# Solving $x|x-1|<3$?

Seems pretty straight-forward but I'm a bit confused with the modulus and inequality.

So I know that if

$|x-1|>3$

because the entire equation on the left is within the modulus we can write

$x-1>3$ or $x-1<-3$

I also know that if $y<0$

$y|x-1|>3$

Can be written as

$|x-1|<3/y$

But what do we do with

$x|x-1|<3$

Can we break it down further without trying to input values?

• What is the range of your solutions? Do you include real numbers, integers or something else? Commented Jul 20, 2018 at 7:15
• Lets assume all integers Commented Jul 20, 2018 at 7:17
• There's actually a smarter way to do this problem – which is to separate it into $x < 0$, $0 < x < 1$, and $1, 2, 3 \cdots$. Commented Jul 20, 2018 at 7:30
• @TobyMak I don't think I understand the smarter way that you are talking about. Do you mean putting values? Commented Jul 20, 2018 at 7:35
• If $x < 0$, $x$ is negative but $|x-1|$ is positive, so $x|x-1|$ is negative and $<0$. If $0<x<1$, $x < 1$ and $|x-1| ≤ 1$, so $x|x-1|$ is at most $1$. Now you can check the cases $1, 2, 3$ separately. Commented Jul 20, 2018 at 7:37

Let's do it like this. For $x>1$ the equation is equivalent to $x(x-1)<3$ this gives $x^2-x-3<0$ Roots of this equation are $$\frac{1+\sqrt{13}}{2}, \frac{1-\sqrt{13}}{2}$$ So this gives the range $x\in(1, \frac{1+\sqrt{13}}{2})$

For $x<1$ the equation is $-x(x-1)<3$ or $x^2-x+3>0$ this always satisfied for any values of $x$. Hence $x\in(-\infty, \frac{1+\sqrt{13}}{2})$, taking $x$ to be integers would make this $x\le2$.

• The inequality trivially satisfied for non-positive $x$.
– A.Γ.
Commented Jul 20, 2018 at 7:25
• @A.Γ. How about when $0 < x < 1$? (I just realised when that happens, $x|x-1| < 1 \cdot 1 < 3$.) Commented Jul 20, 2018 at 7:27
• @PiyushDivyanakar $x=3$ doesn't work as well. Commented Jul 20, 2018 at 7:34
• I have corrected. Commented Jul 20, 2018 at 7:36
• Nitpicking: you're neglecting the case $x=1$. Commented Jul 20, 2018 at 13:44

This method doesn't generalize, but it may be instructive. The inequality surely holds for $x\le0$, so we can concentrate on $x>0$. Dividing both sides by $x$, we obtain $$|x-1|<\frac{3}{x}$$ that's equivalent to $$-\frac{3}{x}<x-1<\frac{3}{x}$$ that in turn becomes \begin{cases} x^2-x+3 > 0 \\[4px] x^2-x-3 < 0 \end{cases} The top one is true for every $x$; the bottom one is satisfied for $$0<x<\frac{1+\sqrt{13}}{2}$$ (recall we're restricting to $x>0$).

Putting back the interval $x\le0$ we discussed before, we can conclude the solution set is $$x<\frac{1+\sqrt{13}}{2}$$

Consider the cases where $x > 1$ and where $x < 1$.

If $x > 1$, we have: $$x(x-1) < 3$$ $$\Rightarrow x^2-x-3 < 0$$

With $x=2$, we have $2^2-2-3 = -1 < 0$, and with $x = 3$ we have $3^2 - 3 - 3 = 3 > 0$, so $x ≤ 2$.

If $x < 1$, we have: $$x(x+1) < 3$$ $$\Rightarrow x^2+x-3 < 0$$

Can you continue?

Consider the graph of $y=x(x-1)$:

$\hspace{2cm}$

If the absolute values are applied to get $y=x|x-1|$, the graph becomes:

$\hspace{2cm}$

The point $A$ has coordinates $x(x-1)=3 \Rightarrow x=\frac{1+\sqrt{13}}{2}$ and $y=3$.

Hence, the solution of the inequality $x|x-1|<3$ is: $$x\in \left(-\infty,\frac{1+\sqrt{13}}{2}\right).$$