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!
$\endgroup$ 12 Answers
$\begingroup$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.
$\endgroup$ 0 $\begingroup$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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