Quadratics: Completing the Square

There are several ways to solve quadratic equations.  The first methods that come to mind are factoring (either factoring out a constant or factoring into binomials) and using the quadratic formula. 

But, many quadratics that cannot be solved by factoring into binomials can be solved by a process called completing the square. 

The process of completing the square relies on your knowledge of special products.

Do you remember that $x^2+6x+9=(x+3)^2$?

Do you remember that $x^2-8x+16=(x-4)^2$?

Quadratics that are equal to the square of a binomial are relatively easy to solve: you square root the squared binomial and then you square root the quantity on the other side of the equation for the answer. 

$\eqalign{(x-3)^2&=7\\\sqrt{(x-3)^2}&=\sqrt{7}\\x-3&=\pm\sqrt{7}\\x&=3\pm\sqrt{7}}$

So, how can you use this process to your advantage?

You can turn most quadratics into perfect squares.  Remember factoring quadratics into polynomials?  You had to find the number that added to make the middle ($b$) term in the quadratic and multipled to make the last ($c$) term in the quadratic.  Quadratics that factor into perfect squares add some number to itself to make the middle ($b$) term and multiplies that same number by itself to make the last ($c$) term.  

Example:

In the equation $x^2+6x+9=$ What number adds to itself to make $6$ (as in $6x$)? It's $3$.  Now multiply $3$ by itself: $9$.  Because $3$ is added to itself to make the middle term and multiplied by itself to make the last term, the quadratic is a special product and can be factored into $(x+3)(x+3)$.  If you aren't convinced, FOIL those binomials and see how to add $3x + 3x$ to make the middle term and multiply $3\times3$ to make the final term. 

Knowing this is helpful when you want to MAKE something into a perfect square. 

Example: 

Let's say you have the equation $x^2-6x-14=0$

This is not a perfect square.  But we can make it into one.  First we want to get rid of the -14, because that doesn't fit into our plans.  We can add that to both sides and move it to the other side of the equation.   When you move it, leave yourself some room to add another last term into that quadratic:

$\eqalign{x^2-6x-14&=0\\ +14 & \qquad +14\\ x^2-6x \text{ ____}&=14}$

Now, take a look at the middle term. To make this a perfect square, we have to find a number that adds to itself to make that middle term.  So, divide that middle term in half. $-6\div2=-3$.  Now, the next step in making this a perfect square is to square half of the $b$, and put that in for $c$. $-3\times-3=9$.  So, add $9$ in as the $c$ term.  And, of course, you must add it to both sides:

$\eqalign{x^2-6x \text{ ____}&=14\\+9 & \qquad +9\\x^2-6x +9&=25}$

Now, you can factor the quadratic into a squared binomial.

$\eqalign{x^2-6x +9&=25\\(x-3)(x-3)&=25\\(x-3)^2&=25}$

Now, to solve, you square root both sides of the equation, and solve for both the positive and negative root of 25:

$\eqalign{\sqrt{(x-3)^2}&=\sqrt{25}\\x-3&=\pm5\\x-3=5 \quad & \quad x-3=-5\\x=8\quad & \quad x=-2}$

And now you have two roots or solutions for the quadratic.

To put it more succinctly, the basic process of completing the square is: 

$\eqalign{x^2-20x+19&=0\\ -19 & \qquad -19 \qquad \qquad \qquad \text{isolate }x^2 \text{ and }bx \text{ on one side}\\ x^2-20x+{(\dfrac{20}{2}})^2&=-19 + {(\dfrac{20}{2}})^2\qquad \qquad \text {add }{(\dfrac{b}{2})}^2 \text{ to each side}\\x^2-20x+100&=-19 +100\\x^2-20x+100&=81\\(x-10)^2&=81\\\sqrt{(x-10)^2}&=\sqrt{81}\qquad \qquad \qquad\qquad\text{ square root both sides}&\\x-10=9 \quad & \quad x-10=-9 \qquad \qquad \text{solve for + and - roots}\\x=19 \quad & \quad x=1}$