Solve the following equation. Check all proposed solutions. Show all work in solving and checking, and state your final answer

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my work

First let's add the fractions on the left. To add any fraction, you must put them over a common denominator, in this case $(x+2)(x^2-4)$

Multiply out the numerator and denominator then combine like terms, but after doing all that I got

$-16=x^3+4x^2-17x $

and I can't get value for $x$

  • $\begingroup$ Did you forget to factorise $(x^2 - 4)$ ? $\endgroup$ – Shailesh Sep 25 '15 at 15:16
  • $\begingroup$ I already did (x-2)(x+2) but still stuck $\endgroup$ – user155971 Sep 25 '15 at 15:17
  • 1
    $\begingroup$ Notice that $x=\pm 2$ cannot be solutions since they make the denominator zero on the left hand side. Therefore, we can divide by $x-2$ and get the below answer. $\endgroup$ – Kevin Sheng Sep 25 '15 at 15:23


You forgot to factorize $(x^2 - 4) = (x + 2)(x -2)$ from the denominator. Then the LCM becomes $(x^2 - 4)$ in which case the numerator gives

$5(x - 2) - 4 = (x-3)(x + 2)$, so you will get $(x^2-6x+8) = 0$ which is $(x- 4)(x-2) = 0$. Looking at the equation $\color{red}{x = 4}$ is the only solution. This is because when you put $x = 2$, the original equation is undefined.


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