# Algebra/Chapter 9/Completing the Square
## Derivation

The purpose of "completing the square" is to either factor a prime quadratic equation or to more easily graph a parabola. The procedure to follow is as follows for a quadratic equation $y=ax^{2}+bx+c$:
1. Divide everything by a, so that the number in front of $x^{2}$ is a perfect square (1):
    ${\frac {y}{a}}=x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}$
2. Now we want to focus on the term in front of the x. Add the quantity $\left({\frac {b}{2a}}\right)^{2}$ to both sides:
    ${\frac {y}{a}}+\left({\frac {b}{2a}}\right)^{2}=x^{2}+{\frac {b}{a}}x+\left({\frac {b}{2a}}\right)^{2}+{\frac {c}{a}}$
3. Now notice that on the right, the first three terms factor into a perfect square:
    $x^{2}+{\frac {b}{a}}x+\left({\frac {b}{2a}}\right)^{2}=\left(x+{\frac {b}{2a}}\right)^{2}$
Multiply this back out to convince yourself that this works.
4. Therefore the completed square form of the quadratic is:
    ${\frac {y}{a}}+\left({\frac {b}{2a}}\right)^{2}=\left(x+{\frac {b}{2a}}\right)^{2}+{\frac {c}{a}}$ or, multiplying through by a,
$y=a\left(x+{\frac {b}{2a}}\right)^{2}+c-{\frac {b^{2}}{4a}}$

## Explanation of Derivation

1. Divide everything by a, so that the number in front of $x^{2}$ is a perfect square (1):
    $x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}={a}$
Think of this as expressing your final result in terms of 1 square x. If your initial equation is
2. Now we want to focus on the term in front of the x. Add the quantity $\left({\frac {b}{2a}}\right)^{2}$ to both sides:
    ${\frac {y}{a}}+\left({\frac {b}{2a}}\right)^{2}=x^{2}+{\frac {b}{a}}x+\left({\frac {b}{2a}}\right)^{2}+{\frac {c}{a}}$
3. Now notice that on the right, the first three terms factor into a perfect square:
    $x^{2}+{\frac {b}{a}}x+\left({\frac {b}{2a}}\right)^{2}=\left(x+{\frac {b}{2a}}\right)^{2}$
Multiply this back out to convince yourself that this works.
4. Therefore the completed square form of the quadratic is:
    ${\frac {y}{a}}+\left({\frac {b}{2a}}\right)^{2}=\left(x+{\frac {b}{2a}}\right)^{2}+{\frac {c}{a}}$ or, multiplying through by a,

## Example

The best way to learn to complete a square is through an example. Suppose you want to solve the following equation for x.
| 2x$^2$ + 24x + 23 = 0 | Does not factor easily, so we complete the square. |
| x$^2$ + 12x + 23/2 = 0 | Make coefficient of x$^2$ a 1, by dividing all terms by 2. |
| x$^2$ + 12x = - 23/2 | Add – 23/2 to both sides. |
| x$^2$ + 12x + 36 = - 23/2 + 36 | Take half of 12 (coefficient of x), and square it. Add to both sides. |
| (x + 6)$^2$ = 49/2 | Factor. Now we can take square roots to easily solve this form of the equation. |
| √(x + 6)$^2$ = √49/√2 | Take the square root. |
| x + 6 = 7/√2 | Simplify. |
| x = -6 + (7√2)/2 | Rationalize the denominator. |

## Square Root Is Not Negative

We often say that ${\sqrt {x}}$ is the positive number, which when squared is equal to $x$. If $x>0$ this is perfectly correct, but if $x=0$ then ${\sqrt {x}}={\sqrt {0}}=0$ which is not positive. So to be technically correct (which is part of the fun of math) we should say that ${\sqrt {x}}$ is the non-negative number, which when squared is equal to $x$.
If it is needed to express that a square root may be both positive and negative, you will see $\pm {\sqrt {x}}$.