$\begingroup$$a$ and $b$ are solutions of $$ \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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