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Basic algebra responsible for brain fart

  1. Jun 24, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm trying to solve for x in a chemical equilibrium problem (college chem). Once x is found I have no problem finding equilibrium concentrations. What's important is below.

    2. Relevant equations

    5.10 = [(1+x)/(1-x)]^2 --------> x=.387 (right answer)

    3. The attempt at a solution

    I've tried to take the square root of the right side and cross multiply and solve for x that way, but I keep getting 1. I also tried to expand the numerator with the common denominator to see if I missed something, but solving for x that way eventually got me 1 also. I know there's something simple I'm missing.
     
  2. jcsd
  3. Jun 24, 2010 #2

    CompuChip

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    Cross multiplying should work. Then you can work out the squares (don't forget that (x + 1)^2 = x^2 + 2x + 1) and bring it to the form
    a x^2 + b x + c = 0.
     
  4. Jun 24, 2010 #3
    That worked awesomely! What I don't understand is why I wasn't able to work it out by taking the square root of both sides. Do you have any insight into that? Not sure if that question's too vague.
     
  5. Jun 24, 2010 #4
    You can do that, but you may need to consider both positive and negative square root of 5.10,
     
  6. Jun 24, 2010 #5

    statdad

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    Physical constraints may tell you that x must be positive - and, if x represents a concentration (I can't tell from your original question) then you know 0 < x < 1, and solving by taking square roots works fine, without the need to consider [itex] -\sqrt{5.10} [/itex]
     
  7. Jun 24, 2010 #6

    CompuChip

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    That's not true statdad. The equation
    [tex]\frac{1 + x}{1 - x} = -\sqrt{5.10}[/tex]
    can still give a positive solution for x (there is a minus sign in the denominator) and, in fact, will.
     
  8. Jun 24, 2010 #7

    statdad

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    Nope: if 0 < x < 1 as I specified, then

    [tex]
    \begin{align*}
    0 > -x & > -1 \\
    1 > 1 - x & > 0
    \end{align*}
    [/tex]

    and no negative term arises.
     
  9. Jul 3, 2010 #8
    This is correct.
    I am guessing this is an equilibrium problem involving an ICE table. If so, then i am going to assume x represents a change in concentration. One of the values obtained for x will not be possible as it will be greater than the initial concentration. Using the positive square-root, the solution should be x=.387. using the negative squareroot, the value for x obtained (if i computed correctly) should be x=2.59. concentrations can be greater than one but if any of the initial concentrations of reagents were smaller than 2.59M, then this value would not be possible, and then the only true solution would be x=.387M (assuming you are working in Molars). MurdocJensen could you post the full question so I could verify this?
    Thanks
    -Theorem
     
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