# Doubt regarding validity of answer in a mod equation question

$$|x^2-2x+2|-|2x^2-5x+2|=|x^2-3x|$$ 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.

$$x^2-2x+2-|(x-2)(2x-1)|-|(x-0)(x-3)|=0$$

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?

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

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

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

So,

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

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

and,

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

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 $$x\in[2,3]$$.

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

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

then

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

and

$$x^2-3x=x (x-3)=0$$ 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.

Finally

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