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$a$ and $b$ are solutions of $$ \frac{1}{x^{2} - 10x-29} + \frac{1}{x^{2} - 10x-45} - \frac{2}{x^{2} - 10x-69} = 0 $$ What is $a+b=?$ $$ $$ Are there better approaches than the one below?


Solution:

By letting $x^{2} - 10x = y$, then we have

$$ \frac{1}{y-29} + \frac{1}{y-45} - \frac{2}{y-69} = 0, \:\: y \notin \{ 29,45,69 \} $$

and $$ (y-45)(y-69) + (y-29)(y-69) - 2(y-29)(y -45) = 0 $$ $$ (y- 69)(y-37) = (y-29)(y-45) $$ $$ y^{2} - 106 y + 69 \cdot 37 = y^{2}-74y + 29 \cdot 45 $$ $$ -32y = 3 (29 \cdot 15 - 23 \cdot 37) = -1248 $$ $$ y = x^{2} - 10x = 39$$ Here are the roots: $$ x^{2} -10x - 39 = 0 \implies (x-13)(x+3) = 0$$ So the answer is $a + b = 13 - 3 = 10$

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    $\begingroup$ As $x^2-10x+k=(x-5)^2+k-25$, it is clear that whenever $x $ is a solution, then so is $10-x$. Hence if we take for granted that the problem statement i snot ill-posed and that an answer to $a+b=?$ can be given at all, then this answer must be $10$ $\endgroup$ – Hagen von Eitzen May 19 '19 at 12:30
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    $\begingroup$ One can avoid most computations in the following way: The equation in $y$ has numerator at most quadratic. It is less than quadratic because $1+1-2=0$ is its quadratic coefficient. It cannot be less than linear because $\frac{1}{y-29}\to+\infty$ as $y\to29^{+}$ and $\frac{1}{y-45}\to-\infty$ as $y\to45^-$. Since the function is continuous in $(29,45)$, then it must have a root. Therefore the equation is equivalent to a linear equation $y=A$, or $x^2-10x-A=0$. But then by Vieta's formulas the sum of the roots of this equation is $10$. $\endgroup$ – logarithm May 19 '19 at 12:50
  • $\begingroup$ @HagenvonEitzen sorry don't understand.. I thought this type of problem is not the type to be answered like this.. $\endgroup$ – Arief May 19 '19 at 13:17

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