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Absolute value |x-3|^2 - 4|x-3|=12

  1. Jul 22, 2012 #1
    Solve |x-3|^2 - 4|x-3|=12

    The solution to this equation is -3, 9. But I'm not sure on the working.

    There is a hint to let u=|x-3|

    So I worked it out the following way.

    u^2 -4u = 12

    u^2 -4u -12 = 0

    (u + 2)(u - 6)=0

    u /= -2 and u /= 6

    |x-3|=-2 (no solution)
    |x-3|= 6, x = 9

    Now to work out negative part.

    u^2 -4u = -12

    u^2 -4u + 12 =0

    (no real roots)

    So would I use |x-3|=-6, x = -3
     
  2. jcsd
  3. Jul 22, 2012 #2

    eumyang

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    Homework Helper

    You're missing something here.
    |x - 3| = 6 means
    x - 3 = 6
    OR
    x - 3 = -6
    So you'll get the 2 solutions here.


    You don't need this, because the "negative part" was taken care of earlier.
     
  4. Jul 22, 2012 #3

    HallsofIvy

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    Staff Emeritus
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    In fact, your original question said '= 12'. It makes no sense to set it equal to -12.
     
  5. Jul 22, 2012 #4
    The u-substitution for |x - 3| should work fine, but another interesting thing to note is that the absolute value signs are not necessary for the first part of the equation. The power of 2 will make anything within the absolute value positive, so |x - 3|^2 is the same things as (x - 3)^2.

    Anyways, if you let u = |x - 3| it should be easier.

    [itex]u^2 - 4u = 12[/itex]

    Then, once you've found u, use the equation u = |x - 3| to solve for x.
     
  6. Jul 22, 2012 #5
    I understand now. I should just have solved the quadratic, then only applied the absolute value +/- to the u function. Thanks for your help.
     
  7. Jul 23, 2012 #6
    don't need the extra work when it's already perfect done imho
     
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