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Problem

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!

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  • $\begingroup$ 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

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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.

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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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