Solve the equation $$\frac{1}{x^2+11x-8} + \frac{1}{x^2+2x-8} + \frac{1}{x^2-13x-8} = 0$$

What I've tried

First I tried factoring the denominators but only the second one can be factored as $(x+4)(x-2)$.

Then I tried substituting $y = x^2 - 8$ but that didn't lead me anywhere.

Where I'm stuck

I don't know how to start this problem. Any hints?

P.S. I would really appreciate it if you give me hints or at least hide the solution. Thanks for all your help in advance!

  • $\begingroup$ While it's only a hint and not a solution, it is always not a bad idea to make a graph (DESMOS) just to see what is happening $\endgroup$
    – imranfat
    Commented Jul 1, 2020 at 21:57

2 Answers 2


Since you require the sum to be zero, all you need is to compute the numerator of the LHS when put over a common denominator, since regardless of the value of the denominator--as long as it is nonzero--the LHS is zero only if the numerator is zero. Then once you solve for the roots of the numerator, check the validity of the solution set by substitution.


You started very well. To make things easier, set $A=x^2+7x-8$ (*) and the equation rewrites $$\frac{1}{A+4x} + \frac{1}{A-5x} + \frac{1}{A-20x} = 0.$$

Denominators are not allowed to be zeros. We solve $$(A-5x)(A-20x)+(A+4x)(A-20x)+(A+4x)(A-5x)=0$$ or equivalently $$3A^2-42Ax=0,$$ or even $$3(x^2+7x-8)(x^2-7x-8)=0,$$ which is easy to finish.

(*) I noticed that $x=1$ satisfies, and decided to make profit from it.


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