Why does this trick for solving this equation work?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

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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 $|b|,|b'|<2\sqrt{ac}$.

Answer 4

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.