Find the number of positive roots of the equation $\left|\frac{x^2-4x}{x-1}\right|=-x.$

Question

Find the number of positive roots of the equation: $$\left|\dfrac{x^2-4x}{x-1}\right|=-x.$$ The absolute value is defined as $$|x|=\begin{cases}x,x\ge0, \\-x, x<0\end{cases}.$$ So the equation is equivalent to $$\dfrac{x^2-4x}{x-1}=-x \\ \text{OR}\\ -\dfrac{x^2-4x}{x-1}=-x.$$ The first equation has roots $0;\dfrac{5}{2}$ and the solution of the second equation is $x=0$. So according to my calculations, the equation has $1$ positive root, which is not true. Thank you in advance!

Answer 1

As the LHS is non-negative, $x$ cannot be positive.

Answer 2

Note that $\frac{5}{2}$ does not satisfy your first equation. For $x=5/2,$ you have

$\frac{(5/2)^2-4(5/2)}{5/2-1}$ is negative.

Answer 3

You have$$\frac{x^2-4x}{x-1}\begin{cases}\leqslant0&\text{ if }x\in(-\infty,0]\cup(1,4]\\\geqslant0&\text{ if }x\in[0,1)\cup[4,\infty)\end{cases}$$and therefore$$\left|\frac{x^2-4x}{x-1}\right|=\begin{cases}-\frac{x^2-4x}{x-1}&\text{ if }x\in(-\infty,0]\cup(1,4]\\\frac{x^2-4x}{x-1}&\text{ if }x\in[0,1)\cup[4,\infty).\end{cases}$$So, solve the equation $-\frac{x^2-4x}{x-1}=-x$ and take its solutions from $(-\infty,0]\cup(1,4]$; also, solve the equation $\frac{x^2-4x}{x-1}=-x$ and take its solutions from $\in[0,1)\cup[4,\infty)$.