# Solving Quadratic Equations by … Factoring_

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```					SOLVING QUADRATIC
EQUATIONS BY …
FACTORING!
Chapter 5.2: Day 2
WAKE-UP CALL!!!
Factor  the following problems:
 8x² – 6x – 5

 -v²   + 2v – 1

 q²   -7q -10
 Factor the following problems:
 8x² – 6x – 5   → (2x+1)(4x-5)

 -v²   + 2v – 1   → -(v-1)(v-1)

 q²   -7q -10     → Cannot be factored
DOUBLE SHOT      OF   ESPRESSO!
 Whatare the “zeros” of a quadratic
equation? How do you find them?

 Solve   the equation: 5x² –13x +6 = 0
DOUBLE SHOT      OF   ESPRESSO: SOLVED
 The zeros are a quadratic equations
are the numbers which make the
equation equal to zero. The zeros of a
factoring the equation, setting each
part equal to zero, and then solving
for the variable.

 X=   3/5 , 2
OBJECTIVES
   Be able to recognize special
factoring patterns.
    Be able to solve quadratic
equations by factoring.
IMPORTANT VOCAB REFRESHER
 Standard   form:

 Can   we factor the following:
2y² – 4y – 8 = y² + y
LESSON INTRODUCTION: PART 1
 Differenceof 2 Squares (no x-term):
 Characteristics
Binomial

No x-term (BONUS: Why is there is x-

term?)
Difference of two perfect squares

 General Example: a² – b² = (a + b)(a - b)
 Specific Example: x² – 4 = (x – 2)(x + 2)
EXAMPLES, EXAMPLES, EXAMPLES …
 4x²   – 25 =
EXAMPLES GALORE!
   64x² – 9 =
LESSON INTRODUCTION: PART 2
 Perfect Square Trinomial:
 Characteristics
Trinomial

Factors into the same thing … twice!

 General Example:
a² + 2ab + b² = (a+b)(a+b) = (a+b)²
a² - 2ab – b² = (a-b)(a-b) = (a-b)²
 Specific Example:
x² + 12x + 36 = (x+6)(x+6) = (x+6)²
x² – 8x + 16 = (x – 4)(x – 4) = (x – 4)²
CAN YOU SAY … EXAMPLES!
 k²   + 24k + 144 =

BONUS: What are the zeros of this
equation?
ONE MORE … EXAMPLE!
 49r²   + 14r + 1 =

BONUS: For perfect trinomials are
there always one or always two
zeros?
ROUND 1
 25y²   – 144 =

BONUS: 16s² – 225=
 25y²   – 144 = (5y – 12)(5y+12)

BONUS: 16s² – 225= (4s+15)(4s – 15)
ROUND 2
 4u²   – 36 =

BONUS: 49r² – 14r + 1 =
 4u²–   36 = 4(u + 3)(u – 3)

BONUS: 49r² – 14r + 1 = (7r – 1)²
ROUND 3
 12x²   + 3x + 3 =

BONUS: 30u² – 57u + 21 =
 12x²   + 3x + 3 = 3(4x^2 + x + 1)

BONUS: 30u² – 57u + 21 = 3(2u – 1)(5u – 7)
ROUND 4
 3v²   – 18v =

BONUS: 36w² + 60w + 25 =
 3v²   – 18v = 3v(v – 6)

BONUS: 36w² + 60w + 25 = (6w + 5)²
ROUND 5
 4z²   – 12z + 9 =

BONUS: x² – 16x + 51 =
 4z²   – 12z + 9 = (2z – 3)²

BONUS: x² – 16x + 51 = Cannot be factored
ROUND 6
 5x²   – 7x + 2 =

BONUS: 4w² – 13w - 27 =
 5x²   – 7x + 2 = (5x – 2)(x – 1)

BONUS: 4w² – 13w - 27 = Cannot be factored
ROUND 7
 2x²   – 2x – 24 =

BONUS: 5p² – 25 = 4p² + 24
 2x²   – 2x – 24 = 2(x – 4)(x + 3)

BONUS: (x – 7)(x + 7)
ROUND 8
 x²   – 16 =

BONUS: q² + q =
 x²   – 16 = (x – 4)(x + 4)

BONUS: q² + q = q (q + 1)
ROUND 9
 p²   – 4p + 4 =

BONUS: 2² – 4y – 8 = -y² + y
 p²   – 4p + 4 = (p – 2)²

BONUS: (y + 1)(3y – 8)
ROUND 10
 2q²   + 4q – 1 = 7q² – 7q + 1

BONUS: (w + 6)²= 3(w + 12) – w²
 (5q   – 1)(q – 2)

BONUS: w (2w + 9)
FILL-IN THE BLANK SUMMARY
 What   on earth did we do today?

 COMING ATTRACTIONS!
CLOSURE QUESTION
equation before you can solve for it
(find the zeros)?
CLOSED!
 Answer: The equation must have a
zero on one side and the other side
must be factored.
HOMEWORK: DUE MONDAY
 Pg.   261 # 53, 55, 57, 61, 63

 Prepare for        Monday’s supply check:
   TI-83 plus or TI-83 plus silver edition calculator
   Pencils
   Folder with pockets
   Spiral Notebook or 3 ring binder with paper
   Loose leaf and graph Paper

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