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Math Help - Absolute Value

  1. #1
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    Absolute Value

    I'm self-teaching, so I'm going back and refreshing myself and I'm trying to figure something out.

    -3|x-2|=3

    I know that the answer is "no real solutions," but my question is why?

    Thanks!
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Absolute Value

    If you divide every side by -3 then you get:
    |x-2|=-1

    Do you have a conclusion now?
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  3. #3
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    Re: Absolute Value

    So, whenever I have a problem like that I have to get the absolute value by itself before I can solve it?
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Absolute Value

    It depends on what the question is, but in this case this looks the easiest way to me, |x-2|=-1 is false because offcourse \forall x \in \mathbb{R}: |x-2|\geq0
    Is this an answer to your question?
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  5. #5
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    Re: Absolute Value

    Yes, thank you.
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: Absolute Value

    You're welcome!
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  7. #7
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    Re: Absolute Value

    Out of curiosity, what about |x^2+2|=4x+2?

    I'm just struggling with the concept of solving this I guess. I thought I had a grasp on the rules, but obviously not. In that problem, how would I solve that?
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  8. #8
    MHF Contributor Siron's Avatar
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    Re: Absolute Value

    When you have to solve equations with absolut value signes then you have to work with the definition mostly (do you know the definition?)
    In this case, what can you say about |x^2+2|? (are the absolut value signs necessary? Why? Why not?)
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  9. #9
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    Re: Absolute Value

    It means that x can either be x or -x, correct?
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  10. #10
    MHF Contributor Siron's Avatar
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    Re: Absolute Value

    Yes, but for which conditions for x? We can say that |x|=x if x\geq0 and |x|=-x if x<0.
    That means for example |x-3|=x-3 if x\geq3 and |x-3|=-(x-3) if x<3

    But in this case, what can you say about |x^2+2|?
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  11. #11
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    Re: Absolute Value

    in general, with absolute value expressions, there are two cases:

    1. the stuff inside the absolute value sign is negative, use the rule |x| = -x when x < 0
    2. the stuff inside the absolute value sign is non-negative, use the rule |x| = x, when x ≥ 0.

    let's look at your two problems in this light, to see if our method gives the same answer as before:

    -3|x - 2| = 3.

    case 1: x - 2 < 0. note this also means x < 2.

    -3(2 - x) = 3 (replacing the stuff inside the absolute value sign by its negative)
    -6 - (-3)x = 3
    -6 + 3x = 3
    3x = 9
    x = 3. but 3 is not less than two, this is a contradiction.

    case 2: |x - 2| ≥ 0, so x ≥ 2.

    -3(x - 2) = 3 (removing the absolute value signs since we are assuming what's inside is non-negative)
    -3x + 6 = 3
    -3x = -3
    x = 1. another contradiction, 1 is not greater than or equal to 2.

    conclusion, no solution for x.

    |x^2 + 2| = 4x + 2

    case 1: x^2 + 2 < 0, so x^2 < -2, this is impossible, a square cannot be negative.

    case 2: x^2 + 2 ≥ 0, so x^2 ≥ -2, true for any x.

    x^2 + 2 = 4x + 2
    x^2 = 4x

    x = 0 will work. let's see if there are any non-zero solutions.

    assuming x ≠ 0:

    x = 4 (dividing by x).

    so case 2 yields the solutions: x = 0,4

    this method, although it can be long-winded (much like myself at times) is the easiest way to "keep the signs straight", and should always work.
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