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Math Help - using quadratic equation

  1. #1
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    using quadratic equation

    a plane goes to an airport that is 4200km away
    on the way back (4200km) it goes 100km/h faster than before
    if the total time taken (there and back) is 13 hours, how fast was the airplane going originally (x)
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  2. #2
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    Use the formula

    r_{avg} = \frac{2r_1r_2}{r_1 + r_2}

    to obtain

    \frac{2(4200)}{13} = \frac{2x(x + 100)}{x + (x + 100)}
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  3. #3
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    and where do you go from there??

    + we never learned that equation
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  4. #4
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    If you would like, I can derive the equation. You would have to derive it (in some sense) in order to solve the problem.

    Continuing on:

    \frac{8400}{13} = \frac{2x^2 + 200x}{2x + 100}

    \frac{8400(2x + 100)}{13} = 2x^2 + 200x

    8400(2x + 100) = 13(2x^2 + 200x)

    16800x + 8400 = 26x^2 + 2600x

    26x^2 - 14200x - 8400 = 0

    x = \frac{14200 \pm \sqrt{14200^2 + 4(26)(8400)}}{52}

    Recognizing that the negative answer for x makes no sense, we have

    x = \frac{14200 + \sqrt{14200^2 + 4(26)(8400)}}{52}

    which is about 546.74 km/h.
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  5. #5
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    Quote Originally Posted by attack4u View Post
    a plane goes to an airport that is 4200km away
    on the way back (4200km) it goes 100km/hr faster than before
    if the total time taken (there and back) is 13 hours, how fast was the airplane going originally (x)
    distance = rate *time,
    so, time = distance /rate -----**

    time in going 4200km away = 4200/x hrs.
    time in going back = 4200 /(x +100) hrs.

    4200/x +4200/(x+100) = 13
    Clear the fractions, multiply both sides by x(x+100),
    4200(x+100) +4200(x) = 13(x)(x+100)
    4200x +420,000 +4200x = 13x^2 +1300x
    13x^2 -7100x -420,000 = 0
    Using the Quadratic formula,
    x = {7100 +,-sqrt[7100^2 -4(13)(-420,000))]} /2(13)
    x = 600 km/hr, or -53.85 km/hr

    Therefore, the plane was going originally at 600 km/hr. -----answer.
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