Results 1 to 3 of 3

Math Help - Prove why the Quadratic is what it is.

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    6

    Prove why the Quadratic is what it is.

    I need to learn why and how the Quadratic is what it is. Could their be anyway someone can show me the steps in it?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    Quote Originally Posted by JohnMon View Post
    I need to learn why and how the Quadratic is what it is. Could their be anyway someone can show me the steps in it?

    Thanks.
    By "quadratic" do you mean you need to derive the quadratic formula?

    This method uses "Completing the squares." If you are not comfortable with this method then I'd suggest you look at some examples first.

    ax^2 + bx + c = 0

    ax^2 + bx = -c

    a \left ( x^2 + \frac{b}{a}x \right ) = -c

    Now complete the square:
    a \left ( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2} \right ) = -c

    a \left ( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} \right ) - a \cdot \frac{b^2}{4a^2} = -c

    a \left ( x + \frac{b}{2a} \right ) ^2 = \frac{b^2}{4a} - c

    a \left ( x + \frac{b}{2a} \right ) ^2 = \frac{b^2 - 4ac}{4a}

    \left ( x + \frac{b}{2a} \right ) ^2 = \frac{b^2 - 4ac}{4a^2}

    x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}

    x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}

    x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    There might be a "cooler" way to derive this, but I think this is the most "transparent" method. (The one that clearly shows where everything comes from.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2008
    Posts
    6

    Thanks

    Quote Originally Posted by topsquark View Post
    By "quadratic" do you mean you need to derive the quadratic formula?

    This method uses "Completing the squares." If you are not comfortable with this method then I'd suggest you look at some examples first.

    ax^2 + bx + c = 0

    ax^2 + bx = -c

    a \left ( x^2 + \frac{b}{a}x \right ) = -c

    Now complete the square:
    a \left ( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2} \right ) = -c

    a \left ( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} \right ) - a \cdot \frac{b^2}{4a^2} = -c

    a \left ( x + \frac{b}{2a} \right ) ^2 = \frac{b^2}{4a} - c

    a \left ( x + \frac{b}{2a} \right ) ^2 = \frac{b^2 - 4ac}{4a}

    \left ( x + \frac{b}{2a} \right ) ^2 = \frac{b^2 - 4ac}{4a^2}

    x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}

    x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}

    x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    There might be a "cooler" way to derive this, but I think this is the most "transparent" method. (The one that clearly shows where everything comes from.)

    -Dan
    Thank you Dan!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Using Rolle's theorem to prove at most one solution of quadratic
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: February 11th 2011, 07:22 AM
  2. use quadratic reciprocity to prove...
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: April 7th 2010, 06:39 AM
  3. Replies: 3
    Last Post: April 13th 2009, 12:00 PM
  4. Replies: 1
    Last Post: January 30th 2009, 02:45 PM
  5. Replies: 1
    Last Post: June 12th 2008, 10:30 PM

Search Tags


/mathhelpforum @mathhelpforum