Completing the square.

  • Thread starter _Mayday_
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  • #1
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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:
 

Answers and Replies

  • #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
symbolipoint
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OKAY, THIS WILL HELP YOU, MAYDAY:

http://faculty.ed.umuc.edu/~swalsh/Math%20Articles/GeomCS.html [Broken]
 
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  • #5
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Thanks you two, I'll check those out. =]
 
  • #6
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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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