# $a$ and $b$ are solutions of $\frac{1}{x^{2} - 10x-29} + \frac{1}{x^{2} - 10x-45} - \frac{2}{x^{2} - 10x-69} = 0$, $a+b=?$

$$a$$ 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 $$a + b = 13 - 3 = 10$$

• 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 $10$ – Hagen von Eitzen May 19 '19 at 12:30
• 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 $10$. – logarithm May 19 '19 at 12:50
• @HagenvonEitzen sorry don't understand.. I thought this type of problem is not the type to be answered like this.. – Arief May 19 '19 at 13:17