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Math Help - Phys Chem blues

  1. #1
    Newbie
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    Phys Chem blues

    Hi all,

    Im having trouble solving for x in this equation.

    x
    4.95 = ______________
    (0.7 - x)(0.7 - x)

    I have tried re-teaching myself the quadratic equation, completing the square etc.. but dont know how to apply it here.

    I know the answer is 0.412 or -1.19.

    I will need this skill for an upcoming exam without a doubt.

    Please help.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Chicago, IL
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    Quote Originally Posted by chem_student View Post
    Hi all,

    Im having trouble solving for x in this equation.

    x
    4.95 = ______________
    (0.7 - x)(0.7 - x)

    I have tried re-teaching myself the quadratic equation, completing the square etc.. but dont know how to apply it here.

    I know the answer is 0.412 or -1.19.

    I will need this skill for an upcoming exam without a doubt.

    Please help.
    Don't worry! I'm here to help!

    4.95=\frac{x}{(0.7-x)^2}

    Multiplying both sides by the denominator, we have the following:

    4.95(0.7-x)^2=x

    Expanding the (0.7-x)^2 term, we get the following:

    4.95(.49-1.4x+x^2)=x

    Distribute the 4.95 to get:

    4.59x^2-6.426x+2.2491=x

    Subtract x from both sides to get a quadratic:

    4.59x^2-7.426x+2.2491=0

    Now solve for x using the quadratic formula:

    x=\frac{-(-7.426)\pm\sqrt{(-7.426)^2-4(4.59)(2.2491)}}{2(4.59)}

    This gives us two solutions: \color{red}\boxed{x=.404} or \color{red}\boxed{x=1.214}.

    If you have any questions feel free to ask!
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  3. #3
    Super Member
    earboth's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Don't worry! I'm here to help!

    4.95=\frac{x}{(0.7-x)^2}

    Multiplying both sides by the denominator, we have the following:

    4.95(0.7-x)^2=x

    Expanding the (0.7-x)^2 term, we get the following:

    4.95(.49-1.4x+x^2)=x

    Distribute the 4.95 to get:

    \underbrace{4.59}_{wrong\ number}x^2-6.426x+2.2491=x *****

    Subtract x from both sides to get a quadratic:

    4.59x^2-7.426x+2.2491=0

    Now solve for x using the quadratic formula:

    x=\frac{-(-7.426)\pm\sqrt{(-7.426)^2-4(4.59)(2.2491)}}{2(4.59)}

    This gives us two solutions: \color{red}\boxed{x=.404} or \color{red}\boxed{x=1.214}.

    If you have any questions feel free to ask!
    Unfortunately you have changed the order of digits when calculating the solution.

    I've got x = 0.41163 or x = 1.19039
    The second solution is positive too
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