Algebra Period 3 (Tues/Wed)

Solving Quadratics by Completing the Square 13-3
Okay, up until now the methods of solving quadratics should have been fairly familiar to you—but this method is a NEW strategy!!

When does the "square root = +/- square root" method work well?
When the side with the variable is a PERFECT SQUARE!
So what if that side is not a perfect BINOMIAL SQUARED?
You can follow steps to make it into one!

This method is great because then you can just square root each side to find the roots!
THIS METHOD ALWAYS WORKS!

EXAMPLE: x² - 10x = 0
Not a TRINOMIAL SQUARE so it would not factor to a BINOMIAL SQUARED. (However, we know that we could just factor out an x to solve—I want you to see how completing the square works—even with easy ones)
Here's how you can make this into a trinomial square

Step 1: b/2
Take half of the b coefficient in this case (- 10/2 = -5)

Step 2: Square b/2
(-5 x -5 = 25)

Step 3: Add (b/2)² to both sides of the equation
(x² - 10x + 25 = +25)

Step 4: Factor to a binomial square
(x - 5)² = 25

Step 5: Square root each side and solve
SQRT (x - 5)² = SQRT 25
x - 5 = + and - 5

ADD 5 TO BOTH SIDES
x = 5 + and - 5
x = 5 + 5 and x = 5 - 5
x = 10 and x = 0
Now the check for this one—just because we are learning this strategy--is to factor and make sure you get the same results.
that is
x² - 10x = 0 we know becomes
x(x-10) = 0
x = 0 and x = 10

x²-4x -7 = 0
add 7 to both sides
x²-4x =7
take –b/2 or -4/2 = -2 now square that (-2),sup>2 = 4
add 4 to both sides
x²-4x + 4 = 7 +4
x²- 4x + 4 = 11
(x -2)² = 11
take the square root of both sides
SQRT(x -2)² = + or – SQRT 11
x - 2 = + or - SQRT 11

IF THERE IS AN "a" COEFFICIENT, YOU MUST DIVIDE EACH TERM BY IT BEFORE YOU CAN COMPLETE THE SQUARE:

Example: 2x² - 3x - 1 = 0
Move the 1 to the other side of the equation:
2x² - 3x = 1
Divide each term by the "a" coefficient:
x² - 3/2 x = 1/2
Now follow the step to find the completing the square term and add it to both sides:
-b/2 = (-3/2)(1/2) remember (instead of dividing by 2, when you have a fraction, multiply by 1/2)= – 3/4
NOTE: REMEMBER this term you will use it again!!
square that [(-3/2)(1/2)]² = (-3/4)²
= 9/16
x² - 3/2 x + 9/16 = 1/2 + 9/16
(x - 3/4)² = 8/16 + 9/16
(x - 3/4)² = 17/16
SQRT[(x - 3/4)2 ] = + or - SQRT [17/16]
x - 3/4 = + or -[SQRT 17] /4
x = 3/4 + or -[SQRT 17] /4
x = 3 + or - [SQRT 17] /4

When our textbook ask you to complete the square
as in x² - 6x
all it is asking is that you follow the steps
-b/2 is -6/2 = -3
then square that (-3)² = 9
so the answer is
x² - 6x + 9