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Math Help - Pretty basic problem

  1. #1
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    Pretty basic problem

    the problem is .055=(.0100-2x)^2/x^2. This becomes .055=(.0001-4x^2)/x^2. I multiplied both sides by x^2 to get .055x^2=.0001-4x^2, then added 4x^2 to each side to get 4.055x^2=.0001, divided both sides by 4.055 to get x^2=2.466*10^-5 and took the square root of both sides to get x=.004965 or 5.0*10^-3. My book however says the answer is 4.4*10^-3 so what have I done wrong? This is actually a chemistry problem so 4.9 to 4.4 is a bigger difference than I am comfortable with.
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  2. #2
    Member integral's Avatar
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    .055=\frac{(.100-2x)^2}{x^2}

    .055=\frac{(.01-2x)^2}{x^2}

    Note:

    .1^2=.01\neq.0001

    that is your mistake.
    Also, (a+b)^2 = a^2+2ab+b^2
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  3. #3
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    it is .01 not .1 in the original problem
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  4. #4
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    Actually, you are both wrong. (a+ b)^2 is NOT equal to a^2+ b^2, it is equal to a^2+ 2ab+ b^2. (.1- 2x)^2 is NOT equal to .01- 4x^2, it is equal to .01- .4x+ 4x^2.
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  5. #5
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    There is no .1!! the original problem is .055=(0.01-2x)^2/x^2
    There is no .1
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  6. #6
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    Quote Originally Posted by mgrade2000 View Post
    There is no .1!! the original problem is .055=(0.01-2x)^2/x^2
    There is no .1
    0.055x^2 = \left(10^{-2} - 2x \right)^2 = 10^{-4} - 4 \times 10^{-2}x + 4x^2

    3.945x^2- 4 \times 10^{-2}x + 10^{-4} = 0

    I suggest using the quadratic formula to solve this one

    I get the roots to be 5.66 \times 10^{-3} and x \approx 4.48 \times 10^{-3}
    Last edited by e^(i*pi); March 28th 2010 at 02:17 PM. Reason: wrong value of a
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  7. #7
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    thanks a lot
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