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Math Help - Solve Equation with Multiple Absolute Values

  1. #1
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    Solve Equation with Multiple Absolute Values

    |x-1|+|x-2|+|x-3|=2
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  2. #2
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    re: Solve Equation with Multiple Absolute Values

    Quote Originally Posted by deathmate View Post
    |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.
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  3. #3
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    Re: Solve Equation with Multiple Absolute Values

    |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.
    Last edited by Ackbeet; September 7th 2011 at 04:10 AM. Reason: Retitled.
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  4. #4
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    Re: Solve Equation with Multiple Absolute Values

    Quote Originally Posted by deathmate View Post
    |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.
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    Re: Solve Equation with Multiple Absolute Values

    Quote Originally Posted by BobP View Post
    |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
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    Re: Solve Equation with Multiple Absolute Values

    Quote Originally Posted by deathmate View Post
    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.
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    Re: Solve Equation with Multiple Absolute Values

    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!
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  8. #8
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    Re: Solve Equation with Multiple Absolute Values

    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.
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    Re: Solve Equation with Multiple Absolute Values

    thank u so much , I stuck on this for a week ,looking around and finally found help (Y)
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    Re: Solve Equation with Multiple Absolute Values

    Quote Originally Posted by Plato View Post
    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??
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    Re: Solve Equation with Multiple Absolute Values

    Quote Originally Posted by psolaki View Post
    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.
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  12. #12
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    Re: Solve Equation with Multiple Absolute Values

    Quote Originally Posted by Plato View Post
    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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