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Solving Quadratic Equations by … Factoring_

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Solving Quadratic Equations by … Factoring_ Powered By Docstoc
					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
WAKE-UP CALL!!! : ANSWERS
 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
 quadratic equation are found by
 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=
ROUND 1: ANSWERS
 25y²   – 144 = (5y – 12)(5y+12)




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




BONUS: 49r² – 14r + 1 =
ROUND 2: ANSWERS
 4u²–   36 = 4(u + 3)(u – 3)




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




BONUS: 30u² – 57u + 21 =
ROUND 3: ANSWERS
 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 =
ROUND 4: ANSWERS
 3v²   – 18v = 3v(v – 6)




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




BONUS: x² – 16x + 51 =
 ROUND 5: ANSWERS
 4z²   – 12z + 9 = (2z – 3)²




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




BONUS: 4w² – 13w - 27 =
  ROUND 6: ANSWERS
 5x²   – 7x + 2 = (5x – 2)(x – 1)




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




BONUS: 5p² – 25 = 4p² + 24
ROUND 7: ANSWERS
 2x²   – 2x – 24 = 2(x – 4)(x + 3)




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




BONUS: q² + q =
ROUND 8: ANSWERS
 x²   – 16 = (x – 4)(x + 4)




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




BONUS: 2² – 4y – 8 = -y² + y
ROUND 9: ANSWERS
 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
ROUND 10: ANSWERS
 (5q   – 1)(q – 2)




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

 COMING ATTRACTIONS!
CLOSURE QUESTION
 What  must be true about a quadratic
 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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posted:11/16/2011
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