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Math Help - A silly problem...

  1. #1
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    A silly problem...

    I know this is a simple problem--I've already done 500 like it. I just can't seem to get the answer.

    The question asks you to solve inequalities.

    ((x-5)/2))^2 < (x-4)/8

    Any help would be much appreciated. I tried to set it equal to zero...
    I just can't figure it out.

    Thanks
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  2. #2
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    Quote Originally Posted by PakalCafe View Post
    I know this is a simple problem--I've already done 500 like it. I just can't seem to get the answer.

    The question asks you to solve inequalities.

    ((x-5)/2))^2 < (x-4)/8

    Any help would be much appreciated. I tried to set it equal to zero...
    I just can't figure it out.

    Thanks
    (\frac{x-5}{2})^2 < \frac{x-4}{8} Multiply by 4 to get:

    (x-5)^2 < \frac{x-4}{2} \Rightarrow x^2 -10x + 25 < \frac{x-4}{2} Multiply by 2 again to get:

    2x^2 -10x + 25 -x + 4 < 0

    Can you finish now?
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  3. #3
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    Quote Originally Posted by Defunkt View Post
    (\frac{x-5}{2})^2 < \frac{x-4}{8} Multiply by 4 to get:

    (x-5)^2 < \frac{x-4}{2} \Rightarrow x^2 -10x + 25 < \frac{x-4}{2} Multiply by 2 again to get:

    2x^2 -10x + 25 -x + 4 < 0

    Can you finish now?
    It's the WHOLE thing times two though...
    that is, 2x^2-20x+50-x+4 < 0 and that's where I got to.
    or am I wrong?
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  4. #4
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    oh wait a second

    i figured it out
    it [I]is[I] 2x^2-20x+50-X+4<0
    Then you factor it out and get
    (2x-9)(x-6)<0
    Then you get
    x<9/2 or x<6

    thanks
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  5. #5
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    oops, yes, forgot to multiply the rest of the left side.
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