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Completing the square.

  1. Jan 25, 2008 #1
    Hey, I've never quite understood this technique of solving quadratic equations. It is something that has started to get to me now, as it restricts some of the things I can do in my work at school. Could anyone give me an in depth explanation of what it entails. My school textbook does not cover it.

    Thank you. :shy:
     
  2. jcsd
  3. Jan 25, 2008 #2

    symbolipoint

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    Hey, Mayday,

    Do NOT feel bad by this comment, but completing the square is very easy to understand especially IF YOU ARE SHOWN THE RIGHT PICTURE to represent it.

    I could send you a figure drawn with msPaint (because I am not technically skilled with this forum to put a picture into a forum message). You can find this derivation shown if you try an internet search. Any way, let x be the side of a square. Extend the length of this square to form a rectangle. Let b be the length that reaches beyond the x. ... this is getting hard to explain for the picture. I really want to show the picture.

    any way, you look at x^2 + bx, and that is what you want to add a square term onto.

    My explanation is not very clear, so I will do my own internet search and post something suitable in this thread. I wish I could be more thorough right now on-forum.
     
  4. Jan 25, 2008 #3

    berkeman

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  5. Jan 25, 2008 #4

    symbolipoint

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  6. Jan 25, 2008 #5
    Thanks you two, I'll check those out. =]
     
  7. Jan 25, 2008 #6
    Here's another way to tell the story, with an example, if it helps.

    If you want to solve an equation like
    [tex]
    3 x^2 + 2 x - 5 = 0
    [/tex]​
    first, you divide the whole thing by 3 (in order to leave the [tex]x^2[/tex] alone), so that now it looks like
    [tex]
    x^2 + \frac 2 3 x - \frac 5 3 = 0
    [/tex]​
    Now, you take half the coefficient of x, that is, the half of [tex]\frac 2 3[/tex] which is [tex]\frac 1 3[/tex]
    and "complete the square" by building this (explanation later):
    [tex]
    \left(x + \frac 1 3 \right)^2 - \left( \frac 1 3 \right)^2 - \frac 5 3 = 0
    [/tex]​
    The [tex]- \left( \frac 1 3 \right)^2 - \frac 5 3[/tex] part is a constant, that you can combine into a single number, which in this case is [tex]-\frac 1 9 - \frac 5 3 = \frac {-1 - 15} 9 = -\frac {16} 9[/tex]
    so that your equation now looks like
    [tex]
    \left(x + \frac 1 3 \right)^2 - \frac {16} 9 = 0
    [/tex]​
    and since x appears only once, now you can solve for it:
    [tex]
    x = \pm \sqrt {\frac {16} 9} - \frac 1 3 = \pm \frac 4 3 - \frac 1 3
    [/tex]​
    giving the two solutions, 1 and -5/3.
    Easy!

    . . .

    The trick was to change this:
    [tex]
    x^2 + \frac 2 3 x
    [/tex]​
    into this:
    [tex]
    \left(x + \frac 1 3 \right)^2 - \left( \frac 1 3 \right)^2
    [/tex]​
    If you expand [tex]\left(x + \frac 1 3 \right)^2[/tex] you will notice why.
     
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