$$|x^2-2x+2|-|2x^2-5x+2|=|x^2-3x|$$ Find the set of values of $x$.

The answer given is $[0,\frac12]\cup [2,3]$.

What I don't get is how is the solution a range? I'm getting the solution as $0,\frac 12, 2 ,3, \frac 25$. That makes sense to me as the first expression is always positive, $$\because x^2-2x+1+1=(x-1)^2+1> 0$$ and the other two expressions give two cases which on being solved gives 4 solutions in total as it's a quadratic.


Using wavy curve on the two expressions inside mod, we get that they are negative for $x\in (\frac 12,2)$ and $x\in [0,3]$ respectively. So, we can conclude that if the first expression is negative then the second expression is negative too as $[\frac 12,2]$ is in $[0,3]$. So we can get 3 cases:

Both are positive, or both are negative, or the first one is positive and the second one is negative.

Solving for each gives the solutions which I said I've received earlier in the post.

So where is the range coming from? Am I interpreting the mod operator wrong? Or is my book wrong?

  • $\begingroup$ "as it's a quadratic" It's not always a quadratic. Recheck your calculations, or post them. $\endgroup$
    – dxiv
    Jun 11, 2022 at 4:23
  • $\begingroup$ @dxiv I found an error which I've rectified and edited in my post $\endgroup$ Jun 11, 2022 at 4:41
  • $\begingroup$ Show your calculations for (at least) one of the intervals, otherwise no one can guess where you went wrong. $\endgroup$
    – dxiv
    Jun 11, 2022 at 4:49
  • $\begingroup$ It is just a matter of grinding through the 5 alternatives, writing the expression with the $|\cdot|$ appropriately removed and checking. $\endgroup$
    – copper.hat
    Jun 11, 2022 at 6:43

3 Answers 3


As you've mentioned, $|x^2-2x+2| = |(x-1)^2+1|$ is always positive.
So $|x^2-2x+2| =x^2-2x+2$.

$|2x^2-5x+2| = \begin{cases}2x^2-5x+2 & \text{for } x \in (-\infty,0.5] \cup [2,\infty)\\ -(2x^{2}-5x+2) &\text{for } x \in [0.5,2]\end{cases}$


$|x^2-2x+2| - |2x^2-5x+2| = \begin{cases}-x^2+3x &\text{for } x \in (-\infty,0.5]\cup [2,\infty)\\3x^2-7x+4&\text{for } x \in [0.5,2]\end{cases}$

and, $|x^2-3x| = \begin{cases}x^2-3x&\text{for } x \in (-\infty,0]\cup [3,\infty) \\ -(x^2-3x) &\text{for } x \in [0,3]\end{cases}$

Now segregating a little bit,
$|x^2-2x+2| - |2x^2-5x+2| = \begin{cases}-x^2+3x &\text{for } x \in (-\infty,0]\cup[0,0.5]\cup [2,3] \cup[3,\infty)\\3x^2-7x+4&\text{for } x \in [0.5,2]\end{cases}$


$|x^2-3x| = \begin{cases}x^2-3x&\text{for } x \in (-\infty,0]\cup [3,\infty) \\ -x^2+3x &\text{for } x \in [0,0.5]\cup[0.5,2]\cup[2,3]\end{cases}$

So, the common range for which, $|x^2-2x+2| - |2x^2-5x+2|=|x^2-3x|$ is $[0,0.5] \cup [2,3]$


Your book is not wrong. You can try plotting the functions $x^2-2x+2$ and $|2x^2-5x+2| + |x^2 - 3x|$ in e.g. Wolfram Alpha and you'll see that they overlap on precisely the segment given by the answer.

As to why the solution is a range, write the equation as $$x^2 - 2x + 2 = |x||x-3|+2|x-\frac 12||x-2|$$ For each of the 4 factors there is a range for which the $| \cdot |$ can be changed to $(\cdot)$, or for which it can be changed to $(\cdot)$ provided that whatever is inside is negated.

For $x\in (-\infty,0)\cup(\frac 12,2)\cup(3,+\infty)$ we simply convert $|\cdot|$ to $(\cdot)$ (because in each term either both factors are negated, or both are positive) and after we move everything to the LHS we end up with a quadratic, hence a finite amount of solutions.

But, for $x\in [0,\frac 12]$ the term $|x||x-3|$ becomes $x(3-x)$ and $2|x-\frac 12||x-2|$ becomes $2(x-\frac 12)(x-2)$ (since both factors have to be negated in that range).

Now the RHS is equal to $-x^2+3x + 2x^2-5x + 2$ which is equal to $x^2-2x+2$. But this is precisely the LHS! So $0=0$ for $x\in [0,\frac 12]$, i.e. on that interval the curves $x^2 - 2x + 2$ and $|x||x-3|+2|x-\frac 12||x-2|$ overlap.

Similar reasoning follows for $x\in[2,3]$.



Since $x^2-2x+2 = x^2-2x+1+1=(x-1)^2 +1 > 0$, $|x^2-2x+2|-|2x^2-5x+2|= x^2-2x+2-|2x^2-5x+2|=|x^2-3x|$


$2x^2-5x+2=(2x-1)(x-2)=0$, $x=\frac{1}{2}$, and $x=2$

$|2x^2-5x+2|= 2x^2-5x+2$ when $-\infty \leq x \leq\frac{1}{2}$ or $2 \leq x \leq\infty$

$|2x^2-5x+2|= -(2x^2-5x+2)$ when $\frac{1}{2}< x < 2$


$x^2-3x=x (x-3)=0$ then $x=0$ or $x=3$

$|x^2-3x|=x^2-3x$ when $-\infty \leq x \leq 0$ or $3 \leq x \leq\infty$

$|x^2-3x|=-(x^2-3x)$ when $0 < x < 3$

After arranging intervals:

$|x^2-2x+2|-|2x^2-5x+2|=|x^2-3x|$ will be

$1.$ $x^2-2x+2-2x^2+5x-2=x^2-3x$, when, $-\infty \leq x \leq 0$ or $3 \leq x \leq\infty$, then, $x=0$ or $x=3$

$2.$ $x^2-2x+2-2x^2+5x-2=-x^2+3x$, when, $0 < x \leq \frac{1}{2}$ or $2 \leq x < 3$ , then, $0=0$ ,so, solution is $\forall x$ in these intervals.

$3.$ $x^2-2x+2+2x^2-5x+2=-x^2+3x$, when, $\frac{1}{2} < x < 2 $ , then, $2x^2-5x+2=0$, $x=\frac{1}{2}$, and $x=2$ ,so, no solution.


$0 \leq x \leq \frac{1}{2}$ or $2 \leq x \leq3$


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