Let us start with why we do this before anything. There is such a thing as a perfect square trinomial. This means that an expression with three terms can be factored to equal one binomial (2 terms) squared. It’s easier to understand with some examples:

\(x^2 + 6x + 9\) is the same as \((x + 3)^2\)

\(3 + 3 = 6\) (the middle term) and \(3\centerdot3 = 9\) (the last term)

\(x^2 - 8x + 16\) is the same as \((x - 4)^2\)

\(-4 + -4 = -8\) (middle term) and \(-4\centerdot-4 = 16\) (last term)

This factoring is helpful when trying to reduce or solve a problem like this.

Ok, so now with that said, completing the square is transforming an expression that’s not perfect, into one that is!

Let’s start with just figuring out the missing piece that would complete the square. Did you notice any patterns in the above examples? Where did they get that 3 from in the first example? How about the 4 in the second?

To complete the square, you divide the number with the “x” by two and then square that number to find the last term.

What would “c” equal?                   \(x^2 + 12x + c\)

12 divided by two is 6, and 6 squared is 36, so \(c = 36!\)

\(x^2 + 12x + 36 = (x + 6)^2\)

Let’s try a tougher one:

What would “c” equal?                   \(x^2 + 5x + c\)

This time, 5 divided by 2 is 2.5 and 2.5 squared is 6.25. To do this in fraction form, leave it as \(\frac{5}{2}\) and \(\frac{5}{2}\) squared is \(\frac{25}{4}\) (square the numerator and the denominator.)

\(x^2 + 5x + \frac{25}{4} = (x + \frac{5}{2})^2\)

Wow. That one was messy. What if a piece wasn’t missing but instead, the trinomial was just not a perfect square? We want to transform it into a perfect square trinomial! We start with:

\(x^2 + 10x - 75 = 0\)

This is not a perfect square trinomial because half of 10 is 5 and 5 squared is 25, but there is a -75 there instead. What we do is move the 75 over to the other side of the equation, so add it to both sides.

\(\array{ x^2 + 10x -75 =&  0 \cr +75 & +75}\)

\(x^2 + 10x = 75\)

We decided that 25 would be the missing piece to make this trinomial perfect, so let’s get this 25 in there, but we must add it in to both sides so that we don’t change the equation.

\(\array{ x^2 + 10x =&  75 \cr +25 & +25}\)

\(x^2 + 10x + 25 = 100\)

We can now use this information to change its form and solve!

\(x^2 + 10x + 25 = 100\)

\((x + 5)^2 = 100\)

Take the square root of both sides. Remember, it splits between a positive AND a negative answer.

\(\array{ x + 5 =& -10 \cr -5 & -5}\)                                                           \(\array{ x + 5 =& 10 \cr -5 & -5}\)

              \(x = 5\)                                                                                \(x = -15\)

Bellow you can download some free math worksheets and practice.