1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor

    The signs are part of the coefficients!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Algebra help
  1. Algebra help (Replies: 1)

  2. Algebra Help (Replies: 1)

  3. Help with Algebra (Replies: 15)

  4. Algebra help (Replies: 6)

  5. Algebra Help (Replies: 2)