Algebra I
Name
Module 3
Lesson 10: Solving Quadratic Equations by Completing the Square
Period
Date
Day #1
Factor the following expressions
#1 x2 + 10x + 25
#2
x2 – 8x + 16
#3
x2 + 20x + 100
#4
#5
x2 – 2x + 1
#6
x2 +24x + 144
x2 – 18x + 81
What do all six of our answers have in common?
When you can factor a trinomial into two binomials that are exactly the same, it is called a
.
[There is a quick and easy way to see if a trinomial is a perfect square trinomial. Find the
middle number and take half of it. Now square that number. Is it the last number of the
problem? If so, you have a perfect square trinomial]
Now, not every trinomial is a perfect square trinomial. In fact, not all trinomials can even
be factored. But what if there was a way that we could FORCE a trinomial to be a perfect
square trinomial?
My friends, there is a way. It is called…
C
T
S
How do we do this? Turn the page --------------------------------------------------------------------->
Steps
1.
Group the first two terms together.
2.
Determine what number you need to complete the square. (Hint: take ½ of b and
square it.)
3.
Add that number and it’s opposite to the expression.
4.
Factor the first three terms and write as the square of a binomial.
5.
Simplify the remaining terms.
Rewrite each expression by completing the square
#7
a2 – 4a + 15
#8
n2 – 2n – 15
#9
c2 + 20c – 40
#10
x2 + 18x + 50
#11
k2 + 12k + 32
#12
m2 – 4m – 5
#13
x2 – 8x + 20
#14
x2 + 8x + 10
#15
c2 – 6c + 11
#16
v2 + 18v + 88
#17
x2 – 14x + 20
#18
n2 + 30n – 100
Day #2
Example #1: Solve the following equations
x2 + 12x + 27 = 0
x2 +10x – 6 = 0
Example #2: Solve the following equation by completing the square
x2 + 12x + 27 = 0
1.)
2.)
3.)
4.)
STEPS
Get all the variable terms on
one side of the equation and the
constant terms on the other side.
Complete the square. Whatever you
added to one side, be sure to add
to the other side. KEEP IT BALANCED!
Take the square root of BOTH sides.
Don’t forget the ±!
If possible, simplify your answer.
So we got the same answers as we did above. Why would anyone use this method if it is so
much more work?
Solving equations by completing the square allows us to solve equations that we aren’t
able to factor. This is one way to solve those “unfactorable” type equations. There is
another method that we will learn about next week.
Solve the following equations by completing the square.
Example #3:
Example #4:
x2 +10x – 6 = 0
x2 + 6x + 12 = 0
REGENTS QUESTION [2 points]
A student was given the equation x2 + 6x – 13 = 0 to solve by completing the square. The
first step that was written is shown below.
x2 + 6x = 13
The next step in the student’s process was x2 + 6x + c = 13 + c.
State the value of c that creates a perfect square trinomial.
Explain how the value of c is determined.
CLASSWORK: Solve the following equations by completing the square
1.)
x2 + 10x = 5
2.)
x2 – 12x = 7
3.)
x2 – 10x + 16 = 0
4.)
x2 – 2x = 12
5.)
x2 – 2x – 7 = 0
6.)
x2 + 14x – 5 = 8
Day #3
Completing the square is useful only under certain conditions. I recommend using
completing the square only when…
1.) You CANNOT
2.) The middle term is an
3.) The leading coefficient is
the trinomial.
number.
.
We will learn how to solve ‘unfactorable’ equations over the next couple of days…but for
now, let’s look at a mixture of problems. Some can be factored, while others cannot. You
can complete the square on all of these, but if there is an easier was to solve the problem
(like factoring), I suggest you do so!
STEP #1:
STEP #2:
STEP #3:
Is your equation equal to zero? If not, get is equal to zero.
Can the equation be factored? If so, do it…and solve!
If it cannot be factored, try completing the square. Since you cannot
factor it, your answers will not be pretty. That’s ok!
#1
x2 – 6x + 8 = 0
#2
x2 – 10x + 6 = 0
#3
x2 + 4x – 9 = 0
#4
x2 + 8x – 4 = 0
#5
x2 + 5x – 14 = 0
#6
x2 + 6x = 40
#7
x2 + 2x = 35
#8
x2 – 9x = 10
TRY THESE…
#1 x2 + 10x – 18 = 0
#2
x2 – 8x – 17 = 0
#3
#4
x2 – 7x = 18
x2 – 6x – 27 = 0
#5
x2 + 2x – 5 = 0
#6
x2 – x – 20 = 0