Completing the square

Using the method of addition squared, we can express the quadratic function ax² + bx + c in the form (x + m)² + n.

The squaring algorithm

The algorithm uses the classical formula for bracket augmentation:

$$(a+b)^2=a^2+2ab+b^2$$

The procedure is as follows. For now, suppose a = 1. Suppose we have a quadratic function f(x) = x² + 6x + 1 and want to convert it to the form (x + m)² + n. The first thing we do is to determine the values from the first parenthesis: (x + m)². What number do we need to put after m? According to the formula, the equality (x + m)² = x² + 2mx + m² must hold . In the function f we have x², that's fine. But in place of the linear term we have 6x. What value must we choose after m for 6x = 2mx to be valid ? Obviously m = b/2 = 6/2 = 3 must be valid . So we always put half the value of b from the quadratic function after m.

This gives us the form (x + 3)² − n. It remains to determine the number n. We now know that (x + m)² = x² + 2mx + m² is valid. The x² + 2mx part is wanted, but the expression m² is left over, there is no equivalent in the function f. Therefore, we subtract m² from the current result to get the form (x + m)² − m², which when multiplied equals x² + 2mx + m² − m². After subtraction, we only have x² + 2mx.

But in the function f we still have the absolute term c. We simply add it. We get the form (x + m)² − m² + c, after multiplying x² + 2mx + m² − m² + c and after simplifying x² + 2mx + c. Because m = b/2, this expression is equal to x² + bx + c, so after multiplying the expression (x + m)² − m² + c we get back the original function, we have calculated correctly. (I repeat that a = 1, nowhere is it missing.)

So for the sample function f, we get an expression in the form of a square (x + 3)² − 9 + 1 = (x + 3)² − 8. We can try to multiply it back:

$$\begin{eqnarray} (x+3)^2-8&=&x^2+6x+9-8\\ &=&x^2+6x+1 \end{eqnarray}$$

The square addition method is used, for example, when you want to draw the graph of a quadratic function.

When a≠1

In the case where we have a function for which it is not true that a = 1, we proceed by plotting the whole function with the number a and proceed as we already know how. Example:

$$2x^2+16x-12=0$$

We output a two from the whole function:

$$2\cdot(x^2+8x-6)=0$$

We can already convert the function inside the parentheses to a square.

$$x^2+8x-6=(x+4)^2-16-6=(x+4)^2-22$$

We plug this result into the previous equation to get:

$$2\cdot(x^2+8x-6)=2\left((x+4)^2-22\right)=2(x+4)^2-44$$