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Need help solving for x

  1. Feb 24, 2005 #1
    Hi,

    I have a mathematical problem coming from chemistry (dealing with pH). I've set everything up but can't do the math, I need the math steps to be shown to me.

    How do you solve for x in this equation?
    (x + 1*10^-8) (x) = 1.0*10^-14

    x should be equal to 9.5 x 10^-8, but what steps do you take to get this value? I have tried this problem for a while and keep getting stuck because there are 2 x's on the left side of the equation. I don't know where to begin anymore or what to do. Any help would be appreciated. Thanks in advance.
     
  2. jcsd
  3. Feb 24, 2005 #2

    Integral

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    please do not double post.

    see this thread.
     
  4. Feb 24, 2005 #3
    (x + 1*10^-8) (x) = 1.0*10^-14
    x^2 + 1*10^-8x=1*10^-14
    x^2 + 1*10^-8x - 1*10^-14 =0

    do you know quatratic formulas?
     
  5. Feb 24, 2005 #4
    Sorry Integral, I posted and about 10 minutes later I thought I posted in the wrong section so I edited my post and decided maybe I should be posting my question here instead. Thank you for pointing me in the right direction for the solution (I looked at the other thread).

    Thanks vincentchan, yes, I know the quadratic formula.
    - b +/- sqrt b^2 - 4(a)(c) over 2a
    I tried this at one point, but got confused and thought it was wrong. I'll try it again.
     
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