|x-1|+|x-2|+|x-3|=2
$\displaystyle \displaystyle \begin{align*} |x - 1| + |x - 2| + |x - 3| &= 2 \\ |x - 1| + |x - 2| &= 2 - |x - 3| \\ \left(|x - 1| + |x - 2|\right)^2 &= \left(2 - |x - 3|\right)^2 \\ (x - 1)^2 + 2|x - 1||x - 2| + (x - 2)^2 &= 4 - 4|x - 3| + (x - 3)^2 \\ (x - 1)^2 + (x - 2)^2 - (x - 3)^2 - 4 &= -4|x - 3| - 2|x - 1||x - 2| \\ x^2 - 8 &= -4|x - 3| - 2|x^2 - 3x + 2| \\ \left(x^2 - 8\right)^2 &= \left(-4|x - 3| - 2|x^2 - 3x + 2|\right)^2 \\ x^4 - 16x^2 + 64 &= (-4)^2(x - 3)^2 + 16|x - 3||x^2 - 3x + 2| + (-2)^2(x^2 - 3x + 2)^2 \\ x^4 - 16x^2 + 64 &= 4x^4-24x^3+68x^2-144x+160 + 16|x^3-6x^2+11x-6| \\ -3x^4 + 24x^3 - 84x^2 + 144x - 96 &= 16|x^3 - 6x^2 + 11x - 6| \\ \left( -3x^4 + 24x^3 - 84x^2 + 144x - 96 \right)^2 &= 16^2 \left( x^3 - 6x^2 + 11x - 6 \right)^2 \\ 9x^8-144x^7+1080x^6-4896x^5+14544x^4-28800x^3+36864x^2-27648x+9216 &= 256x^6-3072x^5+14848x^4-36864x^3+49408x^2-33792x+9216 \\ 9x^8-144x^7+824x^6-1824x^5-304x^4+8064x^3-12544x^2+6144x &= 0 \\ x(x-6)(x-4)(x-2)^2(x+2)(3x-8)(3x-4) &= 0 \end{align*}$
$\displaystyle \displaystyle x = -2\textrm{ or }x = 0\textrm{ or }x = \frac{4}{3}\textrm{ or }x = 2\textrm{ or }x=\frac{8}{3}\textrm{ or }x = 4\textrm{ or }x = 6 $
These are POSSIBLE solutions, as squaring often brings in extraneous solutions. You will need to check them in your original equation to see which ones are true solutions.
$\displaystyle |x-1|=1-x$ if $\displaystyle x<1,$ and $\displaystyle =x-1$ if $\displaystyle x>1$
Similarly,
$\displaystyle |x-2|=2-x$ if $\displaystyle x<2,$ and $\displaystyle =x-2$ if $\displaystyle x>2$,
$\displaystyle |x-3|=3-x$ if $\displaystyle x<3,$ and $\displaystyle =x-3$ if $\displaystyle x>3$.
So, if $\displaystyle x<1$ the LHS will be $\displaystyle (1-x)+(2-x)+(3-x)=6-3x.$
If $\displaystyle 1<x<2,$ the LHS will be $\displaystyle (x-1)+(2-x)+(3-x)=4-x.$
Similarly for the other two ranges.
For the first of these, $\displaystyle 6-3x=2$ if $\displaystyle x=4/3.$ This lies outside the range $\displaystyle x<1,$ so no solution for that one. Repeat for the other three.
now i figured out this , but if i continue like this ,equation has no choice ,none matches ,but in the book choice is =2 , i want to say -> did u forgot to put 1<=x<2 instead 1<x<2 or something like this?? #appreciate your help!
Yes, my apologies, all of the ranges should include equals signs.
You can graph the LHS if you wish.
Draw
$\displaystyle y=6-3x$ for $\displaystyle x\leq1$,
$\displaystyle y=4-x$ for $\displaystyle 1\leq x \leq 2,$
$\displaystyle y=x$ for $\displaystyle 2\leq x\leq 3,$
and
$\displaystyle y=3x-6$ for $\displaystyle 3 \leq x.$