Algebra/Chapter 9/Completing the Square

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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 $y=a{x}^2+bx+c$:

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:

$x^2+\frac{b}{a}x+{\left(\frac{b}{2a}\right)}^2={(x+\frac{b}{2a})}^2$

Multiply this back out to convince yourself that this works.

4. Therefore the completed square form of the quadratic is:

$\frac{y}{a}+{\left(\frac{b}{2a}\right)}^2={(x+\frac{b}{2a})}^2+\frac{c}{a}$ or, multiplying through by a,

Template:TrigBoxOpen $y=a{(x+\frac{b}{2a})}^2+c-\frac{b^2}{4a}$ Template:TrigBoxClose

Explanation of Derivation

Completing The Square

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:

$x^2+\frac{b}{a}x+{\left(\frac{b}{2a}\right)}^2={(x+\frac{b}{2a})}^2$

Multiply this back out to convince yourself that this works.

4. Therefore the completed square form of the quadratic is:

$\frac{y}{a}+{\left(\frac{b}{2a}\right)}^2={(x+\frac{b}{2a})}^2+\frac{c}{a}$ 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.

2x2 + 24x + 23 = 0	Does not factor easily, so we complete the square.
x2 + 12x + 23/2 = 0	Make coefficient of x2 a 1, by dividing all terms by 2.
x2 + 12x = - 23/2	Add – 23/2 to both sides.
x2 + 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 $\pm \sqrt{x}$.