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Math Help - real values for x

  1. #1
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    real values for x

    Hi, I have this part of a question that gets me:
    I did all the previous steps and got to an equation that must =0:

    x^2(-x+12)=0

    I know that making x=12 is one value, but for the x^2 part, do i just ignore it?
    In my answer book it says x=12 and nothing else, so I'm presuming its something to do with real values of x (even though I don't really know what a 'real' value of x is).

    Thanks
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  2. #2
    Super Member wingless's Avatar
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    For x \neq 0, divide both sides by x^2.

    \frac{x^2(12-x)}{x^2} = \frac{0}{x^2}

    12 - x = 0

    x = 12.

    We found x=12, which makes the equation true, but don't forget that we didn't test x for 0. If we do,
    0 (12 - 0) = 0

    Yes, x = 0 is a root too.

    x = \{0,12\}
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  3. #3
    Super Member wingless's Avatar
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    Another way (which you may find easier) to solve the equation:

    If the product of two values is 0, then at least one of them must be 0.

    \underbrace{(x^2)}_{x^2=0}\underbrace{(12-x)}_{12-x=0} = 0

    Either x^2 or 12 - x or both of them must be 0. So,

    x^2 = 0
    x = 0

    And,

    12-x = 0
    x = 12

    Thus,

    x = \{0, 12\}
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