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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!

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    $\begingroup$ $\frac{5}{2}$ is not a solution to the above equation, in fact I think $0$ is the only solution the equation has... $\endgroup$ May 11 at 8:42
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As the LHS is non-negative, $x$ cannot be positive.

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  • $\begingroup$ Yes, I noted that. Since the LHS is positive, then the RHS also must be positive, so $-x\ge0\Leftrightarrow x\le 0$. We can conclude that the equation has no positive roots. But why isn't actually the equation equivalent to the union of the solutions of the two equations I wrote? $\endgroup$
    – Medi
    May 11 at 8:43
  • $\begingroup$ @Medi: did you notice that this completely solves the question ? $\endgroup$
    – user65203
    May 11 at 8:44
  • $\begingroup$ I edited my comment. $\endgroup$
    – Medi
    May 11 at 8:45
  • $\begingroup$ @Medi: because these solutions are not solutions. $\endgroup$
    – user65203
    May 11 at 8:48
  • $\begingroup$ More accurately, the LHS is non-negative, which still implies $x$ cannot be positive. $\endgroup$
    – J.G.
    May 11 at 8:52
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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.

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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 $\in[0,1)\cup[4,\infty)$.

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  • $\begingroup$ A sledgehammer for a problem that can be answered in a blink ! $\endgroup$
    – user65203
    May 11 at 8:47
  • $\begingroup$ Thank you for the response! Can you clarify how do we get the intervals in first place? $\endgroup$
    – Medi
    May 11 at 8:47
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    $\begingroup$ @YvesDaoust, On the contrary! That is exactly what I am asking for. $\endgroup$
    – Medi
    May 11 at 8:48
  • $\begingroup$ For instance, $\frac{x^2-4x}{x-1}\leqslant0$ if and only if $x-1<0\leqslant x^2-4x$ or $x-1>0\geqslant x^2-4x$. $\endgroup$ May 11 at 8:49
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    $\begingroup$ @Medi: understood. There is no need to find the roots. $\endgroup$
    – user65203
    May 11 at 9:03

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