Results 1 to 5 of 5

Math Help - finding x in quadratic equations

  1. #1
    Junior Member
    Joined
    Oct 2009
    From
    Aberystwyth, Wales
    Posts
    36

    finding x in quadratic equations

    i know the equation of x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} to find if at all where a function crosses the x axis but i also found you can just use x = \frac{-b}{2a} which i really dont understand as were does the \sqrt{b^2 - 4ac} go?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    x=\frac{-b}{2a} gives the x coordinate of the vertex of a parabola.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,629
    Thanks
    430
    Quote Originally Posted by renlok View Post
    i know the equation of x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} to find if at all where a function crosses the x axis but i also found you can just use x = \frac{-b}{2a} which i really dont understand as were does the \sqrt{b^2 - 4ac} go?
    x = \frac{-b}{2a} gives the x-value of the vertex for the graph of the quadratic (a parabola)

    the zeros (or roots) of the quadratic function (where it crosses the x-axis) are \pm \frac{\sqrt{b^2-4ac}}{2a} on either side of x = \frac{-b}{2a}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by renlok View Post
    i know the equation of x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} to find if at all where a function crosses the x axis but i also found you can just use x = \frac{-b}{2a} which i really dont understand as were does the \sqrt{b^2 - 4ac} go?
    That is only true when b^2-4ac (which is called the discriminant) is equal to 0

    edit: see above
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2009
    From
    Aberystwyth, Wales
    Posts
    36
    oh right that makes sense thanks guys
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: September 3rd 2011, 08:01 PM
  2. Quadratic Equations
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 14th 2011, 10:08 PM
  3. Replies: 1
    Last Post: August 25th 2009, 11:47 PM
  4. Replies: 1
    Last Post: June 12th 2008, 09:30 PM
  5. Replies: 1
    Last Post: September 1st 2007, 06:35 AM

Search Tags


/mathhelpforum @mathhelpforum