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Math Help - Using the Quadratic Formula Part 2

  1. #1
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    Using the Quadratic Formula Part 2

    I was having difficulty solving a quadratic equation in which one has two different values for x as the numerator. Here is the problem:

    (0.100 + X)(X)/(0.100 - X) = 1.8 x 10^-5

    Any input on the above question will be greatly appreciated. Thank you to the individual who answered my last question.
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  2. #2
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    Quote Originally Posted by confused20 View Post
    I was having difficulty solving a quadratic equation in which one has two different values for x as the numerator. Here is the problem:

    (0.100 + X)(X)/(0.100 - X) = 1.8 x 10^-5

    Any input on the above question will be greatly appreciated. Thank you to the individual who answered my last question.
    Just like the last one you have to do a little work first.
    \frac{(0.1 + x)x}{0.1 - x} = 1.8 \times 10^{-5}

    Mulitply both sides by 0.1 - x:
    (0.1 + x)x = (1.8 \times 10^{-5})(0.1 - x)

    Now expand both sides:
    0.1x + x^2 = 1.8 \times 10^{-6} - (1.8 \times 10^{-5})x

    Get everything onto one sides of the equation:
    x^2 + (0.1 + 1 \times 10^{-5})x - 1.8 \times 10^{-6} = 0

    x^2 + 0.10001x - 0.000018 = 0

    Now use the quadratic formula.

    -Dan
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  3. #3
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    Isn't it .100018x instead of .10001x

    I calculated .100018x instead of .10001x. Am I correct in adding 0.1 to 1x10^-5 instead of adding .1 to 1 x 10^-5 like you did? Thank you in advance for your replay and all your help.
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