Why does this trick for solving this equation work?

The question is to

solve the equation $$\frac{4x}{\left ( 2x-2 \right )^{2}+3} + \frac{12x}{\left ( 4x-5 \right )^{2}+3} = 1.$$

Now if we solve $$\left | 2x-2 \right | = \left | 4x-5 \right |$$ first (that is, setting the denominators equal) we find that $$x=\frac{3}{2},\frac{7}{6}.$$ If we now go back to substitute these values of $$x$$ in LHS of the original equation, we get that $$\frac{ 4x\left ( \left ( 4x-5 \right )^{2}+3 \right ) + 12x\left ( \left ( 2x-2 \right )^{2}+3 \right ) }{\left ( \left ( 4x-5 \right )^{2}+3 \right )\left ( \left ( 2x-2 \right )^{2}+3 \right )} = 6.$$

Thus, we see that $$x=\frac{3}{2},\frac{7}{6}$$ both satisfy the equation $$\frac{4x}{\left ( 2x-2 \right )^{2}+3} + \frac{12x}{\left ( 4x-5 \right )^{2}+3} = 6.$$

However, I don't know how this is relevant to the fact that $$x=\frac{1}{2},\frac{7}{2}$$ are solutions to the original equation $$\frac{4x}{\left ( 2x-2 \right )^{2}+3} + \frac{12x}{\left ( 4x-5 \right )^{2}+3} = 1.$$ That is, what is the relationship between the respective solutions of $$\text{original LHS}=6$$ and $$\text{original LHS}=1$$?

Please tell me the mystery behind this process, and can we use this technique on other equations like this? Thank you.

• I think your question is an interesting one. I am still thinking about it. I think one only needs to understand this geometrically to see why this happens, or else to think about some properties of rationals. Any case let's see what someone comes up with! I'll also like to see an answer to this question (the three answers at the present haven't actually answered your question). Sep 29 '19 at 12:08
• At this stage, I think some clarification from OP is in order: Why do you think there is such a relationship between the equations $\text{LHS}=6$ and $\text{LHS}=1,$ or in other words, how do you deduce the solutions of the latter equation from those of the former. You don't explain this at all. Sep 29 '19 at 13:54
• The mystery is what you call "this trick".
– user65203
Sep 30 '19 at 12:55
• You may be needed here @Quanto. I think you have a way with such tricks. :) Oct 1 '19 at 10:50

In problems like these, the first thing to do is to clear denominators. If we do so and expand everything, we will be left with $$16x^4-100x^3+200x^2-175x+49=0.$$ If you substitute values (with smart guesses using the rational roots theorem), you will be able to find that $$x=1/2$$ and $$x=7/2$$ are solutions. Then by the Factor theorem , $$(2x-1)$$ and $$(2x-7)$$ are factors of the polynomial on the LHS. Then use long division to factor the LHS. You should obtain $$(4x^2-9x+7)(2x-1)(2x-7)=0,$$ and the rest of the solution is easy to complete.

• Can the downvoter please explain? Sep 29 '19 at 14:16
• Sorry for responding late. I'm just seeing this. The reason is clear. You have not addressed OP's question at all. While you have shown them an alternative way to solve the equation (which is commendable in itself), that is primarily not what is bothering OP. Oct 1 '19 at 10:45
• @Allawonder thanks for the response, it's reasonable and I appreciate it. Oct 1 '19 at 10:48

Note that$$\frac{4x}{(2x-2)^2+3}+\frac{12x}{(4x-5)^2+3}-1=\frac{-16 x^4+100 x^3-200 x^2+175 x-49}{\left(4 x^2-10 x+7\right) \left(4x^2-8 x+7\right)},$$the solutions of your equations are the roots of the polynomial$$-16 x^4+100 x^3-200 x^2+175 x-49.$$Using the rational root theorem, you can deduce that $$\frac12$$ and $$\frac72$$ are roots of this polynomial. On the other hand$$\frac{-16 x^4+100 x^3-200 x^2+175 x-49}{\left(x-\frac12\right)\left(x-\frac72\right)}=-4 \left(4 x^2-9 x+7\right).$$So, there are no more real roots and there are two complex non-real roots.

As the LHS is not a homogeneous function, there is no particular relation between the solutions of

$$f(x)=1$$ and $$f(x)=6.$$

By the way, there is no clear relation between $$(\frac32,\frac76)$$ and $$(\frac12,\frac72)$$ besides the coincidental equality of the products.

I doubt that this pseudo-property generalizes.

• Good attempt. I also think OP needs to explain why they think there is a relationship between the solutions. Oct 1 '19 at 10:49

Let $$f(x)=\frac{4x}{(2x-2)^2+3}+\frac{12x}{(4x-5)^2+3}$$. Then, $$f'(x)=4\left(\frac{\big((2x-2)^2+3\big)-2x(4x-4)}{\big((2x-2)^2+3\big)^2}\right)+12\left(\frac{\big((4x-5)^2+3\big)-4x(8x-10)}{\big((4x-5)^2+3\big)^2}\right).$$ So $$f'(x)=-4(4x^2-7)\left(\frac{1}{\big((2x-2)^2+3\big)^2}+\frac{12}{\big((4x-5)^2+3\big)^2}\right).$$ Therefore, $$f$$ is negative and strictly decreasing on $$\left(-\infty,-\frac{\sqrt7}{2}\right)$$, and then it achieves the global minimum at $$-\frac{\sqrt7}{2}$$. After that, it is strictly increasing on $$\left(-\frac{\sqrt7}{2},\frac{\sqrt7}2\right)$$, then achieves the global maximum at $$\frac{\sqrt{7}}{2}$$. Then, it remains positive and strictly decreasing on $$\left(\frac{\sqrt{7}}{2},\infty\right)$$. It follows that for any real number $$y\in\big(f(-\sqrt{7}/2),0\big)\cup\big(0,f(\sqrt{7}/2)\big),$$ there are exactly two values $$x\in \Bbb R$$ such that $$f(x)=y$$. For $$y\in\big\{f(-\sqrt7/2),0,f(\sqrt7/2)\big\},$$ there is exactly one $$x$$ (i.e., $$x=-2/\sqrt7,0,2/\sqrt7$$, resp.) such that $$f(x)=y$$. For $$y\notin \big[f(-\sqrt7/2),f(\sqrt7/2)\big],$$ $$f(x)=y$$ has no real solutions. See this wolframalpha plot. For those who are curious, $$f(\sqrt7/2)=\frac{1}{6}(19+8\sqrt7)=6.694335\ldots$$ and $$f(-\sqrt7/2)=\frac{1}{6}(19-8\sqrt7)=-0.3610017\ldots.$$

Back to the problem, you have found two values of $$x$$ (i.e., $$x=1/2,7/2$$) such that $$f(x)=1$$, then by the analysis above, we see that these are the only values of $$x$$ that work. Similarly, there are only two values of $$x$$ (i.e., $$x=3/2,7/6$$) such that $$f(x)=6$$. However, I do not see a good way to solve equations of the form $$f(x)=y$$ when there are two solutions apart from "divining" some two values of $$x$$ that solve the equation, or brute-force like YiFan's solution. Your method of solving the case $$y=6$$ seems like a coincidence at best.

I found a way to solve this problem without dealing with the quartic equation. Let $$f(x)$$ be as defined in my previous answer. We can rewrite $$f(x)$$ as $$f(x)=\frac{4x}{4x^2-8x+7}+\frac{3x}{4x^2-10x+7}.$$ To solve $$f(x)=y$$ we want to find some $$a,b$$ s.t. $$a+b=y$$ and the numerators of the equation $$a-\frac{4x}{4x^2-8x+7}=\frac{3x}{4x^2-10x+7}-b\ \ \ \ \ (1)$$ are proportional. The numerators are $$a(4x^2-8x+7)-4x$$ and $$-b(4x^2-10x+7)+3x$$. So we can try to find $$a,b$$ such that $$b(8a+4)=a(10b+3)$$ with $$a+b=y$$. That is, $$2ab=4b-3a$$ or $$2a(y-a)=4(y-a)-3a=4y-7a.$$ So $$2a^2-(2y+7)a+4y=0.$$ This gives $$a=\frac{(2y+7)\pm\sqrt{(2y+7)^2-32y}}{4}=\frac{(2y+7)\pm\sqrt{4y^2-4y+49}}{4}.$$ For $$y=1$$, $$a=\frac{9\pm7}{4}$$ or $$a\in\{1/2,4\}$$.

You can use either of this. If you use $$a=1/2$$, (1) is now equivalent to $$\frac{4x^2-16x+7}{4x^2-8x+7}=-\frac{4x^2-16x+7}{4x^2-10x+7}.$$ Thus either $$(2x-1)(2x-7)=4x^2-16x+7=0$$ or $$\frac{1}{4x^2-8x+7}+\frac{1}{4x^2-10x+7}=0$$ but it can be shown easily that both $$4x^2-8x+7$$ and $$4x^2-10x+7$$ are strictly positive for $$x\in\Bbb R$$.

If you use $$a=4$$, (1) is now equivalent to $$\frac{4(4x^2-9x+7)}{4x^2-8x+7}=\frac{3(4x^2-9x+7)}{4x^2-10x+7}.$$ Because $$4x^2-9x+7>0$$ for all $$x\in \Bbb R$$, we get $$\frac{4}{4x^2-8x+7}=\frac{3}{4x^2-10x+7}.$$ So $$16x^2-40x+28=12x^2-24x+21$$ or $$(2x-1)(2x-7)=4x^2-16x+7=0$$.

For $$y=6$$, the suitable values of $$a$$ are $$3/2$$ and $$8$$. The only nice rational values of $$y$$ with rational values of suitable $$a$$ are of the form $$y=\frac{-t^2+2t+48}{4t}$$ for some rational $$t\ne 0$$, which means $$a=\frac{-t+8}{4},\frac{2t+12}{t}.$$ (The case $$y=1$$ is given by $$t\in\{-8,6\}$$.) For example, $$t\in\{2,-24\}$$ produces $$y=6$$, and from either $$a\in\{3/2,8\}$$, we can show that $$x\in\{3/2,7/6\}$$. One last example, $$t\in\{-6,4\}$$ produces $$y=\frac{5}{2}$$ and $$a\in\{1,5\}$$. From this we can easily solve for $$x$$, i.e., $$x=\frac{3\pm\sqrt2}{2}$$.

I would like to note that my solution works simply because

1. each of the two fractions in $$f(x)$$ can be converted into something of the form $$\frac{kx}{ax^2+bx+c}$$ where $$a$$ and $$c$$ are the same in both fractions
2. the denominators of both fractions are always positive.

I believe that solving $$f(x)=y$$ for another $$f(x)$$ of the same form works similarly, i.e., for $$f(x)$$ given by $$f(x)=\frac{kx}{ax^2+bx+c}+\frac{k'x}{ax^2+b'x+c}$$ with $$a,c>0$$ and $$|b|,|b'|<2\sqrt{ac}$$.

• It's better to have one long answer rather than two shorter ones. Sep 29 '19 at 14:14
• @N.Bar Not when MathJaX slows your computer down when you write a long answer. And especially when both answers are long. Sep 29 '19 at 14:14
• What? My computer is almost a decade old and has never slowed from MathJax. Sep 29 '19 at 14:16
• Mine does. And it's very painful to write long answers. Mine is not very new either. Maybe I don't have enough ram. Sep 29 '19 at 14:16