Results 1 to 4 of 4

Math Help - Help !! completing the sqaure

  1. #1
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381

    Help !! completing the sqaure

    How can solve this using completing the squuare?
    x^2 - 5a^2 - 4ax = 0

    WHERE a IS CONSTANT
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,901
    Thanks
    329
    Awards
    1
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    How can solve this using completing the squuare?
    x^2 - 5a^2 - 4ax = 0

    WHERE a IS CONSTANT
    First:
    x^2 - 4ax = 5a^2

    Now, we want to add a constant to this such that the LHS is a perfect square:
    (x + b)^2 = x^2 + 2bx + b^2

    So comparing the linear term in this to the equation above I get that
    -4a = 2b, thus b = -2a

    So we wish to add b^2 = (-2a)^2 = 4a^2 to both sides:
    x^2 - 4ax + 4a^2 = 5a^2 + 4a^2

    (x - 2a)^2 = 9a^2

    x - 2a = (+/-)3a

    x = 2a (+/-) 3a

    So x = 5a or x = -a.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,713
    Thanks
    633
    Hello, ^_^Engineer_Adam^_^!

    How can solve this using completing the square?
    . . x - 5a - 4ax .= .0 . where a is constant.

    The same way as we've been taught.

    . . . . . . . . . . . . . . x - 4ax .= .5a

    Then: . . . . . x - 4ax + 4a .= .5a + 4a

    And: . . . . . . . . . . (x - 2a) .= .9a

    Take square roots: . . x - 2a .= .3a

    So we have: . . . . . . . . . . x .= .2a 3a


    Therefore: . x .= .5a, -a


    Edit: Too fast for me, Dan!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381
    thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Completing the square
    Posted in the Algebra Forum
    Replies: 4
    Last Post: July 3rd 2010, 04:59 AM
  2. 3-4j sqaure roots
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 16th 2010, 03:25 AM
  3. Sqaure roots of complex number
    Posted in the Calculus Forum
    Replies: 7
    Last Post: January 14th 2010, 01:18 PM
  4. sqaure roots
    Posted in the Algebra Forum
    Replies: 5
    Last Post: April 20th 2008, 06:49 PM
  5. Help with Sqaure Roots
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 18th 2007, 10:55 AM

Search Tags


/mathhelpforum @mathhelpforum