# How to complete the square

Usually you have to complete the square when you’re trying to draw a graph of a parabola that has an equation like:

x^{2} + 5x + 3 = 0

You only know where the vertex of the parabola goes if you can get it in the form

(x – a)^{2} + C = 0

The cool thing about “completing the square” is when you do it, you actually will always be able to get any polynomial into that (x – a)^{2} + C = 0 form.

## How to complete the square

Working with our example x^{2} + 5x + 3 = 0

### 1 >> Take the 5 (from the 5x term) and divide it by 2.

5 - 2

### 2 >> Square it.

5^{2}- 2^{2}= 25 -- 4

### 3 >> *Add* AND *subtract* that squared number from the original polynomial

Now, all you have to do is take that value you just got (25/4) and ADD AND SUBTRACT 25/4 FROM THE ORIGINAL x^{2} + 5x + 3 = 0.

Here is how you do it (pay special attention to WHERE I’m putting the 25/4):

x^{2}+ 5x + 3 = 0 x^{2}+ 5x + 25/4 - 25/4 + 3 = 0

Now the important thing to realize is that what I’ve just written is *completely 100%-ly MATHEMATICALLY EQUAL to the original*

x^{2}+ 5x + 25/4 - 25/4 + 3 = x^{2}+ 5x + 3

OK?? Make sure you know that.

### 4 >> “complete the square”

Next, what you do is, you just write this:

x^{2}+ 5x + 25/4 - 25/4 + 3 = 0 (x + 5/2)^{2}- 25/4 + 3 = 0 (x + 5/2)^{2}- 13/4 = 0

Notice the 5/2 value . . . we’ve seen that before (go look where).

And that’s how its done.

We can check that the answer we’ve written is correct by multiplying back out. We should get the original polynomial when we’re done:

(x + 5/2)^{2}- 13/4 = 0 (x + 5/2)(x + 5/2) - 13/4 = 0 x^{2}+ 2(5/2)x + (5/2)^{2}- 13/4 = 0 x^{2}+ 5x + 25/4 - 13/4 = 0 x^{2}+ 5x + 12/4 = 0 x^{2}+ 5x + 3 = 0

And that’s the same as the original polynomial.

### Ok, that was a specific example. What doing this to polynomials in general?

Just apply the pattern. In general, if you have

x^{2} + Bx + C = 0

You complete the square by doing this:

x^{2}+ Bx + C = 0 ( x + (B/2) )^{2}- (B/2)^{2}+ C = 0

Note also that if you have a polynomial of the form

Ax^{2}+ Bx + C = 0

You change it to being in the form:

A(x^{2}+ (B/A)x) + C = 0

So for example, given

5x^{2}+ 7x + 10 = 0 5( x^{2}+ (7/5)x ) + 10 = 0 5( x + (7/10)^{2}- (7/10)^{2}) + 10 = 0 5( x + (7/10) )^{2}- 49/100 ) + 10 = 0 5( x + (7/10) )^{2}- (5)(49)/100 + 10 = 0 5( x + (7/10) )^{2}+ 151/20 = 0

Check its correct by multiplying back out.

5( x + (7/10) )^{2}+ 151/20 = 0 5( x^{2}+ 2(7/10)x + 49/100 ) + 151/20 = 0 5x^{2}+ 7x + 49/20 + 151/20 = 0 5x^{2}+ 7x + 10 = 0

Which is the same as the original.