Equations That Are Quadratic in Form

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