Section 4.6 Rational Equations
When is a ratio zero? When its numerator is zero!
Example 4.27. A rational equation.
Suppose
\begin{equation}
\frac{x^3-4x}{x+1} = 0\tag{4.2}
\end{equation}
then it must be that the numerator is zero. That is,
\begin{equation}
x^3-4x=0.\tag{4.3}
\end{equation}
Factoring,
\begin{equation*}
x(x^2-4) = x(x-2)(x+2) = 0.
\end{equation*}
Thus, the solutions are \(x=0\text{,}\) \(2\text{,}\) and \(-2\text{.}\)
Observe that if
\(x = -1\text{,}\) then the denominator
\(x+1=0\) and the left side of the equation is undefined. Thus,
\(x=-1\) is not in the domain of the equation, but only of our solutions to the polynomial equation
(4.3) are in the domain of the rational equation
(4.2).
In general, a rational equation may be solved by
clearing the fractions. This means to multiply both sides of the equation by an expression which results in canceling all the denominators. But care must be taken to consider the domain of the equation when finished. (See
Warning 4.5.)
Example 4.28. Considering the domain.
Consider
\begin{equation*}
\frac{4}{x+1}+\frac{10}{x^2-1} = \frac{5}{x-1}.
\end{equation*}
We can clear the fractions by multiplying both sides by \(x^2-1 = (x-1)(x+1)\text{.}\) However, we are assuming that \(x\neq 1\) and \(x\neq -1\) as these values are not in the domain of the equation.
\begin{equation*}
(x-1)(x+1)\cdot \left[\frac{4}{x+1}+\frac{10}{x^2-1} \right] = \left[\frac{5}{x-1}\right] \cdot (x-1)(x+1).
\end{equation*}
Distributing through and simplifying,
\begin{equation*}
4(x-1)+10 = 5(x+1)
\end{equation*}
Combining like terms,
\begin{equation*}
4x-4+10 = 5x+5
\end{equation*}
\begin{equation*}
-4+10-5 = 5x-4x
\end{equation*}
\begin{equation*}
x = 1
\end{equation*}
But this value is not in the domain of the original equation and therefore there are no solutions!
Example 4.29. Solving a Rational Equation.
Solve the rational equation
\begin{equation*}
4 + \frac{7}{x-2} = -\frac{1}{x+1}.
\end{equation*}
Solution.
First, observe that \(x = 2\) and \(x=-1\) are not in the domain of the equation and therefore these values are not solutions. Let’s clear the fractions by multiplying both sides by \((x-2)(x+1)\)
\begin{align*}
(x-2)(x+1) \left[4 + \frac{7}{x-2}\right] &= \left[-\frac{1}{x+1}\right](x-2)(x+1)\\
4(x-2)(x+1) + 7(x+1) &= -(x-2) \\
4(x^2-x-2) + 7x+7 &= -x+2 \\
4x^2-4x-8 + 7x+7 + x - 2 &= 0 \\
4x^2+4x-3 &= 0
\end{align*}
This is quadratic equation. Factoring,
\begin{equation*}
(2x+3)(2x-1) = 0
\end{equation*}
So that,
\begin{equation*}
2x+3 = 0 \quad \text{or} \quad 2x - 1 = 0
\end{equation*}
\begin{equation*}
x = -3/2 \quad \text{or} \quad x = 1/2
\end{equation*}
Note that both these values are in the domain of the original equation.
Example 4.30. Solving Rational Equations.
Solve each rational equation. Video solutions follow.
\(\displaystyle \frac{3}{x+1}=\frac{4}{x+2}\)
\(\displaystyle \frac{4x^2-2x}{x+3} =0\)
\(\displaystyle \frac{4}{x+1}+\frac{10}{x^2-1} = \frac{5}{x-1}\)
\(\displaystyle 4 + \frac{7}{x-2} = -\frac{1}{x+1}\)
