MA001: College Algebra

Earlier, we solved rational equations. Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution.

EXAMPLE 11

Solving a Rational Equation Leading to a Quadratic

Solve the following rational equation: $\dfrac{-4 x}{x-1}+\dfrac{4}{x+1}=\dfrac{-8}{x^{2}-1}$.

Solution

We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, $x^{2}-1=(x+1)(x-1)$. Then, the LCD is $(x+1)(x-1)$. Next, we multiply the whole equation by the LCD.

$\begin{aligned}(x+1)(x-1)\left[\dfrac{-4 x}{x-1}+\dfrac{4}{x+1}\right] &=\left[\dfrac{-8}{(x+1)(x-1)}\right](x+1)(x-1) \\-4 x(x+1)+4(x-1) &=-8 \\-4 x^{2}-4 x+4 x-4 &=-8 \\-4 x^{2}+4 &=0 \\-4\left(x^{2}-1\right) &=0 \\-4(x+1)(x-1) &=0 \\ x &=-1 \\ x &=1 \end{aligned}$

In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.

TRY IT #10

Solve $\dfrac{3 x+2}{x-2}+\dfrac{1}{x}=\dfrac{-2}{x^{2}-2 x}$