# Completing the square.

1. Jan 25, 2008

### _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:

2. Jan 25, 2008

### 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.

3. Jan 25, 2008

### Staff: Mentor

4. Jan 25, 2008

### symbolipoint

5. Jan 25, 2008

### _Mayday_

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

6. Jan 25, 2008

### dodo

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.