Rational Equations
What are 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 $Q ( x ) \neq 0$. Recall that a polynomial expression is made up of a combination of monomials that are added or subtracted to form the full expression. A polynomial therefore has the general form:
\[a_{n} x^{n} + a_{n - 1} x^{n - 1} + \hdots + a_{2} x^{2} + a_{1} x + a_{0}\]Where:
- $a x^{n}$ is a monomial;
- $a$ is a real number, known as the coefficient of the term;
- $n$ is a non-negative integer, representing the exponent of the variable.
Distinction between irrational and rational equations
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:
- $\frac{2 s}{2 x - 1}$ is a rational equation since the expression contains only ratios of polynomial terms
- $\frac{1}{\sqrt{2 x - 1}}$ is an irrational since the variable is inside a root.
How to Solve Rational Equations
- 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:
\[x + 1 = 0 x = - 1 \\ x + 2 = 0 x = - 2\]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:
\[& \frac{x + 2}{( x + 1 ) ( x + 2 )} + \frac{x + 1}{( x + 1 ) ( x + 2 )} = 0 \\ & \frac{x + 2 + x + 1}{( x + 1 ) ( x + 2 )} = 0 \\ & \frac{2 x + 3}{( x + 1 ) ( x + 2 )} = 0\]Let’s find solutions that set the numerator $2 x + 3 = 0$ and then check their validity. The equation is reduced to a first-degree linear equation, which admits a unique solution $x = - \frac{3}{2}$. 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.
\[\frac{1}{- \frac{3}{2} + 1} + \frac{1}{- \frac{3}{2} + 2} & = 0 \\ \frac{1}{- \frac{1}{2}} + \frac{1}{\frac{1}{2}} & = 0 \\ + 2 - 2 & = 0\]The equality is verified, therefore $x = ( - \frac{3}{2} )$ is the solution of the equation.
The solution to the equation is: \(x = - \frac{3}{2}\)
Solve the following rational equations
- \(\text{1}. \frac{3 x - 2}{5 - 2 x} = 0\) solution
- \(\text{2}. 1 - \frac{6}{x} = - \frac{8}{x^{2}}\) solution
- \(\text{3}. \frac{2 x + 1}{6} = \frac{1}{x}\) solution
- \(\text{4}. \frac{1}{x + 2} - \frac{1}{x - 1} = \frac{2}{x^{2} - 1}\) solution
- \(\text{5}. \frac{2 x}{x + 1} - \frac{3}{x + 5} = \frac{- 8 x^{2}}{x^{2} + 6 x + 5}\) solution
- \(\text{6}. \frac{4 x - x}{3 x + 2} - \frac{1}{9 x^{2} - 4} = 0\) solution
- \(\text{7}. \frac{1}{x - 4} = \frac{7}{x^{2} + x - 20}\) solution
- \(\text{8}. \frac{x - 5}{x^{3} + 9 x + 27 x + 27} = 0\) solution
- \(\text{9}. \frac{x^{2} - 9}{x - 3} = 4\) solution
- \(\text{10}. \frac{x^{2} + 4 x - 5}{x - 1} = \frac{x - 2}{2}\) solution
The proposed equations are carefully designed to help you consolidate your understanding of irrational equations. Try solving them independently before checking the solutions provided.
GLossary
- Rational equation: an equation featuring at least one fraction in which the numerator and denominator are polynomials.
- Polynomial: an expression made up of a combination of monomials that are added or subtracted.
- Monomial: a term within a polynomial, typically of the form $a x^{n}$, where $a$ is a real number coefficient and $n$ is a non-negative integer exponent.
- Numerator: The top part of a fraction.
- Denominator: The bottom part of a fraction.
- Excluded values: values of the variable that make the denominator of a rational expression zero and are therefore not allowed as solutions.
- Least Common Multiple (LCM): the smallest positive integer that is a multiple of two or more given integers or polynomials. Used to combine fractions in rational equations.
- Irrational equation: an equation that features roots of variables, often represented using radical notation.