**Completing the Square**

Today I would like to discuss the method of completing the square. This is a technique with many useful applications. In particular, there are several types of problems on the revised SAT where this technique can be used.

We complete the square on an expression of the form

*x*^{2} + *bx*

To complete the square we simply take half of *b*, and then square the result. In other words we get (*b*/2)^{2}.

The expression *x*^{2} + *bx +* (*b*/2)^{2} is always a perfect square. In fact,

*x*^{2} + *bx +* (*b*/2)^{2 }= (*x* + *b*/2)^{2}

**Example 1:** Let’s complete the square in the expression *x*^{2} + 6*x*.

Well half of 6 is 3, and when we square 3 we get 9. So the new expression is *x*^{2} + 6*x *+ 9 which factors as (*x* + 3)^{2}.

**Important notes: **(1) When we complete the square we usually get an expression that is __not__ equal to the original expression. For example,

*x*^{2} + 6*x ≠ x ^{2} + *6

*9.*

*x*+(2) The coefficient of *x*^{2} __must__ be 1 before we complete the square. So, for example, we cannot complete the square on the expression

2*x*^{2} + 32*x*.

But we can first factor out the 2 to get 2(*x*^{2} + 16*x*), and then complete the square on the expression *x*^{2} + 16*x* to get

2(*x*^{2} + 16*x + *64).

Note that we increased the expression by 2 ⋅ 64 = 128 .

**Example 2:**

*x*

^{2}– 8

*x*

If the method of completing the square is used to rewrite the expression above in the form (*x* – *h)*^{2} + *k, *then what is the value of *h – k* ?

I will provide you with the solution to this problem next week. In the meantime feel free to try this problem yourself and post your own solution in the comments.

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