Results 1 to 7 of 7

Math Help - Rearranging this equation?

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    10

    Rearranging this equation?

    I don't see how

    100-0.25x^2 =ax^2 - 80ax + 1600a

    becomes

    x^2 + (80a)/(-0.25-a)x + (1600a+100)/(-0.25-a) = 0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,863
    Thanks
    639
    Quote Originally Posted by LandLBoy View Post
    I don't see how

    100-0.25x^2 =ax^2 - 80ax + 1600a

    becomes

    x^2 + (80a)/(-0.25-a)x + (1600a+100)/(-0.25-a) = 0
    it doesn't ...

    100-0.25x^2 =ax^2 - 80ax + 1600a

    -ax^2 -0.25x^2 + 80ax - 1600a + 100 = 0

    -(a + 0.25)x^2 + 80ax - 1600a + 100 = 0

    x^2 + \dfrac{80ax}{-(a+0.25)} - \dfrac{1600a - 100}{-(a+0.25)} = 0

    x^2 - \dfrac{80a}{a+0.25} \cdot x + \dfrac{1600a - 100}{a+0.25} = 0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2011
    Posts
    10
    can you still get this from what you got

    x = - (40a)/(-0.25-a) +- √(1600a^2/(-0.25-a)^2 - (1600a+100)/(-0.25-a))
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,863
    Thanks
    639
    why is this problem being done in a piece-meal fashion?

    mind stating the original problem in its entirety?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2011
    Posts
    10
    http://www.mathhelpforum.com/math-he...em-175040.html

    sorry, we're not supposed to use calculus to solve the problem - i found an answer online and i'm trying to interpret it but i'm having some difficulty as u can see

    ANSWER:

    find the point, where curve 1 and curve 2 touch.

    100-0.25x^2 = a(x-40^2)
    100-0,25x^2 =ax^2 - 80ax + 40^2*a
    ...
    x^2 + (80a)/(-0.25-a)*x + (-40^2a+100)/(-0.25-a) = 0
    x = - (40a)/(-0.25-a) +- √(40^2a^2/(-0.25-a)^2 - (-40^2a+100)/(-0,25-a))

    now to have only 1 tangent point, the term under the root must be = 0
    40^2a^2/(-0,25-a)^2 = (-40^2a+100)/(-0,25-a)

    solution:
    a = 0,083333

    now find x:
    x = - 40*0,08333/(-0,25-0,08333)
    x = 10 m

    The curves touch at x = 10 m,

    and the slope of curve 1 at 10 m is
    y' = -1/2x = -5
    and of curve 2 at 10 m is
    y' = 0,08333(2x-80) = 0,08333*- 60 = - 5

    i.e., the slopes are identical --> nice smooth ride

    I understand what the answerer did but it's the intermediate steps that are confusing me - that is why I posted a part of the answer trying to figure out what the answerer did
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,110
    Thanks
    68
    Quote Originally Posted by LandLBoy View Post
    I don't see how
    100-0.25x^2 =ax^2 - 80ax + 1600a
    becomes
    x^2 + (80a)/(-0.25-a)x + (1600a+100)/(-0.25-a) = 0
    Is there a reason why you're "hanging on" to this weird "-0.25-a"?

    You instead could change the term .25x^2 to x^2 / 4, then multiply through by 4.

    Quite simpler, and you'll need only half the Tylenols
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,863
    Thanks
    639
    100 - .25x^2 = a(x-40)^2

    400 - x^2 = 4a(x^2 - 80x + 1600)<br />

    400 - x^2 = 4ax^2 - 320ax + 6400a

    0 = (4a+1)x^2 - 320ax + (6400a - 400)

    x = \dfrac{320a \pm \sqrt{(-320a)^2 - 4(4a+1)(6400a - 400)}}{2(4a+1)}<br />

    for a single solution, the discriminant = 0 ...

    (-320a)^2 - 4(4a+1)(6400a - 400) = 0<br />

    320^2 a^2 = 4(4a+1)(6400a - 400)

    320^2 a^2 = 1600(4a+1)(16a-1)

    64a^2 = 64a^2 + 12a - 1<br />

    a = \dfrac{1}{12}

    now go back and determine that x = 10
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need help rearranging an equation
    Posted in the Algebra Forum
    Replies: 0
    Last Post: March 22nd 2010, 09:41 AM
  2. Rearranging an equation.
    Posted in the Algebra Forum
    Replies: 9
    Last Post: February 20th 2010, 09:46 PM
  3. Rearranging equation
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: August 30th 2009, 04:45 AM
  4. just rearranging an equation, is it possible?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: July 21st 2009, 07:12 AM
  5. Rearranging an equation
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: November 12th 2007, 08:52 AM

Search Tags


/mathhelpforum @mathhelpforum