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Math Help - Quadratics question... AGAIN!!! :D

  1. #1
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    Quadratics question... AGAIN!!! :D

    Was wondering if someone could please walk me through how you would do this question:

    A ball is thrown into the air. A sensor measures the height of the ball above the ground. The height of the ball is described by the equation y = -5x(x-4), where x represents the time in seconds. Determine the intervals of time when the ball is above a height of 15m.

    Thanks for your help guys!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bliks View Post
    Was wondering if someone could please walk me through how you would do this question:

    A ball is thrown into the air. A sensor measures the height of the ball above the ground. The height of the ball is described by the equation y = -5x(x-4), where x represents the time in seconds. Determine the intervals of time when the ball is above a height of 15m.

    Thanks for your help guys!
    you need to solve -5x(x - 4) > 15 ...........this is where the height is greater than 15

    \Rightarrow -5x^2 + 20x - 15 > 0 or x^2 - 4x + 3 < 0

    can you continue?
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  3. #3
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    I can thank you for your time.
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  4. #4
    Member OnMyWayToBeAMathProffesor's Avatar
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    I think there might be a mistake some where. either with Jhevon or bliks. when you graph \Rightarrow -5x^2 + 20x - 15 > 0 the highest point is 5 when x is 1.999999. Maybe I did it wrong but that is what i got.
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  5. #5
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    I think there might be a mistake some where. either with Jhevon or bliks. when you graph \Rightarrow -5x^2 + 20x - 15 > 0 the highest point is 5 when x is 1.999999. Maybe I did it wrong but that is what i got.
    You are correct that 5 is the maximum of that parabola (but it occurs at x = 2, not 1.999999, as simple differentiation will show), but I fail to see how this affects the problem.
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  6. #6
    Member OnMyWayToBeAMathProffesor's Avatar
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    so wouldn't that mean that the ball never passes over the 15m mark?
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  7. #7
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    so wouldn't that mean that the ball never passes over the 15m mark?
    No. Reread Jhevon's solution.

    The ball's height will be greater than 15 m at those values x which satisfy

    -5x(x - 4) > 15.

    So we solve the inequality:

    -5x(x - 4) > 15

    \Rightarrow -5x^2 + 20x - 15 > 0 (As you can see, we need this parabola to give values greater than 0, not 15)

    \Rightarrow x^2 - 4x + 3 < 0

    \Rightarrow(x - 3)(x - 1) < 0

    \Rightarrow1 < x < 3

    and we have our solution.
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