# $a$ 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 $10$ Commented May 19, 2019 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 $10$. Commented May 19, 2019 at 12:50
• @HagenvonEitzen sorry don't understand.. I thought this type of problem is not the type to be answered like this.. Commented May 19, 2019 at 13:17

We can also solve the exercise without using Vieta’s formulas.

In order to avoid many calculations and big numbers, we can let $$\;y=(x-5)^2-62\,$$ and substitute it in the OP’s equation:

$$\dfrac1{y+8}+\dfrac1{y-8}-\dfrac2{y-32}=0\;\;,$$

$$\dfrac{2y}{y^2-64}-\dfrac2{y-32}=0\;\;,$$

$$\require{cancel}\dfrac{2\left(\!\color{#8888FF}{\cancel{\color{black}{y^2}}}\!-32y-\!\color{#8888FF}{\cancel{\color{black}{y^2}}}\!+64\right)}{(y^2-64)(y-32)}=0\;\;,\quad$$ hence ,

$$y=2\;,\quad$$ consequently ,

$$(x-5)^2=64\;\;,$$

$$x=5\pm8\;\;,$$

so the sum of the solutions is $$\,10\,.$$

Here is my take on the question:

Let $$y=x^2-10x$$. Then simplify the LHS of the given equation. You should get

$$\frac{-64(y-39)}{(y-29)(y-45)(y-69)}=0$$

The above indicates that the numerator equals $$0$$. Therefore

$$-64(y-39)=0$$

However, since $$y=x^2-10x$$ and $$-64$$ is a constant, this means that the original equation becomes

$$x^2-10x-39=0$$

Now, from Viete's formulae for the roots of a polynomial

$$x_1+x_2+\cdots+x_n=-\frac{c_{n-1}}{c_n}$$

(where $$c_n$$ is the $$n$$th coefficient in a polynomial with degree $$k$$ (or $$n$$ or whatever the convention is (I forgot))

Therefore, the sum of our quadratic's roots is $$10$$.

This therefore implies that $$\fbox{a+b=10}$$

Update: Alternatively, from getting $$x^2-10x-39=0$$, you can factorise it and add the solutions, which will still get you $$10$$ as your answer if you want to do without Viete's formulae for the roots of a polynomial.

I had a (very slightly) different solution, but also using Vieta's formulas.

We may assume the denominators are not equal to zero, and multiply everything by the product of the three. We get \begin{align} &(x^2-10x-45)(x^2-10x-69)\\ +\,\ &(x^2-10x-29)(x^2-10x-69)\\ - 2\,&(x^2-10x-29)(x^2-10x-45) = 0 \end{align} We can interpret each product as a difference of squares, in the following fashion: \begin{align} (x^2-10x-45)(x^2-10x-69) &= \big[(x^2-10x-57)+12\big]\big[(x^2-10x-57)-12\big] \\ &= (x^2-10x-57)^2-12^2\\ (x^2-10x-29)(x^2-10x-69) &= \big[(x^2-10x-49)+20\big]\big[(x^2-10x-49)-20\big] \\ &= (x^2-10x-49)^2-20^2\\ (x^2-10x-29)(x^2-10x-45) &= \big[(x^2-10x-37)+8\big]\big[(x^2-10x-37)-8\big] \\ &= (x^2-10x-37)^2-8^2\\ \end{align}

Plugging these three expressions in, we can assign to both $$(x^2-10x-57)^2$$ and $$(x^2-10x-49)^2$$ one copy of $$-(x^2-10x-37)^2$$ (because we have a $$-2$$ coefficient in front of the third product). Applying the difference of squares again (now writing it as a product) and simplifying (if I'm not mistaken, we simply divide by 16), we get the equation $$4x^2-40x-199=0$$. Since $$a$$ and $$b$$ are the roots, their sum equals $$-\frac{-40}{4} = 10$$.