Results 1 to 3 of 3

Math Help - Roots of Equations

  1. #1
    Member
    Joined
    Dec 2007
    Posts
    80

    Roots of Equations

    One root of the equation x^3+ax^2+7=0 is x=-2. I need to find the value of a.
    I tried to do this by using b^2-4ac to find the discriminant, but this didnt work as i dont have a or b just c=7.
    I also rearrnaged the equation to give a on its own as -3, but i substituted this back into the equation, but i didnt get 0.

    Thanks gor your time.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by Chez_ View Post
    One root of the equation x^3+ax^2+7=0 is x=-2. I need to find the value of a.
    I tried to do this by using b^2-4ac to find the discriminant, but this didnt work as i dont have a or b just c=7.
    I also rearrnaged the equation to give a on its own as -3, but i substituted this back into the equation, but i didnt get 0.

    Thanks gor your time.
    x = -2 is a root. That means when x = -2; then f(x) = 0

    (-2)^3 + (a)(-2)^2 + 7 = 0

    -8 + a(4) + 7 = 0

    4a = 1

    a = \frac{1}{4}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,920
    Thanks
    332
    Awards
    1
    Quote Originally Posted by Chez_ View Post
    One root of the equation x^3+ax^2+7=0 is x=-2. I need to find the value of a.
    I tried to do this by using b^2-4ac to find the discriminant, but this didnt work as i dont have a or b just c=7.
    I also rearrnaged the equation to give a on its own as -3, but i substituted this back into the equation, but i didnt get 0.

    Thanks gor your time.
    If the discriminant you are referring to is b^2 - 4ac then you need to look a bit harder at your equation. This discriminant only works for quadratics and you equation is a cubic. It won't work.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Roots of polynomial equations
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: December 26th 2010, 02:04 PM
  2. Roots of polynomial equations
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: October 27th 2009, 01:02 AM
  3. Roots of equations
    Posted in the Math Forum
    Replies: 0
    Last Post: May 14th 2009, 04:01 AM
  4. further maths roots of equations
    Posted in the Algebra Forum
    Replies: 3
    Last Post: September 29th 2008, 11:22 AM
  5. roots of equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: July 10th 2008, 07:28 AM

Search Tags


/mathhelpforum @mathhelpforum