Results 1 to 6 of 6

Math Help - Isolating a variable (hard)

  1. #1
    Junior Member
    Joined
    May 2008
    Posts
    42

    Isolating a variable (hard)

    How do I isolate x here?
    48.99 = 36.34(1+x)^2 + ([(1+x)^2 - 1]/x)

    I get to:
    (1+x)^2 = (48.99x + 1)/(36.34x + 1)
    But I can't simplify my equation further.

    Also, x > 0. I know x = 0.1355 but I can't get to it algebraically.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    \frac{(1+x)^{2}(36.34x+1)-(48.99x+1)}{36.34x+1}=0 \Leftrightarrow 36.34 x^3+73.68 x^2-10.65 x=0 solve for x ,you'll get x=0.13549 for  x> 0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2009
    Posts
    162
    Quote Originally Posted by Volcanicrain View Post
    How do I isolate x here?
    48.99 = 36.34(1+x)^2 + ([(1+x)^2 - 1]/x)

    I get to:
    (1+x)^2 = (48.99x + 1)/(36.34x + 1)
    But I can't simplify my equation further.

    Also, x > 0. I know x = 0.1355 but I can't get to it algebraically.
    48.99 = 36.34(1+x)^2 + ([(1+x)^2 - 1]/x)
    48.99 = 36.34(x^2+2x+1) + ([(1+x)^2 - 1]/x)
    48.99 = 36.34x^2 + 72.68x + 36.34 + (x^2 + 2x / x)
    48.99x = 36.34x^2 + 72.68x + 36.34 + x^2 + 2x
    48.99x = 37.34x^2 + 74.68x + 36.34
    -25.69x = 37.34x^2 + 36.34
    -25.69x = 6.11x + 36.34
    -31.8x = 36.34
    x = -1.14

    Does this help?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    May 2008
    Posts
    42
    Quote Originally Posted by Raoh View Post
    \frac{(1+x)^{2}(36.34x+1)-(48.99x+1)}{36.34x+1}=0 \Leftrightarrow 36.34 x^3+73.68 x^2-10.65 x=0 solve for x ,you'll get x=0.13549 for  x> 0
    Thank you!
    Just a quick question - I can ignore the denominator in this case right? As long as the denominator is not equal to 0, then the equation will be satisfied.

    Thanks a ton!

    Thanks Barthayn too, but I believe you forgot to multiply an x in the middle.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    Quote Originally Posted by Volcanicrain View Post
    Thank you!
    Just a quick question - I can ignore the denominator in this case right? As long as the denominator is not equal to 0, then the equation will be satisfied.

    Thanks a ton!

    Thanks Barthayn too, but I believe you forgot to multiply an x in the middle.
    yes
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,100
    Thanks
    67
    Quote Originally Posted by Volcanicrain View Post
    How do I isolate x here?
    48.99 = 36.34(1+x)^2 + ([(1+x)^2 - 1]/x)
    Try it like this (just a suggestion):

    Let a = 48.99 and b = 36.34

    a - b(1 + x)^2 = [(1 + x)^2 - 1] / x

    ax - bx(1 + x)^2 = (1 + x)^2 - 1

    ax - bx(1 + x)^2 = x^2 + 2x

    a - b(1 + x)^2 = x + 2

    Carry on...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Isolating a Variable
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 26th 2010, 04:12 PM
  2. Isolating a variable
    Posted in the Algebra Forum
    Replies: 9
    Last Post: October 10th 2010, 08:44 PM
  3. Isolating a Variable
    Posted in the Calculus Forum
    Replies: 7
    Last Post: April 12th 2010, 06:21 PM
  4. Help Isolating a Variable
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 2nd 2009, 03:10 PM
  5. Isolating a variable in the numerator
    Posted in the Algebra Forum
    Replies: 3
    Last Post: July 7th 2007, 04:19 PM

Search Tags


/mathhelpforum @mathhelpforum