Results 1 to 4 of 4

Math Help - Intermediate value problem

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    13

    Intermediate value problem

    hey all,

    i got the following problem to solve:

    i got this term: 27*g^2+4*p^3<0

    and i need to proof that there are three real roots to this function:

    x^3+px+q=0

    is it enough to solve it by min/max points of the function?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570

    Condition for a cubic to have real roots

    Hello dannee
    Quote Originally Posted by dannee View Post
    hey all,

    i got the following problem to solve:

    i got this term: 27*g^2+4*p^3<0

    and i need to proof that there are three real roots to this function:

    x^3+px+q=0

    is it enough to solve it by min/max points of the function?
    I'm sure you mean:
    27q^2+4p^3<0
    Then the proof goes like this:

    Every cubic has at least one real root, so let's assume that x^3+px + q = 0 has a real root x = a. Then:
    x^3+px+q \equiv (x-a)(x^2+bx+c), for some constants b and c.
    \equiv x^3 +(b-a)x^2+(c-ab)x-ac
    Comparing coefficients:
    \left\{\begin {array} {l}<br />
0=b-a\\<br />
p=c-ab\\<br />
q=-ac\\<br />
\end{array}\right.

    \Rightarrow \left\{\begin {array} {l}<br />
p=c-b^2\\<br />
q=-bc\\<br />
\end{array}\right.
    So:
    27q^2+4p^3 < 0

    \Rightarrow 27b^2c^2 +4(c-b^2)^3<0
    This can be expanded and factorised to give:
    (4c-b^2)(c+2b^2)^2<0

    \Rightarrow 4c-b^2<0

    \Rightarrow b^2-4c>0
    \Rightarrow x^2+bx+c=0 has two real roots, and hence x^3+px+q=0 has three real roots.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2010
    Posts
    13
    i'm really really sorry, i wrote a bad syntax.... the right term should be:

    27^{(q^2)}+4 p^3<0

    and i need to prove that there are three real roots to this function:

    x^3+px+q=0

    thanks alot and sorry again
    Last edited by CaptainBlack; February 13th 2010 at 07:15 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by dannee View Post
    hey all,

    i got the following problem to solve:

    i got this term: 27*g^2+4*p^3<0

    and i need to proof that there are three real roots to this function:

    x^3+px+q=0

    is it enough to solve it by min/max points of the function?
    Just post the original question, even with your correction this makes no sense.

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Intermediate problem
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: February 19th 2010, 01:38 AM
  2. Intermediate value problem 2
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 13th 2010, 10:55 AM
  3. Is this an intermediate value problem??
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 2nd 2009, 07:36 AM
  4. Intermediate Value Theorem Word Problem
    Posted in the Calculus Forum
    Replies: 0
    Last Post: September 20th 2009, 05:35 PM
  5. Replies: 8
    Last Post: January 24th 2008, 02:49 AM

Search Tags


/mathhelpforum @mathhelpforum