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Homework Help: Another quadratic question

  1. Oct 4, 2006 #1
    i think im pretty much 'in the know' about this one.

    it's an equation about a football player kicking a football..
    h = -4.9t^2 + 10t + 3
    h = height (in meters)
    t = time (in seconds)

    so the vertex is 9.25m height at 1.25 seconds? (i figured this on my graphing calculator).

    and the other question asks how many seconds til the ball reaches the ground and by using the quadratic formula i got 2.3 seconds.

    then the last question has to do about how much time is the ball above the 5 meter mark.

    so this would be:
    5m = -4.9t^2 + 10t + 3?
    i dunno how i'd figure this out. any help is appreciated!

  2. jcsd
  3. Oct 4, 2006 #2
    The vertex is [tex] -\frac{b}{2a} [/tex]. So its [tex] (1.02, 8.10) [/tex]. Your second answer is correct.

    For the third one, you do the following

    [tex] -4.9t^{2} + 10t -2 = 0[/tex]
    [tex] t = 0.224 [/tex]

    Since we know that at time 1.02 seconds the football is at its highest point, then the ball spends [tex] 2(1.02-.224) = 1.592 [/tex] seconds above the 5m mark.
  4. Oct 4, 2006 #3
    im not quite seeing it yet.. where did the -2 come from? nevermind.. 3-5 = -2.. i see it now :) thanks
  5. Oct 4, 2006 #4
    thanks for the help :smile:

    here's another one that's a bit of a pickle. i've attempted it. it has to do with complex numbers.

    "find the equation of a quadratic that has 3 + i and 3- i as it's roots.

    = (x - x(3-i))(x - (3-i))
    = x^2 - x(3-i) - x(3+i) + (3+i)(3-i)
    =x^2 - 6x + 9 - i^2
    x^2 - 6x + 10

    does that look ok?

  6. Oct 4, 2006 #5
    yep, that is correct.
  7. Oct 4, 2006 #6


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    Yes, the situation is symmetric so the time going from the 5 m height to its highest point is equal to the time it spends going back down to the level. A little more direct:
    The quadratic equation [itex]-4.9t^2+ 10t- 2= 0[/itex] has two[\b], the smaller being when it reaches 5 m on the way up, the larger when it passes 5 m on the way down. The time it spends above 5 m is the difference between the two.
  8. Oct 5, 2006 #7
    so to calculate any vertex it's just -B/-2A? and that gives you p or q? and then whats the easy calculation for the other one? (p or q i dunno). so the 'c' value doesnt matter?

    thanks for the tips Hallsofivy

  9. Oct 5, 2006 #8
    yes [tex] -\frac{b}{2a} [/tex] is the vertex. The [tex] p [/tex] and [tex] q [/tex] you are talking about is part of the Rational Root Theorem.

    [tex] p [/tex] is all the factors of the constant term.
    [tex] q [/tex] is all the factors of the leading coefficient.

    So you can have [tex] y = x^{2}-6x+7 [/tex]
    [tex] \frac{p}{q} = \frac{{\pm1,\pm7}}{{\pm1}} [/tex]
  10. Oct 5, 2006 #9
    but how did you end up with two values for the vertex. like in your 2nd post in this thread you did the -b/-2a and got 1.02 and 8.10?

    im not sure what you mean by the root theorem. in my lesson books +p, and q = the vertex.

    y = a(x-P)^2 + q

  11. Oct 5, 2006 #10
    8.10 is the y-coordinate of the vertex. I substituted in the 1.02 into [tex] h = -4.9t^2 + 10t + 3 [/tex].

    So [tex] y = a(x-1.02)^{2} + 8.10 [/tex]
    Last edited: Oct 5, 2006
  12. Oct 5, 2006 #11
    i see now :smile:

    so according to that x (of the vertex) = p, and y(of the vertex) = q?

  13. Oct 5, 2006 #12
    yes, that is correct
  14. Oct 5, 2006 #13
    thanks, that makes things a lot easier for me :wink:

  15. Oct 5, 2006 #14
    Hopefully you still check back to this thread..
    Here's a trick to avoid multiplying a trinomial times a trinomial on such problems..

    Once you simplify the (x-(3+i))(x-(3-1)), you get
    (x-3-i)(x-3+i) Notice that the x-3 is the same in both factors.
    Draw a box around the x-3 in both of the factors. Ignoring what's in the box (but it's the same in both boxes), it looks like
    (box-i)(box+i) which you (hopefully) recognize is a conjugate pair.

    So, the product is (box)^2 - i^2
    or x^2-6x+9 +1 = your answer.

    Since you'll probably be faced with a lot of problems in which either the complex roots or the irrational roots will be a conjugate pair of roots, writing the factors as a conjugate pair in this manner is a great time saver. I avoided showing it more formally, choosing to have you draw a box around the x-3's, rather than a set of parenthesis, simply because that way has "worked" better for students I've had.
  16. Oct 11, 2006 #15
    i see what you're saying and i wrote it in my study notes/cheat sheet :biggrin:. i dont have much in the way of mathematical aptitude, so any quick tricks help :smile:


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