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Algebra help

  1. Nov 5, 2006 #1
    a route up a mountain is 20km
    john followed this route at xkm/h
    he came down a different route
    the length of the route is 25km
    his average speed coming down increase by 2km/h than his average speed going up
    it took john 1.5 hours less to come down than to go up
    write an equation in x and show it simplifies to

    3x^2 + 16x - 80 = 0

    i tried doing it, but i cant reduce it to this equation. can anyone help me

    i basically did this

    20/x - 1.5 = 25/x+2

    it cant reduce to the above equation. please help me. thx
  2. jcsd
  3. Nov 5, 2006 #2
    Use [tex] d = rt [/tex]

    So: [tex] 20 = xt [/tex]

    [tex] 25 = (x+2)(t-1.5) [/tex]
  4. Nov 5, 2006 #3
    thx so much. one more think i need to ask, if i put the conefficients into the quadratic formula, do i have to include the + - signs into it, or do i just use the numbers?
  5. Nov 5, 2006 #4
    I am not sure I understand your question. You have to use the + and - in the quadratic formula.

    [tex] x = \frac{-b\pm \sqrt{b^{2}-4ac}}{2a} [/tex]
    Last edited: Nov 5, 2006
  6. Nov 5, 2006 #5
    if its 5x^2 + 16x - 50 = 0

    and i sub it into the quadratic formula, do i just sub in the coefficients only, or do i have to include the signs.

    for my first question i tried it, but i ended up having

    8x + 1.5x -40 = 0
    i dont know i i did wrong:frown: :frown: :frown:
  7. Nov 5, 2006 #6
    If [tex] 5x^{2} + 16x - 50 = 0 [/tex] then

    [tex] x = \frac{-16\pm \sqrt{16^{2}-4(5)(-50)}}{10} [/tex]
  8. Nov 5, 2006 #7
    okay thx for ur help
  9. Nov 6, 2006 #8


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    The signs are part of the coefficients!
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