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Unit 4: Part 3
Solving Quadratics Day 1 Factoring Day 2 Zero Product Property
Day 3 Small Quiz: Factoring & Solving
Day 4 Quadratic Formula (QF) Day 5 Completing the Square (CTS) Day 6 Review: Solving Quadratics by Factoring, QF, CTS
Day 7 Quiz: Unit 4 Part 3 – Solving Quadratics
Tentative Schedule of Upcoming Classes
Day 1 A Mon 11/ 30
Factoring B Tues 12/1
Day 2 A Wed 12/2
Zero Product Property B Thurs 12/3
Day 3 A Fri 12/4 Quiz: Factoring & Solving Skills Review Assigned B Mon 12/7
Day 4 A Tues 12/8 Quadratic Formula B Wed 12/9
Day 5 A Thurs 12/10 Completing the Square B Fri 12/11
Day 6 A Mon 12/14 Review
Skills Check: Skills Review #2 B Tues 12/15
Day 7 A Wed 12/16 Quiz B Thurs 12/17
Absent?
See Ms. Huelsman AS SOON AS POSSIBLE to get work and any help you need.
Notes are always posted online on the calendar. (If links are not cooperative, try changing to “list” mode)
Handouts and homework keys are posted under assignments
You may also email Ms. Huelsman at Kelsey.huelsman@lcps.org with any questions!
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Need Help?
Ms. Huelsman and Mu Alpha Theta are available to help Monday, Tuesday, Thursday, and Friday mornings in L506 starting at 8:10.
Ms. Huelsman is in L402 on Wednesday mornings.
Need to make up a test/quiz?
Math Make Up Room schedule is posted around the math hallway & in Ms. Huelsman’s classroom
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Day 1 Notes: Factoring Review
Let’s review factoring techniques for quadratic functions so that we can find the x-intercepts of a quadratic function without graphing.
Discuss: What are some things we know about the QUADRATIC function family? How did we find key parts of a Quadratic?
1. Vertex
2. Y-intercept
3. Increasing and decreasing intervals
4. End behavior
5. Zero’s (solutions, roots, x-intercepts)
WHY FACTOR quadratics? To find the zero’s analytically 1. Factoring quadratic TRINOMIAL with leading coefficient of 1 (a = 1)
Steps to factoring:
1. Decide the signs for the parentheses based on the CONSTANT TERM. c is positive same signs
c is negative different signs
2. Find 2 #’s that MULTIPLY to the constant term and ADD to the linear term
3. Write as two binomials. (note: The LARGER # from step 2 should go with the sign of the linear term.)
4. Check your answer (FOIL)
cbxx ++2 Quadratic
Term
Linear Term
T
Constant
Let’s practice: 1. 32142 +− xx 2. 3652 −+ ss
A SPECIAL TRINOMIAL - Perfect Square Trinomials:
Example: x2 – 16x + 64 =
We can summarize this pattern:
a2 + 2ab + b2 = (a + b)(a + b) = (a + b)2
a2 - 2ab + b2 = (a - b)(a - b) = (a - b)2
1. x2 + 10x + 25 = 2. 4x2 + 12x + 9 =
A SPECIAL BINOMIAL - Difference of two squares:
Example: x2 – 25 = x2 + 0x – 25 =
We can summarize this pattern:
a2 – b2 : (a + b)(a – b)
1. x2 – 9 = 2. 49x2 – 121 = 3. x2 + 100 =
What if the leading coefficient is NOT 1?
Factoring out a Greatest Common Factor
1. -4x2 + 196 2. x4 + 13x3 + 30x2
What if we don’t have a GCF and we’re stuck with a coefficient that isn’t 1?
Factoring quadratic TRINOMIAL with leading coefficient ≠ 1
The leading coefficient is _________.
Steps for factoring trinomials by grouping 1. Decide your signs for the parentheses. Suggestion – put your negative first.
2. Multiply CA•
3. Find 2 #’s that multiply to equal CA• and add to the linear term (B).
4. Rewrite Bx as a sum of the two factors. There will be 4 terms.
5. Factor by grouping:
Group the first two terms and the last two terms
Factor the GCF out of each group {the leftovers in parentheses should match}
Use distributive property to write as two binomial
6. Check your answer – FOIL!!
Example: Factor 672 2 ++ xx
Step 1: Decide signs (neg first)
Step 2: Mult A * C
Step 3: M A* C / A B
Step 4: Rewrite Bx as sum of 2 factors
Step 5: Factor by Grouping
Step 6: FOIL to check
CBxAx ++2
Quadratic Term
Linear Term
Constant
Leading Coefficient
Factor
1. 21315 2 ++ xx 2. 73112 2 +− aa
3. 8187 2 +− yy 4. 5129 2 −− xx
5. 26 5 4x x+ − 6.
23 18 15x x− +
Factoring Practice: x4 – 25 (x + 1)2 – 4 5x2 – 125
3x2 + 18x - 48
x2 + 24x + 144 2x2 + 5x + 3
4x4 – 4x2 + 1 x6 – y4 x2 – 14x + 49
5136 2 −− yy X2 + 9 4X2 + 100
Day 2 Notes: Using the Zero Product Property
Let’s learn and apply the zero product property to solve a quadratic equation so that we can find the solutions (x-intercepts) of quadratics that can't be solved with the square root method.
Key Ideas About SOLUTIONS to Quadratic Equations
• Quadratics can have 1, 2, or 0 REAL solutions. • The REAL solution is the x-coordinate where the parabola crosses the x-axis. • The y-coordinate of ANY point on the x-axis is 0, so…….
to find solutions, we set our quadratic = 0.
We've solved quadratics using the SQUARE ROOT METHOD…
1. x2 – 10 = 0
2. Solve the equation (x – 6)2 = 4 (Why can you use the square root method on this?)
BUT WHAT IF we are stuck with an x term? We must solve the quadratic
BY FACTORING and USING THE ZERO PRODUCT PROPERTY
Zero Product Property: If the product of two expressions is zero then one or both of the expressions must be zero.
If A·B = 0, then either A = 0 or B = 0
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Notice that the
FACTORS and the SOLUTIONS switch signs!
We can apply this concept to quadratic equations to find the solutions:
If x2 – 6x – 5 = 0
(x – 1)(x – 5) = 0
Then either (x – 1) = 0 or (x – 5) = 0
Therefore x = _____ or x = ______ (Graph this to check the intercepts)
1. Solve the quadratic equation x2 – 6x = 0 (Discuss: How is this different from our example?)
2. Solve the quadratic equation x2 – 4x = 45
Discuss: How is using the Zero Product Property to solve a quadratic equation related to the intercept form of a quadratic function?
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3. Find the solutions of 2x2 - 11x + 12 = 0
4. Find the solutions to the quadratic equation –4x2 – 8x – 3 = 3 – 5x2
5. Find the x-intercepts of f(x) = –2x2 – 14x – 10 (What must you do first?)
6. Solve the equation 6x2 – 13x + 3 = –3
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Day 4 Notes: The Quadratic Formula
Let’s learn a new method of solving quadratics – the quadratic formula so that we can solve a quadratic equation even when it won't factor. Part 1: The Discriminant –
How can we tell quickly whether a quadratic has 0, 1, or 2 real solutions? Is there an analytic way to determine this?
1. f(x) = x2 – 2x 2. f(x) = –x2 + 2x
a = ______ b = ______ c = ______ a = ______ b = ______ c = ______
Now calculate: (b)2 – 4(a)(c) = ________ Now calculate: (b)2 – 4(a)(c) = ________
** Discuss: Why are the parentheses important?
3. f(x) =–x2 + 2x - 1 4. F(x) = x2 – 2x + 2
a = ______ b = ______ c = ______ a = ______ b = ______ c = ______
Now calculate: (b)2 – 4(a)(c) = ________ Now calculate: (b)2 – 4(a)(c) = ________
Discuss: What are your observations?
The DISCRIMINANT (b)2 – 4(a)(c) : if DISCRIMINANT > 0, then
DISCRIMINANT = 0, then
DISCRIMINANT < 0, then
Why do I even need that if I can use my calculator?
Try this one -- How many real solutions for y = x2 + 2x – 143 ?
Where does this come from?
1. Describe the nature of the solution set and determine the solutions to:
x2 – 5x = 4
Get this ready: Can this be factored?
a. DESCRIBE the nature of the solutions (or solution set)
a = ______ b = ______ c = ______ Discriminant: (b)2 – 4(a)(c) = ________
b. SOLVE the quadratic
Use the quadratic formula: x =
2. Describe the nature of the solution set and determine the solutions to:
4x2 + 10x = -10x - 25
Get this ready: Can this be factored?
a. DESCRIBE the solutions (or solution set)
a = ______ b = ______ c = ______ Discriminant: (b)2 – 4(a)(c) = ________
b. SOLVE the quadratic
Use the quadratic formula: x =
3. Describe the nature of the solution set and determine the solutions to:
x2 - 6x = -10
Get this ready: Can this be factored?
a. DESCRIBE the solutions (or solution set)
a = ______ b = ______ c = ______ Discriminant: (b)2 – 4(a)(c) = ________
b. SOLVE the quadratic
Use the quadratic formula: x =
How about these? Can you SOLVE these quadratics? Discuss: Could you use factoring?
1. x2 + 2x + 1 = 0
2. 2x2 + 4x – 1 = 0
3. 3x2 – 4x = -5
Day 5 Notes: Completing the Square
Let’s learn ANOTHER method of solving quadratics to use when we can’t factor… so that later, we can find the vertex and center of important shapes.
We’ve been having FUN with quadratics – SOLVING them, & GRAPHING them.
1. Solve & Graph y = x2 – x - 12
2. Solve & Graph y = (x – 2)(x + 2)
3. y = x2 + 10x + 25
Solve it:
Graph it:
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Dealing with perfect square trinomials is NICE! Our next method takes advantage of this and FORCES quadratics into being a perfect square trinomial.
Background for Completing the Square – Making a perfect square trinomial
Trinomial = ax2 + bx + c Constant term (or c) = 2
2
b
Fill in c to make these quadratics a perfect square trinomial
y = x2 – 4x + ____
y = x2 + 4x + ____
y = x2 – 6x + ____
y = x2 + 6x + ____
y = x2 – 8x + ____
y = x2 + 8x + ____
How were these similar? How are they different?
Steps to Solving Quadratics by COMPLETING THE SQUARE
1. Move constant to one side of the equation 2. Calculate the “c” needed to complete the square 3. Add this value to BOTH sides 4. Rewrite your trinomial as a perfect square 5. Use the square root method to solve the equation
1. x2 – 10x + 1 = 0
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Discuss: How is completing the square related to the vertex form of a quadratic?
2. x2 – 10x = –22
SOLVE this quadratic by completing the square. Note – COULD we factor these?
3. x2 – 12x + 4 = 0 Where is the vertex? __________
4. x2 – 4x + 8 = 0
Where is the vertex? __________
Will this parabola cross the x-axis? Why?
5. x2 + 6x + 4 = 0 Where is the vertex? __________
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How to: Solve by Completing the Square when a≠1 3x2 – 36x + 150 = 0 Divide each side (every term) by the coefficient of x2
YOU TRY. Solve by completing the square.
1. 2x2 + 8x + 14 = 0
Discuss: Why would you use the method of …. Give an example of a quadratic for each method.
1. FACTORING
2. SQUARE ROOT METHOD
3. COMPLETING THE SQUARE
What is your favorite method of solving a quadratic and WHY?
Review: Solving Quadratics – Putting It All Together
Factor completely.
1. 9x 2 −25 2. x2−2x −15
3. 3x 2 +5x −12 4. 9x 2 −30x +25
5. 8x 2 −18 6. 5x 2 −30x +45
7. Give 3 synonyms for “solution”: ___________, ___________, _____________
Our QUADRATIC TOOLBOX
We know 4 ways to solve quadratic equations. List them. Explain each technique and WHEN you would use it.
1.
2.
3.
4.
Using the QUADRATIC TOOLBOX
Solve the following using TWO techniques. Specify each one & explain why you chose it.
1. x2 – 7x + 12 = 0 Method 1:____________________ Reason:
Work:
Method 2:____________________ Reason:
Work:
2. x2 = 1 - x Method 1:____________________ Reason:
Work:
Method 2:____________________ Reason:
Work:
3. x2 = 3x + 15 Method 1:____________________ Reason:
Work:
Method 2:____________________ Reason:
Work:
4. x2 + 8x – 13 = 0 Method 1:____________________ Reason:
Work:
Method 2:____________________ Reason:
Work:
Solve by the Zero Product Property (Factoring)
x2 – 2x – 8 = 0
Solutions:____________
Solve by Completing the Square
x2 – 2x – 8 = 0
Solutions:____________
Solve Using the Quadratic Formula
x2 – 2x – 8 = 0
Solutions:____________
Rewrite to Intercept Form y = x2 – 2x – 8
Intercept form:_____________
Identify the x-intercepts
( , ) and ( , )
Rewrite to Vertex Form y = x2 – 2x – 8
Vertex Form:_______________
Identify the Vertex: ________
Describe the Root
Find the Discriminant: ______
Identify the Number and Type of Solutions: _________________
Given y = x2 – 2x – 8
Find the Vertex: ________
Axis of symmetry: _______
Direction: ______
Size (Vertical Stretch, Vertical Shrink or Standard): _____________________
Graph: y = x2 – 2x – 8
For the quadratic x2 – 2x – 8 = 0
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