Unit 4 Part 3: Methods of Solving Quadratic Equations

Day 1: The Square Root Method

Day 2: Factoring

Day 3: Zero Product Property

Day 4: Completing the Square

Day 5: Quadratic Formula

Day 6: Review ALL methods of solving quadratics

Day 7: Quiz on Solving Quadratics

Day 8: Systems of Equations and review

Day 9: Review for Unit 4 Test

Day 10: UNIT 4 Test (all 3 notes packets)

Tentative Schedule of Upcoming Classes

Day 1

A

Thurs11/17

Solving quadratics using the square root method.

B

Fri 11/18

Day 2

A

Mon 11/21

Factoring review

B

Tues11/22

Day 3

A

Mon 11/28

Solving quadratics using the zero product property.

B

Tues11/29

Day 4

A

Wed 11/30

Solving quadratics by completing the square.

B

Thurs 12/1

Day 5

A

Fri 12/2

Solving quadratics using the quadratic formula.

B

Mon 12/5

Day 6

A

Tues 12/6

Skills check 1 and review

B

Wed 12/7

Day 7

A

Thurs 12/8

Quiz

Skills review #2

B

Fri 12/9

Day 8

A

Mon 12/12

Solving Systems - review

B

Tues 12/13

Day 9

A

Wed 12/14

Unit 4 test review (all four parts)

B

Thurs 12/15

Day 10

A

Fri 12/16

Unit 4 Test

B

Mon 12/19

*Skills Check #1 will be on Review Day!

Absent?

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Monday, Tuesday, Thursday, and Friday mornings in L506 starting at 8:10.

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Day 1: Solving Quadratics by Square Root Method

Graph the following equations on the grid provided and estimate the zeroes.

Solve the equation algebraically by substituting a “0” for y.

Today we will begin solving quadratic equations algebraically. Remember that another name for the x-intercepts in an equation is the “solutions”.

Graph the following equations on the grid provided.

Solve the equation algebraically by substituting a “0” for y.

What can we conclude about the solutions to a quadratic equation and the graph?

Solve the following quadratic equations using the square root method:

1. Isolate the squared term or binomial

2. Take the square root of both sides

3. Solve the resulting equation(s)*

*Don’t forget that there are two answers when we square root both sides of an equation +/-.

1.2.

3. 4.

5. 6.

7. 8.

9. 10.

Applications:

11. The product of two positive numbers is 140. Determine the numbers if the larger is 5.6 times the smaller.

Let _________= _________

Then _________=_________

Equation:

Solution:

12. The area of a rectangle is 30 square meters. Find the length and width if the length is 1.45 times the width.

Let =

Then _________=_________________

Equation:

Solution:

Day 2 Notes: Factoring Review

The Goal: To review factoring techniques for quadratic expressionsThe Reason: So that we can find the x-intercepts of a quadratic function without graphing.

1. Factoring TRINOMIALS with leading coefficient of 1 (a = 1)

Quadratic Term

Term

Constant term

Linear Term

Term

Steps to factoring:

1. Find 2 #’s that MULTIPLY to the constant term and COMBINE to the linear term

2. Write as two binomials.

3. Check your answer (FOIL)

1. x2 – 14x - 32 2. x2 + 5x – 36 3. x2 +7x +30

2. A SPECIAL TRINOMIAL - Perfect Square Trinomials:

1. x2 – 16x + 64 2. x2 + 10x + 25 3. x2 + 24x + 144

4. x2 – 14x + 49 5. 4x2 + 12x + 96. 4x4 – 4x2 + 1

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

3. A SPECIAL BINOMIAL - Difference of two squares:

1. x2 – 25 = x2 + 0x – 25 2. x2 – 9

3. 49x2 – 121 4. x2 + 100

5. x4 – 256. x6 – y4

We can summarize this pattern:

a2 – b2 :(a + b)(a – b)

4. Factoring out a Greatest Common Factor (GCF)

1. 5x2 – 1252. -4x2 + 196

3. x4 + 13x3 + 30x24. 3x2 + 9x

What if we don’t have a GCF and the coefficient ≠1?

5. Factoring trinomials with a leading coefficient ≠ 1

Leading Coefficient

Coefficient

Constant

rm

Linear Term

Quadratic Term

Steps for factoring trinomials by grouping

1.

Multiply

2.

Find 2 #’s that multiply to equal and combine to the linear term (B).

3. Rewrite Bx as a sum of the two factors. There will be 4 terms.

4. Factor by grouping:

Group the first two terms and the last two terms

Factor the GCF out of each group {the parentheses should match}

Use distributive property to write as two binomials

5. Check your answer by multiplying (FOIL).

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11.

Day 3 Notes: Using the Zero Product Property

The Goal: Learn and apply the zero product property to solve a quadratic equation

The Reason: Once you factor, you are now in INTERCEPT form. You can now find your x-intercepts and graph in INTERCEPT form (instead of standard form)

Key Ideas About SOLUTIONS to Quadratic Equations

· Quadratics can have 0, 1, or 2 REAL solutions.

· The REAL solution is the x-coordinate where the parabola crosses/touches 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. Solve: x2 – 10 = 0

2. Solve: (x – 6)2 = 4

BUT WHAT IF… the equation is in the form:

We can solve the quadratic by factoring and using the zero product property.

Zero Product Property: If the product of two expressions equals zero then one or both of the expressions gives us a zero. If A·B = 0, then either A = 0 or B = 0

We can apply this concept to quadratic equations to find the solutions:

Ifx2 – 6x + 5 = 0

Notice that the

FACTORS and the SOLUTIONS switch signs!

(x – 1)(x – 5) = 0

Then either (x – 1) = 0 or (x – 5) = 0

Therefore x = _____ or x = ______ (Graph this to check the intercepts)

Find the solutions of the given polynomial by FACTORING.

x2 - 16x + 48 = 0

x2 – 4 = 0

x2 + x – 12 = 0

4x2 + 7x – 2 = 0

5x2 + 20x = 0

x2 + 6x + 9 = 0

x2 – 81 = 0

x2 – 5x = 36

SOLVING QUADRATIC EQUATIONS means:

Finding SOLUTIONS = Finding X-INTERCEPTS = Finding ROOTS = Finding ZEROS

Quadratic equation

(standard form)

Quadratic equation

(intercept form)

Solve by factoring

Solve by graphing

y = x2 – 11x + 24

Vertex ___________

y = x2 - 9

Vertex ___________

Y = x2 – 4x + 4

Vertex ___________

Day 4 Notes – Completing the Square

The Goal: To learn ANOTHER method of solving quadratics to use when we can’t factorThe Reason: Once you CTS, you now have a quadratic in VERTEX form. You can now graph using the vertex form (instead of the standard form) and you can solve by the square root method.

Review:

Factor the following perfect square trinomials.

1. x2 + 6x + 92. x2 + 10x + 253. x2 – 8x + 16

4. x2 – 18x + 81 5. x2 + 4x + 4

Try to figure out what the last term will be in order to create a perfect square trinomial:

5. x2 + 12x + _______6. x2 – 14x + _______

7. x2 - 10x + _______8. x2 + 22x + _______

9. How did you determine what the last term would be?

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) =

Perfect Square TrinomialSquare of a Binomial

1. x2 + 18x + c c = ____= (x + )2

2. x2 − 2x + c c = ____= (x − )2

Solve a quadratic equation by taking a square root: x2 + 20x + 100 = 81

Write the trinomial as a binomial squared,

then solve by the square root method:

What if you don’t have a perfect square trinomial?...

Steps to Solving Quadratics by COMPLETING THE SQUARE

1. Move the constant to one side of the equation

2. Calculate the “c” needed to complete the square

3. Add this value to BOTH sides of the equation

4. Rewrite your trinomial as a perfect binomial squared

5. Use the square root method to solve the equation

Solve by completing the square:

x2 – 10x + 1 = 0 x2 – 4x + 8 = 0

Solve by Completing the Square when a1

Divide each side (every term) by the leading coefficient (coefficient of x2)

3x2 – 36x + 150 = 0 2x2 - 24x + 22 = 0

Write a quadratic function in Vertex Form y = a(x – h)2 + k

y = x2 + 18x + 22

SOLVING QUADRATIC EQUATIONS using multiple methods.

Quadratic equation

(standard form)

Quadratic equation

in Vertex form

(complete the square)

Solve by taking the square root

Solve by graphing

y = x2 – 6x - 1

Vertex ___________

y = x2 + 2x + 4

Vertex ___________

Day 5 Notes – The Quadratic Formula

The Goal: To learn a new method of solving quadratics – the quadratic formula.The Reason: So that we can solve a quadratic equation even if it doesn’t factor and it is in standard form.

Standard Form: y = ax2+ bx + c

Part 1: The Discriminant – is the radicand of the quadratic formula.

We can quickly tell whether a quadratic has 0, 1, or 2 real solutions using the discriminant

1. f(x) = x2 – 2x2. f(x) = –x2 + 2x

a = ______ b = ______ c = ______a = ______ b = ______ c = ______

Now calculate: (b)2 – 4(a)(c) = ________Now calculate: (b)2 – 4(a)(c) = ________

3. f(x) =–x2 + 2x - 14. 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) = ________

Observations: What type of a number do you get for (b)2 – 4(a)(c) regarding the number of real solutions?

Part 2: The Quadratic Formula:

Practice: How many real solutions for

y = x2 + 2x – 143 ?

Example 1. Solve: x2 – 5x = 4

Set equal to zero:

a. DESCRIBE the nature of the roots or solutions (or solution set)

a = ______ b = ______ c = ______Discriminant: (b)2 – 4(a)(c) = ________

b. SOLVE the quadratic

Use the quadratic formula: x =

Example 2. Solve: 4x2 + 10x = -10x - 25

Set equal to zero:

a. DESCRIBE the nature of the roots or solutions (or solution set)

a = ______ b = ______ c = ______Discriminant: (b)2 – 4(a)(c) = ________

b. SOLVE the quadratic

Use the quadratic formula: x =

Example3. Solve: x2 - 6x = -10

Set equal to zero:

a. DESCRIBE the nature of the roots or solutions (or solution set)

a = ______ b = ______ c = ______Discriminant: (b)2 – 4(a)(c) = ________

b. SOLVE the quadratic

Use the quadratic formula: x =

SOLVING QUADRATIC EQUATIONS

Finding SOLUTIONS = Finding X-INTERCEPTS = Finding ROOTS = Finding ZEROS

Quadratic equation

(standard form)

Solve by quadratic formula

Solve by graphing

y = x2 – x - 6

Vertex ___________

y = x2 - 9

Vertex ___________

Y = x2 + 3x + 6

Vertex ___________

Day 6: Solving Quadratics Review

Name 3 synonyms for “solution”: ___________, ___________, _____________

Our QUADRATIC TOOLBOX

We know 4 ways to solve quadratic equations. List them.

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 + 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: ______

Vertical Stretch, Vertical Shrink or

Standard: _____________________

Graph: y = x2 – 2x – 8

Day 8: Notes – Solving SYSTEMS of Equations

The Goal: To solve SYSTEMS of equations (more than one equation at a time)

The Reason: To find where two functions intersect. These are the ordered pairs that satisfy each equation.

Facts to remember –

When you are solving ONE equation your solutions are the X-INTERCEPTS.

When you are solving a SYSTEM of equations your solutions are the POINTS Of INTERSECTION. Write your answers as (x, y).

The easiest way to solve a system of equations will always be graphing, but this is not always accurate.

When graphing, the solutions to a system of equations are the points of intersection!

1. 2.

Graph the following systems of equations. The solution is the set of all ordered pairs that are intersections of the two functions.

Ask yourself what KINDS of functions you have in your system….

1. 2.

We can also solve systems algebraically by substitution.

Steps:

1. Solve one of the equations for a variable.

2. Substitute the expression into the other equation.

3. Solve the equation.

4. Substitute the value found into one of the equations. Solve for the remaining variable.

5. Name the point (x, y).

Solve by substitution and Graph to verify your solution.

1. y = x2 + 6x + 12. y = x2 - 2x - 1

y = -2x – 14y = - x2 + 4x - 1

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