# Completing the square.

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
Homework Helper
Gold Member
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.

symbolipoint
Homework Helper
Gold Member

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

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

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

If you want to solve an equation like
$$3 x^2 + 2 x - 5 = 0$$​
first, you divide the whole thing by 3 (in order to leave the $$x^2$$ alone), so that now it looks like
$$x^2 + \frac 2 3 x - \frac 5 3 = 0$$​
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):
$$\left(x + \frac 1 3 \right)^2 - \left( \frac 1 3 \right)^2 - \frac 5 3 = 0$$​
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
$$\left(x + \frac 1 3 \right)^2 - \frac {16} 9 = 0$$​
and since x appears only once, now you can solve for it:
$$x = \pm \sqrt {\frac {16} 9} - \frac 1 3 = \pm \frac 4 3 - \frac 1 3$$​
giving the two solutions, 1 and -5/3.
Easy!

. . .

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