Download Algebra II

Algebra 2
4.7 Completing the Square
Name_________________________ Date ________
Foil several square binomials to become more familiar with the pattern:
(x + 3)(x + 3)
(x + 7)(x + 7)
(x − 9)(x − 9)
Let’s see how this pattern can help us create perfect squares. Say you had this
trinomial:
x2 + 8x + ______
What constant term would make it a perfect square? ___________
What are the factors? ________________
Try another!!
x2 + 10x + ______.
Do you see a pattern here?
The factors are ____________________.
Divide the linear coefficient by 2 . . . then square the result!!!
Try these:
1. x2 + 18x + ______ = (x +
)2
2. x2 − 20x + ______ = (x −
)2
What if the linear coefficient isn’t even? That’s OK!! It’s easy to square a fraction!!
7
49
Watch: x2 + 7x +_____. Divide 7 by 2 and square!! ( )2 =
2
4
(x +
7 2
) !!
2
Now, we have to get into some hard stuff. What if there is already a constant in the
trinomial, but it’s not the one we want?? And, what if we don’t want to change the value of
the polynomial?
The answer is simple: we will temporarily ignore the given constant!! And then we will just add
the equivalent of 0!!
Consider x2 + 14x + 10. Move the 10 to the right and temporarily ignore it . . .
x2 + 14x + ________ + 10.
Calculate the constant you want!! Divide 14 by 2 and square: 49!!
Now, for the tricky part: Balance the equation! Add and subtract 49!!!! (What is the net
effect if you add and subtract the same number???)
x2 + 14x + 49 + 10 − 49
Rewrite:
(x + 7)2 − 39
You essentially add 0, not changing the value.
Factor the perfect square trinomial and add the leftover.
If you foil and combine like terms, you should get x2 + 14x + 10!!
Try.
x2 − 8x − 2
Just one more thing . . . .
What if a is a number other than 1?
2x2 + 20x + 3
Factor a 2 out of the first two terms only!! 2(x2 + 10x) + 3
Calculate the constant you want . . . Divide 10 by 2 and square: 25. Now, think!! When you
put 25 in the ( ), the ( ) is being multiplied by 2 . . . so you are really adding 50!! So to
counteract that, you have to subtract 50!!
2(x2 + 10x+ 25) + 3 − 50
Rewrite: 2(x + 5)2 − 47 To convince yourself that you still have the same expression,
foil, distribute and combine like terms!! You should end up with 2x2 + 20x + 3.
Why would anyone want to go through all this just to end up with the same expression?
Two reasons!!
Reason one: you might want to write a quadratic equation in “vertex form” instead of
standard form.
Vertex Form: y = a (x – h) 2 + k
The vertex of the parabola is (h, k) !
y = x2 + 6x + 2
standard form
y = x2 + 6x + 9 + 2 − 9
9 is the constant you need for a perfect square
y = (x + 3)2 − 7
vertex form
--Put both functions into your calculator and show they are the same!!!
--Foil the vertex form and get standard form back again . . . more proof!!
Reason 2: you might want to solve a quadratic equation by using square roots.
3x2 + 6x − 2 = 0
It’s easier to move the 2 to the other side.
3(x2 + 2x) = 2
Factor out the 3 . . . you need x2 to have a
coefficient of 1.
3(x2 + 2x + 1) = 2 + 3
Add 3(1) to both sides!!
3(x + 1)2 = 5
(x + 1)2 =
5
3
x + 1 = 
5
3
x = −1 +
5
,−1 −
3
5
3
Solve.
1. x2 + 20x + 104 = 0
3.
6x2 + 84x + 300 = 0
2. x2 - 17x + 200 = 13x - 43
4.
2x2 - 12x = -14