Results 1 to 8 of 8

Math Help - Solve: X^4-13x^2+36=0

  1. #1
    Junior Member
    Joined
    Jul 2011
    Posts
    48

    Solve: X^4-13x^2+36=0

    Solve: X^4-13x^2+36=0.

    I can't figure this out.

    Also,

    Y^(2/3)-2y^(1/3)-8=0

    for the second one, can I add 8 to the right side, cube both sides and end up with

    Y^2-2y=512, then use the quadratic formula?
    Last edited by Ackbeet; August 6th 2011 at 10:30 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1

    Re: Solve: X^4-13x^2+36=0

    Quote Originally Posted by StudentMCCS View Post
    Solve: X^4-13x^2+36=0

    I can't figure this out.

    Also,

    Y^(2/3)-2y^(1/3)-8=0

    for the second one, can I add 8 to the right side, cube both sides and end up with

    Y^2-2y=512, then use the quadratic formula?

    These are both forms of quadratics

    1. Let u = x^2

    2. Let u = y^{1/3}
    Last edited by e^(i*pi); August 6th 2011 at 09:47 AM. Reason: quoting OP
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2011
    Posts
    48

    Re: Solve: X^4-13x^2+36=0

    Ahhh! Thanks.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2011
    Posts
    48

    Re: Solve: X^4-13x^2+36=0

    Quote Originally Posted by e^(i*pi) View Post
    These are both forms of quadratics

    1. Let u = x^2

    2. Let u = y^{1/3}
    I tried two ways now and for some reason am still having a hard time solving.

    The first way, was to factor X^2 from the first term, (a), and X from the second, (b). And so I have:

    X^2(X^2)-13x(x)+36=0 A=x^2 B=13x, and C=36

    The second way was to factor change the first term, x^4, into X^3(x). And so I have:

    -13x^2+X^3(x)+36=0 A=-13 B=x^3, and C=36

    ---------------------------------------------------------

    When I plug these values into the quadratic equation:A=x^2, B=13x, C=36

    x=-13+or-[(169-144x^2)/2x^2]^-2 I solve as:

    X=-13+or-13-12(2^-2) X=-13+or-(-3.970...)

    ---------------------------------------------------------

    When I use the other values: A=-13 B=X^3 and C=36

    x=-x^3+or-[(X^6+1872)/-26]^-2 I have a negative inside a square root.

    ---------------------------------------------------------

    Am I doing this wrong? Or am I doing it right, but making algebraic errors when simplifying?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jul 2011
    Posts
    48

    Re: Solve: X^4-13x^2+36=0

    Actually I just noticed a mistake, 13 instead of -13. I'll recheck.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Solve: X^4-13x^2+36=0

    A,B,C are constant coefficients of the quadratic equation (in general). If you substitute, let x^2=u then you get:
    u^2-13u+36=0, solve this equation to u and afterwards do the back-substitution.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jul 2011
    Posts
    15

    Re: Solve: X^4-13x^2+36=0

    X^4-13x^2+36=0

    (X^2)^2 - 13X^2 + 36 = 0

    {instead "x" you have square of it, take it easy}

    (X^2 - 9) (X^2 - 4) = 0

    X^2 - 9 = 0 OR  X^2 - 4 = 0

    X^2 = 9 OR X^2 = 4

    X = 3, -3, 2, -2
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    911
    Thanks
    27

    Re: Solve: X^4-13x^2+36=0

    Hello StudentMCCS,
    Looking at your post in which you were having trouble solving quadratics it is best to use the factoring method before going to the quadratic formula. In these two problems each can be solved this way.You should learn it if not already known.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need to solve summation equation to solve sum(x2)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: July 16th 2010, 10:29 PM
  2. Replies: 1
    Last Post: June 9th 2009, 10:37 PM
  3. how do i solve this?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 2nd 2008, 02:58 PM
  4. how to solve ..
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: August 2nd 2008, 08:17 AM
  5. please help me solve this
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 1st 2008, 09:50 PM

Search Tags


/mathhelpforum @mathhelpforum