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

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!

• $\frac{5}{2}$ is not a solution to the above equation, in fact I think $0$ is the only solution the equation has... May 11 at 8:42

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

• 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?
– Medi
May 11 at 8:43
• @Medi: did you notice that this completely solves the question ?
– user65203
May 11 at 8:44
• I edited my comment.
– Medi
May 11 at 8:45
• @Medi: because these solutions are not solutions.
– user65203
May 11 at 8:48
• More accurately, the LHS is non-negative, which still implies $x$ cannot be positive.
– J.G.
May 11 at 8:52

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.

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

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