Results 1 to 8 of 8

Math Help - Solving -4 = x(x^2 + 1)

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    4

    Solving -4 = x(x^2 + 1)

    Ok, so I have a bit of a quandry. I was trying to find the x-intercept(s) of y = x + [4/(x^2 + 1)]. After setting y = 0, I get the equation -4 = x(x^2 + 1). That's where the problem sets in. I suddenly realized that I have no idea how to solve this. Does anybody else have an idea? I can put the equation in a calculator and get a good approximation, but I'd like a conceptual way of getting at the solution if possible, as I'm now really interested in this problem.

    Thanks so much!

    James
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    820
    Thanks
    31
    Looks like you got yourself a *cubic*.

    There is a standard algorithmic way of solving this, but it's unwieldy and long-winded (although very pretty).

    Your best way of tackling one of these is (assuming you're given one of these as an exercise at elementary level) is seeing whether there's a factor.

    That is, given an equation in the form:
    x^3 + a x^2 + b x + c = 0
    (which is effectively what you've got when you rearrange your equation above)

    ... you want to try and get it in the form:
    (x + p)(x^2 + qx + r) = 0

    Any of this familiar? If not, you've got a way to go before you're up to this - you will need to struggle through solving quadratics (which is like cubics but not as bad).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
    4
    I'm up to the task of factoring. Unfortunately, I don't think this one does. One solution is x=0 (which I found by factoring out that first x.) The other solution (the calculator showed me there was only one other) is irrational.

    I can employ "completing the square" when I have to find solutions to quadratics that don't factor; are there any similar methods for cubics?

    Thanks so much for the help!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2009
    Posts
    4
    Ok, I'm an idiot :-P I said x=0, but that's only if the equation is set to equal zero, which it's not here. Nevermind. So yeah, still stuck since the only solution the calculator gave me was irrational :-P

    And yeah, there was only one solution, not two.

    Thanks again!

    James
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    I was going to say use rational zero theorem and Descarte's rule of signs but your calculator already told you it irrational. Using Descarte's rule of signs you can see there is only 1 negative real zero.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    820
    Thanks
    31
    Quote Originally Posted by crazydrclaw View Post
    I'm up to the task of factoring. Unfortunately, I don't think this one does. One solution is x=0 (which I found by factoring out that first x.) The other solution (the calculator showed me there was only one other) is irrational.

    I can employ "completing the square" when I have to find solutions to quadratics that don't factor; are there any similar methods for cubics?

    Thanks so much for the help!
    Yes there is. Hang onto your hat:

    Cardan's Formula - ProofWiki

    ... and there's also one for quartics:

    Ferrari's Method - ProofWiki

    ... but there is no such formula for higher order polynomials.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Sep 2009
    Posts
    4
    Thanks!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by crazydrclaw View Post
    Ok, so I have a bit of a quandry. I was trying to find the x-intercept(s) of y = x + [4/(x^2 + 1)]. After setting y = 0, I get the equation -4 = x(x^2 + 1). That's where the problem sets in. I suddenly realized that I have no idea how to solve this. Does anybody else have an idea? I can put the equation in a calculator and get a good approximation, but I'd like a conceptual way of getting at the solution if possible, as I'm now really interested in this problem.

    Thanks so much!

    James
    Descartes' rule of signs tells you that this has one negative root and zero positive roots.

    Now you can find its approximate value numerically (it is easy enough to show the root is between -1 and -2, ..)

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. solving for x
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 7th 2009, 04:13 PM
  2. help solving for y
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 1st 2009, 10:25 AM
  3. help me in solving this
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: December 13th 2008, 01:51 PM
  4. Solving x'Px = v
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: December 11th 2008, 03:21 PM
  5. Replies: 3
    Last Post: October 11th 2006, 09:15 PM

Search Tags


/mathhelpforum @mathhelpforum