Results 1 to 3 of 3

Thread: Fantastic system

  1. #1
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    582
    Thanks
    3

    Fantastic system

    Find all reals numbers solutions of system :
    $\displaystyle \left\{ \begin{array}{l}
    {\rm a + b + c = 4} \\
    {\rm a}^{\rm 2} + b^2 + c^2 = 14 \\
    a^3 + b^3 + c^3 = 34 \\
    \end{array} \right.
    $
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    $\displaystyle a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)\Rightarrow ab+ac+bc=1$

    $\displaystyle a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)+3abc\Rightarrow abc=-6$

    So we have

    $\displaystyle \left\{\begin{array}{ll}a+b+c=4\\ab+ac+bc=1\\abc=-6\end{array}\right.$

    Then $\displaystyle a,b,c$ are the roots of the equation $\displaystyle x^3-4x^2+x+6=0$

    The equation has the solutions $\displaystyle x_1=-1, \ x_2=2, \ x_3=3$

    Then $\displaystyle \left\{\begin{array}{ll}a=-1\\b=2\\c=3\end{array}\right.$ or $\displaystyle \left\{\begin{array}{ll}a=2\\b=-1\\c=3\end{array}\right.$ or.....etc.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    582
    Thanks
    3
    Quote Originally Posted by red_dog View Post
    $\displaystyle a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)\Rightarrow ab+ac+bc=1$

    $\displaystyle a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)+3abc\Rightarrow abc=-6$

    So we have

    $\displaystyle \left\{\begin{array}{ll}a+b+c=4\\ab+ac+bc=1\\abc=-6\end{array}\right.$

    Then $\displaystyle a,b,c$ are the roots of the equation $\displaystyle x^3-4x^2+x+6=0$

    The equation has the solutions $\displaystyle x_1=-1, \ x_2=2, \ x_3=3$

    Then $\displaystyle \left\{\begin{array}{ll}a=-1\\b=2\\c=3\end{array}\right.$ or $\displaystyle \left\{\begin{array}{ll}a=2\\b=-1\\c=3\end{array}\right.$ or.....etc.
    Thank you
    The system has the solutions
    $\displaystyle
    \begin{array}{l}
    \left\{ \begin{array}{l}
    a = 2 \\
    b = 3 \\
    c = - 1 \\
    \end{array} \right.\left\{ \begin{array}{l}
    a = 3 \\
    b = - 1 \\
    c = 2 \\
    \end{array} \right.\left\{ \begin{array}{l}
    a = 3 \\
    b = 2 \\
    c = - 1 \\
    \end{array} \right. \\
    \\
    \end{array}

    $
    And :
    $\displaystyle \left\{ \begin{array}{l}
    a = - 1 \\
    b = 2 \\
    c = 3 \\
    \end{array} \right.\left\{ \begin{array}{l}
    a = - 1 \\
    b = 3 \\
    c = 2 \\
    \end{array} \right.\left\{ \begin{array}{l}
    a = 2 \\
    b = - 1 \\
    c = 3 \\
    \end{array} \right.$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Fantastic-integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Sep 12th 2009, 08:36 AM
  2. Fantastic limit
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Jun 21st 2009, 11:51 AM

Search Tags


/mathhelpforum @mathhelpforum