$a$ 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 $a + b = 13 - 3 = 10$
$\endgroup$ 3 Reset to default
