Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - completing the square

  1. #1
    Member GAdams's Avatar
    Joined
    Apr 2007
    Posts
    171
    Awards
    1

    completing the square

    The second last sentence says that we can only use completing the square when the x coefficient is 1. But the equation has an x coefficient of 6??
    Attached Thumbnails Attached Thumbnails completing the square-completing-square2.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by GAdams View Post
    The second last sentence says that we can only use completing the square when the x coefficient is 1. But the equation has an x coefficient of 6??
    Hello,

    that's a typo. It should say: The technique is only valid if 1 is the factor of x^2
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member GAdams's Avatar
    Joined
    Apr 2007
    Posts
    171
    Awards
    1
    OK. Thanks.

    So if x^2 has a coefficient of more than one, then we can't use this method?

    It's only that I have an equation that I was told could be done via this method. Hmmm.

    2x^2 + 3x - 2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member GAdams's Avatar
    Joined
    Apr 2007
    Posts
    171
    Awards
    1
    Does anybody know a weblink that explains completing the square with explanations of why they do what they do or something?

    I just can't get my head around it.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432
    Nah, you still can, you just need to factor out the coefficient of x^2 so you get 1 as the coefficient.

    2x^2 + 3x - 2

    =2(x^2+\frac{3}{2}x-1)

    Continue normally from here.

    Here are some links you might like to try:
    Completing the Square
    Completing the Square: Solving Quadratic Equations
    Last edited by DivideBy0; September 27th 2007 at 11:35 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member GAdams's Avatar
    Joined
    Apr 2007
    Posts
    171
    Awards
    1
    I tried an easier one first. Does this look ok?
    Attached Thumbnails Attached Thumbnails completing the square-quadratic-completin-gsquare.jpg  
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member GAdams's Avatar
    Joined
    Apr 2007
    Posts
    171
    Awards
    1
    This is as far as i got...I seem to complicate it more if I try to simplify further
    Attached Thumbnails Attached Thumbnails completing the square-comp-square.jpg  
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432
    You don't need to get all confused with a,b,c when completing the square. a,b,c are just used in the theory of completing the square. They are good for when you want to understand what's actually going on, but don't worry about them when you're actually doing a problem.
    When I was learning how to factorise with completing the square I repeated this mantra:
    "Add and subtract the square of half the coefficient of the x term". That's practically all there is to it.

    Quote Originally Posted by GAdams View Post
    I tried an easier one first. Does this look ok?
    x^2+6x-7=0

    x^2+6x+\left(\frac{6}{2} \right)^2-\left(\frac{6}{2} \right)^2-7=0

    (x+3)^2-\left(\frac{6}{2}\right)^2-7=0

    (x+3)^2-16=0

    (x+3)^2=16

    x+3=\pm 4

    x=-3 \pm 4

    x=-7 or x=1

    (You could probably do this faster with trial and error factorisation)

    Quote Originally Posted by GAdams View Post
    This is as far as i got...I seem to complicate it more if I try to simplify further
    For this, I'll just factorise it,

    2x^2+3x-2

    =2\left(x^2+\frac{3}{2}x-1\right)

    =2\left(x^2+\frac{3}{2}x+\left(\frac{3}{4} \right)^2-\left(\frac{3}{4} \right)^2-1\right)

    =2\left(\left(x+\frac{3}{4}\right)^2-\frac{25}{16}\right)

    Since this is the same as

    =2\left(\left(x+\frac{3}{4}\right)^2-\left(\sqrt{\frac{25}{16}}\right)^2\right)

    Now you can use difference of perfect squares a^2-b^2=(a-b)(a+b) to factorise this further:

    =2\left(x+\frac{3}{4}+\sqrt{\frac{25}{16}}\right)\  left(x+\frac{3}{4}-\sqrt{\frac{25}{16}}\right)

    =2\left(x+\frac{3}{4}+\frac{5}{4}\right)\left(x+\f  rac{3}{4}-\frac{5}{4}\right)

    =2(x+2)(x-\frac{1}{2})
    Last edited by DivideBy0; September 28th 2007 at 03:37 AM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Another attempt

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

    Now

    2x(x+2)-(x+2)=(x+2)(2x-1)
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member GAdams's Avatar
    Joined
    Apr 2007
    Posts
    171
    Awards
    1
    OK. I got the first one. Thanks.

    The second one is still confusing me, I can't see how you got the 3/4.

    Here's what I tried. I have gone wrong somewhere....not sure where:
    Attached Thumbnails Attached Thumbnails completing the square-eq.jpg  
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by GAdams View Post
    The second one is still confusing me, I can't see how you got the 3/4.

    Here's what I tried. I have gone wrong somewhere....not sure where:
    Hello,

    If you've got an equation and you want to use your method then you have to divide first the complete equation by the coefficient of the x. (Remember: The coefficient of x must be 1):

    2x^2+3x-2 = 0~\iff~x^2+\frac32 x-1=0

    Now use your method:

    x^2+\frac32 x + \left(\frac34 \right)^2-\left(\frac34 \right)^2-1=0~\iff ~\left(x+\frac34 \right)^2-\left(\frac9{16} + \frac{16}{16}\right)=0

    You've got a difference of squares which can be factored easily:
    <br />
\left(x+\frac34 \right)^2-\left(\frac9{16} + \frac{16}{16}\right)=0~\iff~\left(x+\frac34 +\frac54\right)\left(x+\frac34 -\frac54\right)=0~ \iff~(x+2)(x-\frac12)=0

    I'll leave the rest for you.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,856
    Thanks
    321
    Awards
    1
    GAdams: You seem to be missing the point of this method. What you are trying to do here is turn the expression
    x^2 + bx
    into
    x^2 + bx + \frac{b^2}{4} - \frac{b^2}{4}

    Why are you doing this? Because looking at just the first three terms:
    x^2 + bx + \frac{b^2}{4} = \left ( x + \frac{b}{2} \right )^2

    This process gives us an equation that is much easier to solve than the original.

    As an example, solve
    2x^2 + 7x - 11 = 0

    First group the first two terms:
    (2x^2 + 7x) - 11 = 0

    Now factor the coefficient of the x^2 term:
    2 \left ( x^2 + \frac{7}{2} x \right ) - 11 = 0

    Now let's complete the square. The coefficient of the linear term is \frac{7}{2}. We want to take half of that and square it. But we want to do so in such a way that we aren't adding anything to the LHS of the equation. So we're going to add a term and subtract the same to give a net change of 0:
    2 \left ( x^2 + \frac{7}{2} x + \left ( \frac{7}{4} \right )^2 - \left ( \frac{7}{4} \right )^2 \right ) - 11 = 0

    Now rearrange a bit:
    2 \left ( x^2 + \frac{7}{2} x + \left ( \frac{7}{4} \right )^2 \right ) - 2 \left ( \frac{7}{4} \right )^2 - 11 = 0

    Now the first three terms is a perfect square:
    2 \left ( x + \frac{7}{4} \right ) ^2 - \frac{49}{8} - 11 = 0

    Now to neaten a few things up:
    2 \left ( x + \frac{7}{4} \right ) ^2 - \frac{137}{8} = 0

    2 \left ( x + \frac{7}{4} \right ) ^2 = \frac{137}{8}

    \left ( x + \frac{7}{4} \right ) ^2 = \frac{137}{16}

    x + \frac{7}{4} = \pm \sqrt{\frac{137}{16}}

    x = -\frac{7}{4} \pm \frac{\sqrt{137}}{4}

    -Dan
    Last edited by topsquark; September 29th 2007 at 03:11 AM.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member GAdams's Avatar
    Joined
    Apr 2007
    Posts
    171
    Awards
    1

    Thumbs up

    Quote Originally Posted by topsquark View Post
    GAdams: You seem to be missing the point of this method. What you are trying to do here is turn the expression
    x^2 + bx
    into
    x^2 + bx + \frac{b^2}{4} - \frac{b^2}{4}

    Why are you doing this? Because looking at just the first three terms:
    x^2 + bx + \frac{b^2}{4} = \left ( x + \frac{b}{2} \right )^2

    This process gives us an equation that is much easier to solve than the original.

    As an example, solve
    2x^2 + 7x - 11 = 0

    First group the first two terms:
    (2x^2 + 7x) - 11 = 0

    Now factor the coefficient of the x^2 term:
    2 \left ( x^2 + \frac{7}{2} x \right ) - 11 = 0

    Now let's complete the square. The coefficient of the linear term is \frac{7}{2}. We want to take half of that and square it. But we want to do so in such a way that we aren't adding anything to the LHS of the equation. So we're going to add a term and subtract the same to give a net change of 0:
    2 \left ( x^2 + \frac{7}{2} x + \left ( \frac{7}{2} \right )^2 - \left ( \frac{7}{2} \right )^2 \right ) - 11 = 0


    -Dan
    OK. You said 'take half of that and square it. But you haven't halved it. It's still 7/2 -you just squared it.

    Did you mean half the original coefficient of 7?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,856
    Thanks
    321
    Awards
    1
    Quote Originally Posted by GAdams View Post
    OK. You said 'take half of that and square it. But you haven't halved it. It's still 7/2 -you just squared it.

    Did you mean half the original coefficient of 7?
    Nope. I did a typo. I have fixed it in my original post. Sorry about that. (But hey! You were able to follow it and spotted the mistake! )

    -Dan
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member GAdams's Avatar
    Joined
    Apr 2007
    Posts
    171
    Awards
    1
    Quote Originally Posted by topsquark View Post

    So we're going to add a term and subtract the same to give a net change of 0:
    2 \left ( x^2 + \frac{7}{2} x + \left ( \frac{7}{4} \right )^2 - \left ( \frac{7}{4} \right )^2 \right ) - 11 = 0

    Now rearrange a bit:
    2 \left ( x^2 + \frac{7}{2} x + \left ( \frac{7}{4} \right )^2 \right ) - 2 \left ( \frac{7}{4} \right )^2 - 11 = 0

    Now the first three terms is a perfect square:
    2 \left ( x + \frac{7}{4} \right ) ^2 - \frac{49}{8} - 11 = 0



    -Dan
    I can't see how you got teh step after you re-arranged.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

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