What is Completing the Square and How Does it Solve Quadratic Equations?

[_Mayday_]

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:

 

[symbolipoint]

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.

 

[berkeman]

Here you go, complete with the figures:

http://en.wikipedia.org/wiki/Completing_the_square

 

[symbolipoint]

OKAY, THIS WILL HELP YOU, MAYDAY:

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

 

[_Mayday_]

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

 

[dodo]

Here's another way to tell the story, with an example, if it helps.

If you want to solve an equation like

<br /> 3 x^2 + 2 x - 5 = 0<br />​

first, you divide the whole thing by 3 (in order to leave the x^2 alone), so that now it looks like

<br /> x^2 + \frac 2 3 x - \frac 5 3 = 0<br />​

Now, you take half the coefficient of x, that is, the half of \frac 2 3 which is \frac 1 3
and "complete the square" by building this (explanation later):

<br /> \left(x + \frac 1 3 \right)^2 - \left( \frac 1 3 \right)^2 - \frac 5 3 = 0<br />​

The - \left( \frac 1 3 \right)^2 - \frac 5 3 part is a constant, that you can combine into a single number, which in this case is -\frac 1 9 - \frac 5 3 = \frac {-1 - 15} 9 = -\frac {16} 9
so that your equation now looks like

<br /> \left(x + \frac 1 3 \right)^2 - \frac {16} 9 = 0<br />​

and since x appears only once, now you can solve for it:

<br /> x = \pm \sqrt {\frac {16} 9} - \frac 1 3 = \pm \frac 4 3 - \frac 1 3<br />​

giving the two solutions, 1 and -5/3.
Easy!

. . .

The trick was to change this:

<br /> x^2 + \frac 2 3 x<br />​

into this:

<br /> \left(x + \frac 1 3 \right)^2 - \left( \frac 1 3 \right)^2<br />​

If you expand \left(x + \frac 1 3 \right)^2 you will notice why.