Notes from class on completing the square!

1. PC - Functions Name ________________________
Completing the Square - Notes
Vocabulary:
Completing the square is the process of adding a constant """ to the expression
# $ + &# to make it a perfect square trinomial.
Examples of perfect square trinomials:
# $ + 8# + 16 = (# + 4)$
# $ + 6# + 9 = (# + 3)$
# $ + 4# + 4 = (# + 2)$
The quadratics above already have a “c” value that makes it a perfect square trinomial. Look
for a pattern between the “b” and “c” value of each. What do you notice?
2
When the “a” of a quadratic is 1, we can find a perfect “c” value by ( )$ . If you notice, half of
$
the “b” value squared always gives us the perfect “c.”
Try it on your own for these examples. Find the perfect “c” and factor.
# $ + 12# + ____ =
# $ + 16# + ____ =
# $ − 18# + ____ = Did this work with a negative “b?”
So, now you know how to come up with the perfect “c” value. We can use completing the
square to either solve a quadratic that’s not factorable (or is for that matter) or to find vertex
form. The rule is, whatever you add to one side of an equation, you must
__________________________________.

2. Finding Vertex Form:
1. Notice that we have a “c,” but it’s not
5(#) = # $ + 10# + 13 perfect. First step is to always move the “c”
-13 -13 to the other side.
2. Find the perfect “c” for the remaining
5(#) − 13 = # $ + 10# + ________ terms. Add this to both sides to balance the
equation.
5(#) − 13 + 25 = # $ + 10# + 25
3. Factor the trinomial into a perfect square
5(#) + 12 = # $ + 10# + 25 binomial.
4. Finish writing vertex form by getting f(x)
5(#) + 12 = (# + 5)$ -12 by itself.
-12
5(#) = (# + 5)$ − 12
Solving using completing the square:
52 = # $ + 14# − 26 1. Notice that we have a “c,” but it’s not
perfect. First step is to always move the “c”
+26 +26
to the other side.
78 = # $ + 14# + ________ 2. Find the perfect “c” for the remaining
terms. Add this to both sides to balance the
78 + 49 = # $ + 14# + 49 equation.
3. Factor the trinomial into a perfect square
127 = # $ + 14# + 49 binomial.
127 = (# + 7)$ 4. This time we solve for “x” by undoing all
the operations surrounding “x.”
±√127 = ;(# + 7)$ 5. Reduce the radical if possible.
±√127 = # + 7
Example √8 = 2√2
-7 -7
# = −7 ± √127
Modify the above work if “a” is not 1. What do you think you have to do? Try it on the
following examples. If you just see f(x), this means write a new function. Otherwise, solve.
5(#) = 2# $ + 24# − 36
77 = 3# $ − 18# + 33