Results 1 to 5 of 5

Math Help - Completing the square.

  1. #1
    Newbie
    Joined
    Nov 2007
    Posts
    10

    Completing the square.

    Well, I am working on my quadratics at the moment, and one of my main problems with it is completing the square. I have a basic idea on how to do it, but could anyone explain it from start to finish in detail. That would be great, or if you just have a link to a good site that'd be great. Cheers
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Apr 2007
    Posts
    16
    Quote Originally Posted by tumbleweed View Post
    Well, I am working on my quadratics at the moment, and one of my main problems with it is completing the square. I have a basic idea on how to do it, but could anyone explain it from start to finish in detail. That would be great, or if you just have a link to a good site that'd be great. Cheers
    Here's an example.

    Complete the square: y^2-4y-3=x

    When completing the square, you have to think a little bit about what the left side of this equation looks like. You need to add something to the left side to make it a perfect square. A little bit of thought leads us to the conclusion that you need to have y^2-4y+4 on the left side because that can be factored down to (y-2)^2. We want to take all the constants on the left side over to the other side to give us a blank slate to work with on the left side.

    y^2-4y-3+3=x+3
    y^2-4y=x+3
    Now, add 4 to both sides to "complete the square":
    y^2-4y+4=x+3+4
    y^2-4y+4=x+7
    Factor.
    (y-2)^2=x+7

    The trickiest part is knowing what constant has to be added to the left side in order to "complete the square". You can figure it out by recognizing that the expansion of (y-c)^2 is y^2-2cy+c^2. Since you know the middle term (2cy) corresponds to -4y, you know that 2cy=-4y. Cancel the y's and divide both sides by 2. c=-2. Since the third term in the expansion is c^2, you know it must be +4. That's how you know it has to be y^2-4y+4.

    That's about all there is to it. I hope this was clear enough...

    Pulsar
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2007
    Posts
    18
    Quote Originally Posted by tumbleweed View Post
    Well, I am working on my quadratics at the moment, and one of my main problems with it is completing the square. I have a basic idea on how to do it, but could anyone explain it from start to finish in detail. That would be great, or if you just have a link to a good site that'd be great. Cheers
    Another way to do it is this:

    y^2 - 4y - 3 = x

    Now we need to take half of b (4) and square it. Thats how I was taught to do it. This makes sense because if you think about it, in the end it would workout to be a perfect square. So "b" is taken from ax^2 + bx + c = 0. The first step is to always take half of b and square it. [(1/2b)^2] In this question, you get 4. Now we must add and subtract 4 on the left side. We need that 4 in there to get a perfect square, but we cannot randomly add it. Therefore, we add and subtract it, as if we didn't do anything at all, because 4-4 is just 0. Also, we leave the "c" value right where it is and don't include it in the brackets. So we get :

    (y^2 - 4y + 4 - 4 ) - 3 = x

    Now we can have a perfect square using y^2 - 4y + 4, giving us:

    [(y-2)^2 -4)] - 3 = x

    Then, we add the -4 and the -3 together giving us:

    (y-2)^2 - 7 = x

    And thats your perfect square. Hope that made sense.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539
    Hello, tumbleweed!

    I assume you're solving a quadratic equation.

    An additional problem arises if the leading coefficient is not 1 (one).


    For example: . 4x^2 - 24x + 11 \:=\:0

    "Move" the constant to the right side: . 4x^2 - 24x \:=\:-11

    Divide through by the leading coefficient: . x^2 - 6x\qquad \;=\;-\frac{11}{4}

    Complete the square: take half the x-coefficient and square.
    . . . . \frac{1}{2}(\text{-}6) \;=\;-3\quad\Rightarrow\quad(\text{-}3)^2 \;=\;9

    Add to both sides: . x^2 - 6x\,{\color{blue} +\,9} \;=\;-\frac{11}{4}\,{\color{blue}+\,9}

    And we have: . (x - 3)^2 \;=\;\frac{25}{4}


    Now we can solve for x.

    Take the square root of both sides: . x - 3 \;=\;\pm\sqrt{\frac{25}{4}}
    . . and we have: . x - 3 \;=\;\pm\frac{5}{2}

    Therefore: . x \;=\;3\pm\frac{5}{2} \;=\;\begin{Bmatrix}3 + \frac{5}{2} & = & \frac{11}{2} \\ 3 - \frac{5}{2} & = & \frac{1}{2} \end{Bmatrix}


    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    11
    Quote Originally Posted by Soroban View Post
    "Move" the constant to the right side: . 4x^2 - 24x \:=\:-11
    Here's another way:

    (2x-6)^2-36=-11\implies|2x-6|=5.

    From here 2x-6=5 & 6-2x=5,

    which gives Soroban's solutions.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Completing the square
    Posted in the Algebra Forum
    Replies: 5
    Last Post: June 19th 2010, 11:36 AM
  2. Completing the square
    Posted in the Algebra Forum
    Replies: 5
    Last Post: December 21st 2009, 11:00 PM
  3. Completing The Square
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 4th 2007, 05:18 AM
  4. Completing the Square
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: September 3rd 2007, 11:40 AM
  5. Completing the Square
    Posted in the Algebra Forum
    Replies: 7
    Last Post: August 31st 2007, 10:36 PM

Search Tags


/mathhelpforum @mathhelpforum