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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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