Results 1 to 3 of 3

Math Help - Quadratic Equation

  1. #1
    Junior Member Mist's Avatar
    Joined
    Oct 2009
    From
    Yorkshire
    Posts
    26

    Quadratic Equation

    I'm finding it difficult to solve this question
    It's solving quadratic eqaution by completing the square

    x^2+6x+1=0

    Any help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Quote Originally Posted by Mist View Post
    I'm finding it difficult to solve this question
    It's solving quadratic eqaution by completing the square

    x^2+6x+1=0

    Any help?
    Here's how I'd proceed.

    When solving the square of a quadratic in the form x^2 + 2bx +c = 0, first you put into brackets, (x+b)^2 + c = 0.

    So in the case of your question you get (x+3)^2 +1 = 0. However, if you multiply out (x+3)^2,you find that you get an extra 9 in your equation, to compensate for this you also place a -9 in your equation, giving you (x+3)^2 - 9 +1 = 0.

    A little bit of rearranging should give you (x+3)^2 = 8, which in turn leads to x+3 = \pm2\sqrt{2}.

    I presume you can continue from here to find the two solutions.

    Craig
    Follow Math Help Forum on Facebook and Google+

  3. #3
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    11
    Awards
    1
    Quote Originally Posted by Mist View Post
    I'm finding it difficult to solve this question
    It's solving quadratic eqaution by completing the square

    x^2+6x+1=0

    Any help?
    Hi Mist,

    Step 1: Transpose the constant to the right hand side.

    x^2+6x=-1

    Step 2: Take half the coefficent of the linear term (x), square it, and add the result to both sides of the equation.

    x^2+6x+9=-1+9

    Step 3: You have just made the left side a perfect square trinomial.

    (x+3)^2=8

    Step 4: Take the square root of both sides.

    x+3=\pm \sqrt{8}

    I'll bet you can finish up, can't you?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: April 25th 2010, 03:53 PM
  2. Quadratic Equation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 18th 2010, 02:58 AM
  3. Quadratic Equation Help!
    Posted in the Algebra Forum
    Replies: 4
    Last Post: March 23rd 2009, 02:43 AM
  4. Quadratic equation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 19th 2008, 05:38 PM
  5. Quadratic Equation
    Posted in the Algebra Forum
    Replies: 7
    Last Post: May 22nd 2008, 04:34 PM

Search Tags


/mathhelpforum @mathhelpforum