equation with absolute values

Question

I am stuck with solving this equation: $$ |x^2 - 4x +3 | + |x-1| + |x-2| -2x=0 $$

I tryed to raise it by power of 2 (including moving some of the factors to the right side, as well as factor the square polynomial) but it did not help. Any suggestions? thanks!

Answer 1

You need to split the domain of definition of the function ($\mathbb{R}$), with the values where one (or several) absolute values are $0$. Then, for every interval, get rid of the absolute values, knowing the sign of what is inside.

For example, between $-\infty$ and $1$ :

• $x^2-4x+3$ is positive ;
• $x-1$ is negative ;
• $x-2$ is negative.

So the equation becomes $(x^2-4x+3)-(x-1)-(x-2)-2x = 0$ and then $x^2-8x+6 = 0$. You then have to solve the equation and if the solutions lie between $[-\infty, 1]$, they are admissible.

Now, you need to do this for the other intervals !