Results 1 to 5 of 5

Math Help - Complete the Square

  1. #1
    Junior Member phgao's Avatar
    Joined
    May 2005
    Posts
    39

    Complete the Square

    1. x^2 + 2x + k
    2. kx^2 + 2x + k
    3. x^2 - 3kx + 1

    For the first I get: -1+- root(1+k)
    Second : [-1+- root(1-k)]/k

    But I'm told im wrong. Help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member jacs's Avatar
    Joined
    Jan 2006
    From
    Sydney
    Posts
    107
    are you supposed to just find k? or solve for k?

    x + 2x + k
    x + 2x + 1 + k - 1
    (x + 1) + k -1

    not sure if that is what you mean

    since there is no equation, there is nothing to solve

    hope that helps maybe?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member phgao's Avatar
    Joined
    May 2005
    Posts
    39
    solve for x,
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by phgao
    1. x^2 + 2x + k
    2. kx^2 + 2x + k
    3. x^2 - 3kx + 1

    For the first I get: -1+- root(1+k)
    Second : [-1+- root(1-k)]/k

    But I'm told im wrong. Help!
    Hello,

    the title of your message suggests, that you are looking for a value of k so you got a square. Is this right? If so:

    to 1) k=1 \ \mbox{then you get} \  (x+1)^2
    to 2) k=1 \ \mbox{then you get} \  (x+1)^2
    to 3) k=\frac{2}{3} \ \mbox{then you get} \  (x-1)^2

    I hope that this here is of some help.

    Bye
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by phgao
    1. x^2 + 2x + k
    2. kx^2 + 2x + k
    3. x^2 - 3kx + 1

    For the first I get: -1+- root(1+k)
    Second : [-1+- root(1-k)]/k

    But I'm told im wrong. Help!
    "Complete the square" means complete the perfect square. Perfect square here is in the form (x+a)^2, where "a" is plus or minus, depending on the given data.

    Before attempting completing the square, make sure first that the coeff(icient of the x^2 is 1 only. Meaning, it it were originally b(x^2), then divide it by "b"......

    Completing the square of (x^2 +bx) is adding, and subtracting at the same time, the square of half of the coefficient of linear x.
    x^2 +bx +(b/2)^2 -(b/2)^2 ------------***

    ------------------------
    x^2 +2x +k = 0
    x^2 +2x +(2/2)^2 -(2/2)^2 +k = 0
    x^2 +2x +1 -1 +k = 0
    (x+1)^2 -1 +k = 0
    (x+1)^2 = 1-k
    Take the square roots of both sides,
    x+1 = +,-sqrt(1-k)
    So,
    x = -1 +,-sqrt(1-k) ----------answer.

    ----------------------
    kx^2 +2x +k = 0

    The x^2 has a coefficient k which is not 1, so divide kx^2 by k. In so doing, divide also all of the other terms of the original equation by k to retain the equality of the original equation.
    x^2 +(2/k)x +1 = 0
    Now that the coefficient of the x^2 is 1 only, we can start completing the square,
    x^2 +(2/k)x +(1/k)^2 -(1/k)^2 +1 = 0
    (x +1/k)^2 -1/(k^2) +1 = 0
    (x +1/k)^2 = 1/(k^2) -1
    x +1/k = +,-sqrt[1/(k^2) -1]
    x = -1/k +,-sqrt[1/(k^2) -1] ------------answer.

    --------------------------------
    x^2 -3kx +1 = 0
    x^2 -3kx +(-3k/2)^2 -(-3k/2)^2 +1 = 0
    (x -3k/2)^2 -(9/4)k^2 +1 = 0
    (x -3k/2)^2 = (9/4)k^2 -1
    x -3k/2 = +,-sqrt[(9/4)k^2 -1]
    x = 3k/2 +,-sqrt[(9/4)k^2 -1] ----------answer.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. complete the square
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: December 1st 2008, 05:13 PM
  2. complete the square
    Posted in the Algebra Forum
    Replies: 4
    Last Post: June 8th 2008, 07:10 PM
  3. Complete the square of ax^2+bx+c where b=0
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 15th 2008, 03:03 PM
  4. Complete the Square??
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 30th 2007, 07:32 PM
  5. complete the square
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 16th 2007, 04:48 PM

Search Tags


/mathhelpforum @mathhelpforum