. Specifically, rational equations have the following form:

\[\frac{P(x)}{Q(x)}=0\]

where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) \neq 0\).

The key distinction between rational and irrational equations lies in their structure. Rational equations consist of fractions with polynomials in both numerator and denominator and can always be expressed as a ratio of two polynomials. Irrational equations feature roots of various orders and solutions that cannot be described as rational numbers. These roots are represented using radical notation and an index that indicates their order. This fundamental difference is crucial to understanding the nature of these equations. For example:

\[\frac{1}{2x-1}\]

is a *rational equation* since the variable \(x\) appears in a polynomial of the first degree in the denominator.

\[\frac{1}{\sqrt{2x-1}}\]

is an irrational since the variable is inside a root.

## Solving method

The first step to solving rational equations is determining the values that make the denominators zero. These values are not allowable solutions because they lead to an indeterminate form.

The sequent step entails determining the least common multiple (LCM) of the polynomials in all the denominators and finding the solutions for the polynomials in the numerator.

In the final stage of the process, removing the values that nullify the denominators and, subsequently, validating the acceptability of the remaining solutions to evaluate their admissibility in light of the given conditions is necessary.

## Example

Solve the rational equation: \[\frac{1}{x+1} + \frac{1}{x+2} = 0\]

The first step to solving these equations is determining the values that make the denominators zero. These values are not allowable solutions because they lead to an indeterminate form:

\begin{align} x+1 = 0 \quad\quad x = -1\\[0.5em] x+2 = 0 \quad\quad x = -2\\[0.5em] \end{align}

Values for which \(x = -1\) and \(x = -2\) must be excluded from the solutions because they would make the denominators zero.

Now let’s proceed with the calculations and obtain:

\begin{align} &\frac{x+2}{(x+1)(x+2)} + \frac{x+1}{(x+1)(x+2)} = 0\\[1em] & \frac{x+2+x+1}{(x+1)(x+2)} = 0\\[1em] &\frac{2x+3}{(x+1)(x+2)} = 0 \end{align}

Let’s find solutions that set the numerator \(2x+3 = 0\) and then check their validity. The equation is reduced to a first-degree linear equation, which admits a unique solution \(\large{x = -\frac{3}{2}}\). The solution is not among the values that nullify \(x\) in the denominator; therefore, it is an admissible solution. Finally, we substitute the solution into the initial equation and verify whether equality holds.

\begin{align*} \frac{1}{{-\frac{3}{2}+1}} + \frac{1}{{-\frac{3}{2}+2}} &= 0\\[1em] \frac{1}{{-\frac{1}{2}}} + \frac{1}{{\frac{1}{2}}} &= 0\\[1em] +2-2 &=0 \end{align*}

The equality is verified, therefore \(x = \large(-\frac{3}{2})\) is the solution of the equation.

The solution to the equation is: \[x= – \frac{3}{2}\]

That shown is a very simple example of resolution. Generally, the resolution principle works as outlined above, but there can be more complex cases. The key is to ensure that the solutions found are admissible by excluding any values of the unknown that nullify the denominator, regardless of its form.

Practice solving the rational equations proposed in the exercise module.