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Math Help - Does not compute!

  1. #1
    Junior Member
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    Jul 2007
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    Does not compute!

    A word problem is given which I've answered correctly, but a second question doesn't add up. The problem is:

    A North-South highway intersects an East-West highway at a point P. An automobile crosses P at 10:00 AM, traveling east at a constant rate of 20 mph. At that same instant another automobile is two miles north of P, traveling south at 50 mph. Find a formula which expresses the distance d between the automobiles at time t (hours) after 10:00 AM. At what time will the automobiles be 104 miles apart?

    My answer, which the book confirms, is

    d = \sqrt{2900t^2 - 100t + 4}

    To answer the second question, I started with

    104 = \sqrt{2900t^2 - 100t + 4}

    \Rightarrow 10816 = 2900t^2 -100t + 4

    \Rightarrow 2704 = 725t^2 - 25t + 1

    \Rightarrow 725t^2 - 25t - 2703 = 0

    Using the Quadratic Formula, I get

    \frac {25 \pm \sqrt {625 + 7838700}} {1450}

    which I can only reduce to

    \frac {5 \pm \sqrt {313573}} {290}

    But the answer given in the book is:

    \frac {1 + \sqrt 30} {29}

    So how do I get there? Did I make a mistake along the way?
    Last edited by earachefl; July 16th 2007 at 06:19 AM. Reason: mistake
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello,earachefl!

    A North-South highway intersects an East-West highway at a point P.
    An automobile crosses P at 10:00 AM, traveling east at a constant rate of 20 mph.
    At that same instant another automobile is two miles north of P, traveling south at 50 mph.
    (a) Find a formula which expresses the distance d between the automobiles
    . . .at time t (hours) after 10:00 AM.
    (b) At what time will the automobiles be 104 miles apart?

    I don't agree with either of their answers . . .
    Code:
              Q *
                |
            50t |
                |
              B *
                |  *   d
          2-50t |     *
                |        *
          - - P * - - - - - * - -
                |    20t    A
                |

    At 10 AM, the first car is at point P.
    . . t hours later, it is at point A\!:\;PA = 20t

    At 10 AM, the second car is at point Q\!:\;QP = 2.
    . . t hours later, it is at point B\!:\;QB = 50t.
    . . Hence: . BP = 2-50t

    Pythagorus says: . d^2\:=\:(20t)^2 + (2-50t)^2

    (a) Then we have: . \boxed{d \:=\:\sqrt{2900t^2 - {\color{red}200}t + 4}}


    If d = 104\!:\;\;\sqrt{2900t^2 - 200t + 4} \:=\:104

    Square both sides: . 2900t^2 - 200t + 4 \:=\:10816\quad\Rightarrow\quad 2900t^2 - 200t - 10812 \:=\:0

    Divide by 4: . 725t^2 - 50t - 2703 \:=\:0

    Quadratic Formula: . t \;=\;\frac{-(-50) \pm\sqrt{(-50)^2 - 4(725)(-2703)}}{2(725)}

    . . t \;=\;\frac{50\pm\sqrt{7841200}}{1450} \;=\;\frac{50 \pm 20\sqrt{19603}}{1450} \;=\;\frac{5 \pm 2\sqrt{19603}}{145}


    And the positive root is: . \boxed{t \:=\:\frac{5 + 2\sqrt{19603}}{145} \:\approx\:1.966\text{ hours.}}

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