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Thank you. :shy:

- Thread starter _Mayday_
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- #1

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Thank you. :shy:

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symbolipoint

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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.

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berkeman

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symbolipoint

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OKAY, THIS WILL HELP YOU, MAYDAY:

http://faculty.ed.umuc.edu/~swalsh/Math%20Articles/GeomCS.html [Broken]

http://faculty.ed.umuc.edu/~swalsh/Math%20Articles/GeomCS.html [Broken]

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Thanks you two, I'll check those out. =]

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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 like3 x^2 + 2 x - 5 = 0

[/tex]

[tex]

x^2 + \frac 2 3 x - \frac 5 3 = 0

[/tex]

Now, you take x^2 + \frac 2 3 x - \frac 5 3 = 0

[/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]\left(x + \frac 1 3 \right)^2 - \left( \frac 1 3 \right)^2 - \frac 5 3 = 0

[/tex]

so that your equation now looks like

[tex]

\left(x + \frac 1 3 \right)^2 - \frac {16} 9 = 0

[/tex]

and since \left(x + \frac 1 3 \right)^2 - \frac {16} 9 = 0

[/tex]

[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.x = \pm \sqrt {\frac {16} 9} - \frac 1 3 = \pm \frac 4 3 - \frac 1 3

[/tex]

Easy!

. . .

The trick was to change this:

[tex]

x^2 + \frac 2 3 x

[/tex]

into this:x^2 + \frac 2 3 x

[/tex]

[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.\left(x + \frac 1 3 \right)^2 - \left( \frac 1 3 \right)^2

[/tex]

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