Solve Equation with Multiple Absolute Values

**deathmate** · September 7th 2011, 12:23 AM

|x-1|+|x-2|+|x-3|=2

**Prove It** · September 7th 2011, 12:59 AM

Originally Posted by deathmate

|x-1|+|x-2|+|x-3|=2

$\displaystyle \begin{align*} |x - 1| + |x - 2| + |x - 3| &= 2 \\ |x - 1| + |x - 2| &= 2 - |x - 3| \\ \left(|x - 1| + |x - 2|\right)^2 &= \left(2 - |x - 3|\right)^2 \\ (x - 1)^2 + 2|x - 1||x - 2| + (x - 2)^2 &= 4 - 4|x - 3| + (x - 3)^2 \\ (x - 1)^2 + (x - 2)^2 - (x - 3)^2 - 4 &= -4|x - 3| - 2|x - 1||x - 2| \\ x^2 - 8 &= -4|x - 3| - 2|x^2 - 3x + 2| \\ \left(x^2 - 8\right)^2 &= \left(-4|x - 3| - 2|x^2 - 3x + 2|\right)^2 \\ x^4 - 16x^2 + 64 &= (-4)^2(x - 3)^2 + 16|x - 3||x^2 - 3x + 2| + (-2)^2(x^2 - 3x + 2)^2 \\ x^4 - 16x^2 + 64 &= 4x^4-24x^3+68x^2-144x+160 + 16|x^3-6x^2+11x-6| \\ -3x^4 + 24x^3 - 84x^2 + 144x - 96 &= 16|x^3 - 6x^2 + 11x - 6| \\ \left( -3x^4 + 24x^3 - 84x^2 + 144x - 96 \right)^2 &= 16^2 \left( x^3 - 6x^2 + 11x - 6 \right)^2 \\ 9x^8-144x^7+1080x^6-4896x^5+14544x^4-28800x^3+36864x^2-27648x+9216 &= 256x^6-3072x^5+14848x^4-36864x^3+49408x^2-33792x+9216 \\ 9x^8-144x^7+824x^6-1824x^5-304x^4+8064x^3-12544x^2+6144x &= 0 \\ x(x-6)(x-4)(x-2)^2(x+2)(3x-8)(3x-4) &= 0 \end{align*}$

$\displaystyle x = -2\textrm{ or }x = 0\textrm{ or }x = \frac{4}{3}\textrm{ or }x = 2\textrm{ or }x=\frac{8}{3}\textrm{ or }x = 4\textrm{ or }x = 6$

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.

**BobP** · September 7th 2011, 04:08 AM

$|x-1|=1-x$ if $x<1,$ and $=x-1$ if $x>1$
Similarly,
$|x-2|=2-x$ if $x<2,$ and $=x-2$ if $x>2$ ,
$|x-3|=3-x$ if $x<3,$ and $=x-3$ if $x>3$ .

So, if $x<1$ the LHS will be $(1-x)+(2-x)+(3-x)=6-3x.$
If $1<x<2,$ the LHS will be $(x-1)+(2-x)+(3-x)=4-x.$
Similarly for the other two ranges.

For the first of these, $6-3x=2$ if $x=4/3.$ This lies outside the range $x<1,$ so no solution for that one. Repeat for the other three.

**Plato** · September 7th 2011, 04:35 AM

Originally Posted by deathmate

|x-1|+|x-2|+|x-3|=2

This is more of a concept question than it is a question to be worked out.
The value of $|x-1|$ is the distance from x to 1.
So what number is such the its distances from 1 and from 2 and from 3 add up 2? There is just one.

**deathmate** · September 7th 2011, 05:21 AM

Originally Posted by BobP

$|x-1|=1-x$ if $x<1,$ and $=x-1$ if $x>1$
Similarly,
$|x-2|=2-x$ if $x<2,$ and $=x-2$ if $x>2$ ,
$|x-3|=3-x$ if $x<3,$ and $=x-3$ if $x>3$ .

So, if $x<1$ the LHS will be $(1-x)+(2-x)+(3-x)=6-3x.$
If $1<x<2,$ the LHS will be $(x-1)+(2-x)+(3-x)=4-x.$
Similarly for the other two ranges.

For the first of these, $6-3x=2$ if $x=4/3.$ This lies outside the range $x<1,$ so no solution for that one. Repeat for the other three.

dude ,could you solve this all to the end because i cant understand the way of solving , and the other thing i dont get it , why you solved this like |1-x|+|2-x|+|3-x|=2 , when the equation is |x-1|+|x-2|+|x-3|=2 ///thank you for all

**Plato** · September 7th 2011, 06:03 AM

Originally Posted by deathmate

dude ,could you solve this all to the end because i cant understand the way of solving

Did you read reply #4?
There is only one number that is the solution to that equation/
If you figure it out, then you will have a fuller understanding of absolute value.

**deathmate** · September 7th 2011, 07:30 AM

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!

**BobP** · September 7th 2011, 09:22 AM

Yes, my apologies, all of the ranges should include equals signs.

You can graph the LHS if you wish.

Draw

$y=6-3x$ for $x\leq1$ ,

$y=4-x$ for $1\leq x \leq 2,$

$y=x$ for $2\leq x\leq 3,$

and

$y=3x-6$ for $3 \leq x.$

**deathmate** · September 7th 2011, 10:34 AM

thank u so much , I stuck on this for a week ,looking around and finally found help (Y)

**psolaki** · September 7th 2011, 11:47 AM

Originally Posted by Plato

Did you read reply #4?
There is only one number that is the solution to that equation/
If you figure it out, then you will have a fuller understanding of absolute value.

And that no should be :x-1 +x-2 +x-3 =2 => 3x-6 =2 => x= 8/3 .

So this N0 is the only solution to the equation?? is that it??

**Plato** · September 7th 2011, 11:56 AM

Originally Posted by psolaki

And that no should be :x-1 +x-2 +x-3 =2 => 3x-6 =2 => x= 8/3 .
So this N0 is the only solution to the equation?? is that it??

I have absolutely no idea what you are doing there.
The answer is $x=2$
The distance from 1 to 2 is 1.
The distance from 2 to 2 is 0.
The distance from 3 to 2 is 1.
AND $1+0+1=2$ .

**deathmate** · September 8th 2011, 12:21 AM

Originally Posted by Plato

I have absolutely no idea what you are doing there.
The answer is $x=2$
The distance from 1 to 2 is 1.
The distance from 2 to 2 is 0.
The distance from 3 to 2 is 1.
AND $1+0+1=2$ .

thats it PLATO , the answer is x=2

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