Results 1 to 2 of 2

Math Help - Need Some Guidance

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    6

    Need Some Guidance

    Solve.
    15x^4 - 28x^2 + 5 = 0
    substitution should result in a quadratic equation in the variable u, ie

    I let u = x^2

    15u^4 - 28x^2 + 5= 0
    15u^2 - 28u + 5 = 0

    I used the quadratic formula

    -(-28) + or - sqrt -28^2 - 4(15)(5) / 2(15)
    28 + or - sqrt 784+300/30
    28 + or - sqrt 1084/30

    how'd I do?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,110
    Thanks
    982
    Quote Originally Posted by Just4 View Post
    Solve.
    15x^4 - 28x^2 + 5 = 0
    substitution should result in a quadratic equation in the variable u, ie

    I let u = x^2

    15u^4 - 28x^2 + 5= 0
    15u^2 - 28u + 5 = 0

    I used the quadratic formula

    -(-28) + or - sqrt -28^2 - 4(15)(5) / 2(15)
    28 + or - sqrt 784+300/30
    28 + or - sqrt 1084/30

    how'd I do?
    made a mistake and didn't finish ...

    u = \frac{-28 \pm \sqrt{(-28)^2 - 4(15)(5)}}{2(15)}

    u = \frac{-28 \pm \sqrt{784 - 300}}{30}

    u = \frac{-28 \pm \sqrt{484}}{30}

    u = \frac{-14 \pm 11}{15}

    you're solving for x, remember?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integration guidance please
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 1st 2011, 04:37 PM
  2. Inductin guidance
    Posted in the Discrete Math Forum
    Replies: 22
    Last Post: July 6th 2010, 08:58 PM
  3. Replies: 0
    Last Post: February 1st 2010, 06:09 AM
  4. Need Guidance...
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 27th 2009, 08:12 PM
  5. Need some derivatives guidance
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 28th 2007, 03:48 PM

Search Tags


/mathhelpforum @mathhelpforum