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Slide 1 / 175 Slide 2 / 175 Algebra I Quadratics 2015-11-04 www.njctl.org Slide 3 / 175 · Characteristics of Quadratic Equations · Graphing Quadratic Equations · Transforming Quadratic Equations · Solve Quadratic Equations by Graphing · Solve Quadratic Equations by Factoring · Solve Quadratic Equations Using Square Roots · Solve Quadratic Equations by Completing the Square · Solve Quadratic Equations by Using the Quadratic Formula · Key Terms · Solving Application Problems · The Discriminant Table of Contents Click on the topic to go to that section Slide 4 / 175 Key Terms Return to Table of Contents Slide 5 / 175 Axis of symmetry: The vertical line that divides a parabola into two symmetrical halves Axis of Symmetry Slide 6 / 175 Maximum: The y-value of the vertex if a < 0 and the parabola opens downward Minimum: The y-value of the vertex if a > 0 and the parabola opens upward Parabola: The curve result of graphing a quadratic equation (+ a) Max Min (- a) Parabolas

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Page 1: Algebra Icontent.njctl.org/courses/math/algebra-i/quadratic...2015/11/04  · Quadratic Equations Return to Table of Contents Slide 30 / 175 x x2-3 9-2 4-1 1 0 0 1 1 2 4 3 9 The quadratic

Slide 1 / 175 Slide 2 / 175

Algebra I

Quadratics

2015-11-04

www.njctl.org

Slide 3 / 175

· Characteristics of Quadratic Equations

· Graphing Quadratic Equations

· Transforming Quadratic Equations

· Solve Quadratic Equations by Graphing

· Solve Quadratic Equations by Factoring

· Solve Quadratic Equations Using Square Roots

· Solve Quadratic Equations by Completing the Square

· Solve Quadratic Equations by Using the Quadratic Formula

· Key Terms

· Solving Application Problems

· The Discriminant

Table of ContentsClick on the topic to go to that section

Slide 4 / 175

Key Terms

Return to Tableof Contents

Slide 5 / 175

Axis of symmetry: The vertical line thatdivides a parabola intotwo symmetrical halves

Axis of Symmetry

Slide 6 / 175

Maximum: The y-value of thevertex if a < 0 and the parabola opens downward

Minimum: The y-value of thevertex if a > 0 and theparabola opens upward

Parabola: The curve resultof graphing aquadratic equation

(+ a)Max

Min(- a)

Parabolas

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Slide 7 / 175

Quadratic Equation: An equation that can be written in the standard form ax 2 + bx + c = 0. Where a, b and c are real numbers and a does not = 0.

Vertex: The highest or lowest point ona parabola.

Zero of a Function: An x value that makes the function equal zero.

Quadratics

Slide 8 / 175

Characteristics of Quadratic

Equations

Return to Tableof Contents

Slide 9 / 175

A quadratic equation is an equation of the form ax2 + bx + c = 0 , where a is not equal to 0.

Quadratics

The form ax2 + bx + c = 0 is called the standard form of the quadratic equation.

The standard form is not unique. For example, x2 - x + 1 = 0 can be written as the equivalent equation -x2 + x - 1 = 0.

Also, 4x2 - 2x + 2 = 0 can be written as the equivalent equation 2x2 - x + 1 = 0. Why is this equivalent?

Slide 10 / 175

Practice writing quadratic equations in standard form: (Simplify if possible.)

Write 2x2 = x + 4 in standard form:

Writing Quadratic Equations

Slide 11 / 175

1 Write 3x = -x2 + 7 in standard form:

A. x2 + 3x-7= 0

B. x2 -3x +7=0

C. -x2 -3x -7= 0

Slide 12 / 175

2 Write 6x2 - 6x = 12 in standard form:

A. 6x2 - 6x -12 = 0

B. x2 - x - 2 = 0

C. -x2 + x + 2 = 0

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Slide 13 / 175

3 Write 3x - 2 = 5x in standard form:

A. 2x + 2 = 0

B. -2x - 2 = 0

C. not a quadratic equation

Slide 14 / 175

← upward

→downward

The graph of a quadratic is a parabola, a u-shaped figure.

The parabola will open upward or downward.

Characteristics of Quadratic Functions

Slide 15 / 175

A parabola that opens upward contains a vertex that is a minimum point. A parabola that opens downward contains a vertex that is a maximum point.

vertex

vertex

Characteristics of Quadratic Functions

Slide 16 / 175

The domain of a quadratic function is all real numbers.

Characteristics of Quadratic Functions

D = Reals

Slide 17 / 175

To determine the range of a quadratic function,ask yourself two questions:

> Is the vertex a minimum or maximum? > What is the y-value of the vertex?

If the vertex is a minimum, then the range is all real numbers greater than or equal to the y-value.

The range of this quadratic is [–6,∞)

Characteristics of Quadratic Functions

Slide 18 / 175

If the vertex is a maximum, then the range is all real numbers less than or equal to the y-value.

The range of this quadratic is (–∞,10]

Characteristics of Quadratic Functions

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Slide 19 / 175

7. An axis of symmetry (also known as a line of symmetry) will divide the parabola into mirror images. The line of symmetry is always a vertical line of the form

x=2

Characteristics of Quadratic Functions

x = –b2a

x = –(–8) 2(2)

= 2 y = 2x2 – 8x + 2

Slide 20 / 175

To find the axis of symmetry simply plug the values of a and b into the equation: Remember the form ax 2 + bx + c. In this example a = 2, b = -8 and c =2

Characteristics of Quadratic Functions

x=2x = –b2a

x = –(–8) 2(2)

= 2 y = 2x2 – 8x + 2

a b c

Slide 21 / 175

The x-intercepts are the points at which a parabola intersects the x-axis. These points are also knownas zeroes, roots or solutions and solution sets. Each quadratic equation will have two, one or no real x-intercepts.

Characteristics of Quadratic Functions

Slide 22 / 175

4 The vertical line that divides a parabola into two symmetrical halves is called...

A discriminant

B perfect square

C axis of symmetry

D vertex

E slice

Slide 23 / 175 Slide 24 / 175

6 The equation y = x2 + 3x − 18 isgraphed on the set of axes below.

−3 and 60 and −183 and −6 3 and −18

ABC

D

Based on this graph, what are the roots of the equation x2 + 3x − 18 = 0?

From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from

www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

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Slide 25 / 175

7 The equation y = − x 2 − 2x + 8 isgraphed on the set of axes below.

Based on this graph, what are the roots of the equation −x2 − 2x + 8 = 0?

A 8 and 0

B 2 and –4

C 9 and –1

D 4 and –2

From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from

www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

Slide 26 / 175

8 What is an equation of the axis of symmetry of the parabola represented by y = −x2 + 6x − 4?

A x = 3B y = 3C x = 6D y = 6

From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from

www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

Slide 27 / 175

9 The height, y, of a ball tossed into the air can be represented by the equation y = –x2 + 10x + 3, where x is the elapsed time. What is the equation of the axis of symmetry of this parabola?

A y = 5

B y = –5C x = 5D x = –5

From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from

www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

Slide 28 / 175

Slide 29 / 175

TransformingQuadratic Equations

Return to Tableof Contents

Slide 30 / 175

x x2

-3 9-2 4-1 10 01 12 43 9

The quadratic parent equation is y = x2. The graph of all other quadratic equations are transformations of the graph of y= x2.

Quadratic Parent Equation

y = x2

y = – – x223

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Slide 31 / 175

The quadratic parent equation is y = x 2. How is y = x2 changed into y = 2x2?

x 2

-3 18

-2 8

-1 2

0 0

1 2

2 8

3 18

Quadratic Parent Equation

y = 2x2

y = x2

Slide 32 / 175

x 0.5

-3 4.5

-2 2

-1 0.5

0 0

1 0.5

2 2

3 4.5

The quadratic parent equation is y = x 2. How is y = x2 changed into y = .5x2?

Quadratic Parent Equation

y = x2

y = – x212

Slide 33 / 175

How does a > 0 affect the parabola? How does a < 0 affect the parabola?

What does "a" do in y = ax2+ bx + c ?What Does "A" Do?

y = x2 y = –x2

Slide 34 / 175

What does "a" also do in y =ax2 + bx +c ?

How does your conclusion about "a" changeas "a" changes?

What Does "A" Do?

y = x2y = – x212

y = 3x2

y = –1x2 y = –3x2 y = – – x212

Slide 35 / 175

If the absolute value of a is > 1, then the graph of the equation is narrower than the graph of the parent equation.

If the absolute value of a is < 1, then the graph of the equation is wider than the graph of the parent equation.

If a > 0, the graph opens up.

If a < 0, the graph opens down.

What does "a" do in y = ax2 + bx + c ?

What Does "A" Do?

Slide 36 / 175

11 Without graphing determine which direction does the parabola open and if the graph is wider or narrower than the parent equation.

A up, wider

B up, narrower

C down, wider

D down, narrower

y = .3x2

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12 Without graphing determine which direction does the parabola open and if the graph is wider or narrower than the parent function.

A up, wider

B up, narrower

C down, wider

D down, narrower

y = –4x2

Slide 38 / 175

13 Without graphing determine which direction does the parabola open and if the graph is wider or narrower than the parent equation.

A up, wider

B up, narrower

C down, wider

D down, narrower

y = –2x2 + 100x + 45

Slide 39 / 175

14 Without graphing determine which direction does the parabola open and if the graph is wider or narrower than the parent function.

A up, wider

B up, narrower

C down, wider

D down, narrower

y = – – x223

Slide 40 / 175

15 Without graphing determine which direction does the parabola open and if the graph is wider or narrower than the parent function.

A up, wider

B up, narrower

C down, wider

D down, narrower

y = – – x275

Slide 41 / 175

What does "c" do in y = ax2 + bx + c ?

What Does "C" Do?

y = x2 + 6

y = x2 + 3

y = x2

y = x2 – 2

y = x2 – 5

y = x2 – 9

Slide 42 / 175

"c" moves the graph up or down the same value as "c."

"c" is the y- intercept.

What does "c" do in y = ax2 + bx + c ?

What Does "C" Do?

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Slide 43 / 175

16 Without graphing, what is the y- intercept of the the given parabola?

y = x2 + 17

Slide 44 / 175

17 Without graphing, what is the y- intercept of the the given parabola?

y = –x2 –6

Slide 45 / 175

18 Without graphing, what is the y- intercept of the the given parabola?

y = –3x2 + 13x – 9

Slide 46 / 175

19 Without graphing, what is the y- intercept of the the given parabola?

y = 2x2 + 5x

Slide 47 / 175

20 Choose all that apply to the following quadratic:

opens up

opens down

wider than parent function

narrower than parent

function

A

B

C

D

y-intercept of y = –4

y-intercept of y = –2

y-intercept of y = 0

y-intercept of y = 2

y-intercept of y = 4

y-intercept of y = 6

A

B

C

D

E

F

f(x) = –.7x2 –4

Slide 48 / 175

21 Choose all that apply to the following quadratic:

A opens up

B opens down

C wider than parent function

D narrower than parent function

E y-intercept of y = –4

F y-intercept of y = –2

G y-intercept of y = 0

H y-intercept of y = 2I y-intercept of y = 4

J y-intercept of y = 6

f(x) = – – x2 –6x43

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Slide 49 / 175 Slide 50 / 175

Graphing Quadratic Equations

Return to Tableof Contents

Slide 51 / 175

Graph by Following Six Steps:

Step 1 - Find Axis of Symmetry

Step 2 - Find Vertex

Step 3 - Find Y intercept

Step 4 - Find two more points

Step 5 - Partially graph

Step 6 - Reflect

Slide 52 / 175

Axis of Symmetry

Axis of SymmetryStep 1 - Find Axis of Symmetry

What is the Axis of Symmetry?

Slide 53 / 175 Slide 54 / 175

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Slide 55 / 175

Step 3 - Find y interceptWhat is the y-intercept?

y- intercept

Slide 56 / 175

The y- intercept is always the c value, because x = 0.

c = 1

The y-intercept is 1 and the graph passes through (0,1).

y = ax2 + bx + c

y = 3x2 – 6x + 1

Step 3 - Find y intercept

Graph y = 3x2 – 6x + 1

Slide 57 / 175

Choose different values of x and plug in to find points.

Step 4 - Find Two More Points

Find two more points on the parabola.

Graph y = 3x2 – 6x + 1

Let's pick x = –1 and x = –2

y = 3x2 – 6x + 1

y = 3(–1)2 – 6(–1) + 1

y = 3 + 6 + 1y = 10

(–1,10)

Slide 58 / 175

Step 4 - Find Two More Points (continued)

Graph y = 3x2 – 6x + 1

y = 3x2 – 6x + 1

y = 3(–2)2 – 6(–2) + 1

y = 3(4) + 12 + 1

y = 25

(–2, 25)

Slide 59 / 175

Step 5 - Graph the Axis of Symmetry

Graph the axis of symmetry, the vertex, the point containing the y-intercept and two other points.

Slide 60 / 175

(4,25)

Step 6 - Reflect the Points

Reflect the points across the axis of symmetry. Connect the points with a smooth curve.

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23 What is the axis of symmetry for y = x2 + 2x - 3 (Step 1)?

Slide 62 / 175

24 What is the vertex for y = x2 + 2x - 3 (Step 2)?

A (-1, -4)

B (1, -4)

C (-1, 4)

Slide 63 / 175

25 What is the y-intercept for y = x2 + 2x - 3 (Step 3)?

A -3

B 3

Slide 64 / 175

axis of symmetry = –1

vertex = –1, –4

y intercept = –3

2 other points (step 4)

(1,0)

(2,5)

Partially graph (step 5)

Reflect (step 6)

Graphy= x2 + 2x – 3

Slide 65 / 175

Graphy = 2x2 – 6x + 4

Slide 66 / 175

Graph

y = –x2 – 4x + 5

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Slide 67 / 175

Graphy = 3x2 – 7

Slide 68 / 175

Solve Quadratic Equations by Graphing

Return to Tableof Contents

Slide 69 / 175

Find the Zeros

One way to solve a quadratic equation in standard form is find the zeros by graphing.

A zero is the point at which the parabola intersects the x-axis.

A quadratic may have one, two or no zeros.

Slide 70 / 175

No zeroes(doesn't crossthe "x" axis)

2 zeroes; x = -1 and x=3

1 zero;x=1

Find the ZerosHow many zeros do the parabolas have?

What are the values of the zeros?

click click click

Slide 71 / 175

Review

Step 1 - Find Axis of Symmetry

Step 2 - Find Vertex

Step 3 - Find Y intercept

Step 4 - Find two more points

Step 5 - Partially graph

Step 6 - Reflect

To solve a quadratic equation by graphing follow the6 step process we already learned.

Slide 72 / 175

26

Which of these is in standard form?

y = 2x2 – 12x + 18

Solve the equation by graphing.

–12x + 18 = –2x2

y = –2x2 – 12x + 18

y = –2x2 + 12x – 18

B

A

C

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27 What is the axis of symmetry?

y = –2x2 + 12x – 18

A –3B 3C 4D –5

Slide 74 / 175

28 y = –2x2 + 12x – 18

What is the vertex?

A (3,0)

B (–3,0)

C (4,0)

D (–5,0)

Slide 75 / 175

29

What is the y- intercept?

A (0, 0)

B (0, 18)

C (0, –18)

D (0, 12)

y = –2x2 + 12x – 18

Slide 76 / 175

30

AB

C D

If two other points are (5, –8) and (4 ,–2),what doesthe graph of y = –2x2 + 12x – 18 look like?

Slide 77 / 175

31 y = –2x2 + 12x – 18

What is(are) the zero(s)?

A –18

B 4

C 3

D –8

click for graph of answer

Slide 78 / 175

Solve Quadratic Equations by Factoring

Return to Tableof Contents

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Review of factoring - To factor a quadratic trinomial of the form x2 + bx + c, find two factors of c whose sum is b.

Example - To factor x2 + 9x + 18, look for factors whose sum is 9.

Factors of 18 Sum

1 and 18 19

2 and 9 11

3 and 6 9

Solving Quadratic Equationsby Factoring

x2 + 9x + 18 = (x + 3)(x + 6)

Slide 80 / 175

When c is positive, it's factors have the same sign.The sign of b tells you whether the factors are positive or negative.When b is positive, the factors are positive.When b is negative, the factors are negative.

Solving Quadratic Equationsby Factoring

Slide 81 / 175

4. Multiply the Last terms (x + 3)(x + 2) 3 2 = 6

3. Multiply the Inner terms (x + 3)(x + 2) 3 x = 3x

2. Multiply the Outer terms (x + 3)(x + 2) x 2 = 2x

1. Multiply the First terms (x + 3)(x + 2) x x = x2

F O I L

Remember the FOIL method for multiplying binomials

Solving Quadratic Equationsby Factoring

(x + 3)(x + 2) = x2 + 2x + 3x + 6 = x2 + 5x + 6

Slide 82 / 175

For all real numbers a and b, if the product of two quantities equals zero, at least one of the quantities equals zero.

Numbers Algebra

3(0) = 0 If ab = 0,

4(0) = 0 Then a = 0 or b = 0

Zero Product Property

Slide 83 / 175

Example 1: Solve x2 + 4x – 12 = 0

x + 6 = 0 or x – 2 = 0 –6 –6 + 2 +2 x = –6 x = 2

–62 + 4(–6) – 12 = 0 –62 + (–24) – 12 = 0 36 – 24 – 12 = 0 0 = 0 or22 + 4(2) – 12 = 0 4 + 8 – 12 = 0 0 = 0

Use "FUSE" !

Zero Product Property

Factor the trinomial using the FOIL method.

Use the Zero property

Substitue found value into original equation

Equal - problem solved! The solutions are -6 and 2.

(x + 6) (x – 2) = 0

Slide 84 / 175

Example 2: Solve x2 + 36 = 12x –12x –12x

The equation has to be written in standard form (ax2 + bx + c). So subtract 12x from both sides.

Zero Product Property

Factor the trinomial using the FOIL method.

Use the Zero property

Substitue found value into original equation

Equal - problem solved!

x2 – 12x + 36 = 0

(x – 6)(x – 6) = 0 x – 6 = 0 +6 +6 x = 6

62 + 36 = 12(6)

36 + 36 = 72

72 = 72

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Example 3: Solve x2 – 16x + 48= 0

(x – 4)(x – 12) = 0

x – 4 = 0 x –12 = 0 +4 +4 +12 +12 x = 4 x = 12

Zero Product Property

Factor the trinomial using the FOIL method. Use the Zero property

Substitue found value into original equation

Equal - problem solved!

42 – 16(4) + 48 = 0 16 – 64 + 48 = 0 –48+48 = 0 0 = 0

122 – 16(12) + 48 = 0 144 –192 + 48 = 0 –48 + 48 = 0 0 = 0

–48

Slide 86 / 175

32 Solve

A –7

B –5

C –3

D –2

E 2

F 3

G 5

H 6

I 7

J 15

x2 – 5x + 6 = 0

Slide 87 / 175

33 Solve m2 + 10m + 25 = 0

A –7

B –5

C –3

D –2

E 2

F 3

G 5

H 6

I 7

J 15

Slide 88 / 175

34 Solve h2 – h = 12

A –12

B –4

C –3

D –2

E 2

F 3

G 4

H 6

I 8J 12

Slide 89 / 175

35 Solve d2 – 35d = 2d

A –7

B –5

C –3

D 35E 12

F 0

G 5

H 6

I 7J 37

Slide 90 / 175

36 Solve 8y2 + 2y = 3

A –3/4

B –1/2

C –4/3D –2

E 2

F 3/4

G 1/2

H 4/3

I –3

J 3

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37 Which equation has roots of −3 and 5?

A x2 + 2x − 15 = 0B x2 − 2x − 15 = 0C x2 + 2x + 15 = 0 D x2 − 2x + 15 = 0

Slide 92 / 175

Solve Quadratic Equations Using Square Roots

Return to Tableof Contents

Slide 93 / 175

You can solve a quadratic equation by the square root method if you can write it in the form: x² = c

If x and c are algebraic expressions, then: x = c or x = – c

written as: x = ± c

Square Root Method

√ √

Slide 94 / 175

Solve for z: z² = 49z = ± 49z = ±7

The solution set is 7 and –7

Square Root Method

Slide 95 / 175

The solution set is and –

A quadratic equation of the form x2 = c can be solved using the Square Root Property.

Example: Solve 4x2 = 20

x = ±

Square Root Method

4x2 = 20 4 4 x2 = 5

Divide both sides by 4 to isolate x²

5√ 5√

5√

Slide 96 / 175

5x2 = 20 5 5

x2 = 4

x = or x = – x = ± 2

4 4

Square Root Method

Solve 5x² = 20 using the square root method:

√ √

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2x – 1 = 202x – 1 = (4)(5)2x – 1 = 2 52x = 1 + 2 5 1 + 2 5x = 2

Solve (2x – 1)² = 20 using the square root method.

or

Square Root Method

2x – 1 = – 202x – 1 = – (4)(5)2x – 1 = –2 52x = 1 – 2 5 1 – 2 5x = 2

solution: x = 1 ± 2 5

2

√√√√

√√

√√

√click

click click

Slide 98 / 175

38 When you take the square root of a real number, your answer will always be positive.

True

False

Slide 99 / 175

39 If x2 = 16, then x =

A 4

B 2

C –2

D 26

E –4

Slide 100 / 175

40 If y2 = 4, then y =

A 4

B 2

C –2

D 26

E –4

Slide 101 / 175

41 If 8j2 = 96, then j =

A – 3 2

B – 2 3

C 2 3

D 3 2

E ±12

Slide 102 / 175

42 If 4h2 –10= 30, then h =

A – 10

B – 2 5

C 2 5

D 10

E ±10

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43 If (3g – 9)2 + 7= 43, then g =

A 1

B 9 – 5 2

C 9 + 5 2

D 5

E ±3

3

3

Slide 104 / 175

Solving Quadratic Equations by

Completing the Square

Return to Tableof Contents

Slide 105 / 175

x2 + 8x + ___

x2 + 20x + 100

x2 – 16x + 64

x2 – 2x + 1

Before we can solve the quadratic equation, we first have to find the missing value of C. To do this, simply take the value of b, divide it in 2 and then square the result.

Find the value that completes the square.

(b/2)2

8/2 = 4

42 = 16

ax2+bx+c

Find the Missing Value of "C"

Slide 106 / 175

44 Find (b/2)2 if b = 14

Slide 107 / 175

45 Find (b/2)2 if b = –12

Slide 108 / 175

46 Complete the square to form a perfect square trinomial

x2 + 18x + ?

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Slide 109 / 175

47 Complete the square to form a perfect square trinomial

x2 – 6x + ?

Slide 110 / 175

Solving Quadratic Equationsby Completing the Square

Step 1 - Write the equation in the form x 2 + bx = c

Step 2 - Find (b ÷ 2) 2

Step 3 - Complete the square by adding (b ÷ 2)2 to both sides of the equation.

Step 4 - Factor the perfect square trinomial.

Step 5 - Take the square root of both sides

Step 6 - Write two equations, using both the positive and negative square root and solve each equation.

Slide 111 / 175

Let's look at an example to solve: x2 + 14x = 15

x2 + 14x = 15 Step 1 - Already done!

(14 ÷ 2)2 = 49 Step 2 - Find (b÷2) 2

x2 + 14x + 49 = 15 + 49 Step 3 - Add 49 to both sides

(x + 7)2 = 64 Step 4 - Factor and simplify

x + 7 = ±8 Step 5 - Take the square root of both sides

x + 7 = 8 or x + 7 = –8 Step 6 - Write and solve two equations

x = 1 or x = –15

Solving Quadratic Equationsby Completing the Square

Slide 112 / 175

Another example to solve: x 2 – 2x – 2 = 0

x2 – 2x – 2 = 0 Step 1 - Write as x 2+bx=c +2 +2x2 – 2x = 2

(–2 ÷ 2)2 = (–1)2 = 1 Step 2 - Find (b÷2)2

x2 – 2x + 1 = 2 + 1 Step 3 - Add 1 to both sides

(x – 1)2 = 3 Step 4 - Factor and simplify

x – 1 = ± 3 Step 5 - Take the square root of both sides

x – 1 = 3 or x – 1 = – 3 Step 6 - Write and solve two equationsx = 1 + 3 or x = 1 – 3

Solving Quadratic Equationsby Completing the Square

√ √

√ √

Slide 113 / 175

48 Solve the following by completing the square :

x2 + 6x = –5

A –5

B –2C –1D 5E 2

Slide 114 / 175

49 Solve the following by completing the square:

x2 – 8x = 20

A –10B –2C –1D 10E 2

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Slide 115 / 175

50 Solve the following by completing the square :

–36x = 3x2 + 108

A –6

B 6C 0D 6E – 6

Slide 116 / 175

x – = ±5 43 3

10x 3

x2 – = –1

Write as x 2+bx=c

Find (b÷2)2

Add 25/9 to both sidesFactor and simplify

A more difficult example:Solve3x2 – 10x = –3

3x2 10x = –3 3 3 3

10 3

÷ 2 = x = = –10 1 –5 25 3 2 3 9)( ( ) ( )2 2 2

10x 25 25 3 9 9

x2 – + = –1 +

x – =5 163 9)( 2

√ Take the square root of both sides

Slide 117 / 175

– 254

51 Solve the following by completing the square:

4x2 – 7x – 2 = 0

– 14

A

B

C

D

E

14

– 254

2

Slide 118 / 175

The Discriminant

Return to Tableof Contents

Slide 119 / 175

x = –b ± √b2 – 4ac 2a

Discriminant - the part of the equation under the radical sign in a quadratic equation.

The Discriminant

b2 – 4ac is the discriminant

Slide 120 / 175

ax2 + bx + c = 0

The discriminant, b2 – 4ac, or the part of the equation under the radical sign, may be used to determine the number of

real solutions there are to a quadratic equation.

The Discriminant

If b2 – 4ac > 0, the equation has two real solutionsIf b2 – 4ac = 0, the equation has one real solutionIf b2 – 4ac < 0, the equation has no real solutions

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Slide 121 / 175

The Discriminant

Remember:The square root of a positive number has two solutions.The square root of zero is 0.

The square root of a negative number has no real solution.

Slide 122 / 175

Example

√4 = ± 2

(2) (2) = 4 and (–2)(–2) = 4So BOTH 2 and –2 are solutions

The Discriminant

Slide 123 / 175

What is the relationship between the discriminant of a quadratic and its graph?

Discriminant(8)2 – 4(1)(10) = 64 – 40 = 24 (–6)2 –4(3)(–4) = 36 + 48 = 84

The Discriminant

y = x2 – 8x + 10 y = 3x2 + 8x – 4

Slide 124 / 175

What is the relationship between the discriminant of a quadratic and its graph?

The Discriminant

Discriminant(–4)2 – 4(2)(2) = 16 – 16 = 0 (6)2 –4(1)(9) = 36 – 36 = 0

y = 2x2 – 4x + 2 y = x2 + 6x + 9

Slide 125 / 175

What is the relationship between the discriminant of a quadratic and its graph?

Discriminant

(5)2 – 4(1)(9) = 25 – 36 = –11 (–3)2 –4(3)(4) = 9 – 48 = –39

The Discriminant

y = x2 + 5x + 9 y = 3x2 – 3x + 4

Slide 126 / 175

52 What is value of the discriminant of 2x2 – 3x + 5 = 0 ?

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Slide 127 / 175

53 Find the number of solutions using the discriminant for 2x2 – 3x + 5 = 0

A 0

B 1

C 2

Slide 128 / 175

54 What is value of the discriminant of x2 – 8x + 4 = 0 ?

Slide 129 / 175

55 Find the number of solutions using the discriminant for x2 – 8x + 4 = 0

A 0

B 1

C 2

Slide 130 / 175

Solve Quadratic Equations by Using

the Quadratic Formula

Return to Tableof Contents

Slide 131 / 175

At this point you have learned how to solve quadratic equations by:· graphing· factoring· using square roots and · completing the square

Today we will be given a tool to solve ANY quadratic equation.

It ALWAYS works.

Many quadratic equations may be solved using these methods; however, some cannot be solved using any of these methods.

Solve Any Quadratic Equation

Slide 132 / 175

"x equals the opposite of b, plus or minus the square root of b squared minus 4ac, all divided by 2a."

The Quadratic Formula

The solutions of ax2 + bx + c = 0, where a ≠ 0, are:

x = –b ± b2 – 4ac√2a

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Slide 133 / 175

x = –3 ± √32 –4(2)(–5) 2(2)

x = –b ± √b2 –4ac 2a

continued on next slide

Write the Quadratic Formula

Identify values of a, b and c

Substitute the values of a, b and c

2x2 + 3x + (–5) = 0

2x2 + 3x – 5 = 0

Example 1

The Quadratic Formula

Slide 134 / 175

x = –3 – 7 4

= –3 ± 7 4

x = –3 ± √49 4

x = –3 ± √9 – (–40) 4

x = –3 + 7 4 or

The Quadratic Formula

Simplify

Write as two equations

Solve each equationx = 1 or x = –5 2

Slide 135 / 175

Solution on next slide

The Quadratic FormulaExample 2 2x = x2 – 3

Remember - In order to use the Quadratic Formula, the equation must be in standard form (ax2 + bx +c = 0).

First, rewrite the equation in standard form.

2x = x2 – 3–2x –2x

0 = x2 + (-2x) + (–3)

x2 + (–2x) + (–3) = 0

Use only addition for standard form

Flip the equation

Now you are ready to use the Quadratic Formula

Slide 136 / 175

x = –(–2) ± √(–2)2 –4(1)(–3) 2(1)

x = –b ± √b2 –4ac 2a

Continued on next slide

The Quadratic Formulax2 + (–2x) + (–3) = 0

1x2 + (–2x) + (–3) = 0 Identify values of a, b and c

Write the Quadratic Formula

Substitute the values of a, b and c

Slide 137 / 175

x = 2 ± √162

= 2 ± 42

x = 2 ± √4 – (–12)2a

Simplify

x = 2 ± 42

or x = 2 - 42

x = 3 or x = –1

Write as two equations

Solve each equation

The Quadratic Formula

Slide 138 / 175

56 Solve the following equation using the quadratic formula:

A -5

B -4

C -3

D -2

E -1

F 1

G 2

H 3

I 4

J 5

x2 – 5x + 4 = 0

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Slide 139 / 175

57 Solve the following equation using the quadratic formula:

A –5

B –4

C –3

D –2

E –1

F 1

G 2

H 3

I 4J 5

x2 = x + 20

Slide 140 / 175

–3 2

58 Solve the following equation using the quadratic formula:

A –5

B –4

C

D –2

E –1

F 1

G 2

H

I 4

J 5

2x2 + 12 = 11x

32

Slide 141 / 175

x = -b ± √b2 -4ac

2a

Continued on next slide

The Quadratic FormulaExample 3

x2 – 2x – 4 = 0

1x2 + (–2x) + (–4) = 0 Identify values of a, b and c

Write the Quadratic Formula

Substitute the values of a, b and c

x = –(–2) ± √(–2)2 –4(1)(–4) 2(1)

Slide 142 / 175

The Quadratic Formula

x = 2 ± √202

x = 2 ± √4 – (–16)2

Simplify

x = 2 ± 2√52

Write as two equationsor x = 2 - 2√52

orx = 2 ± √202

x = 2 - √202

x = 1 + √5 or x = 1 – √5

x ≈ 3.24 or x ≈ –1.24 Use a calculator to estimate x

Slide 143 / 175

59 Find the larger solution to

x2 + 6x – 1 = 0

Slide 144 / 175

60 Find the smaller solution to

x2 + 6x – 1 = 0

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Slide 145 / 175

Application Problems

Return to Tableof Contents

Slide 146 / 175

A sampling of applied problems that lend themselves to being solved by quadratic equations:

Number Reasoning

Free Falling Objects

DistancesGeometry: Dimensions

Height of a Projectile

Quadratic Equations and Applications

Slide 147 / 175

The product of two consecutive negative integers is 1,122. What are the numbers?

Remember that consecutive integers are one unit apart, so the numbers are n and n + 1.

Multiplying to get the product:n(n + 1) = 1122 n2 + n = 1122 n2 + n – 1122 = 0 (n + 34)(n - 33) = 0

n = –34 and n = 33.

The solution is either –34 and –33 or 33 and 34, since the directionask for negative integers –34 and –33 are the correct pair.

→ STANDARD Form→ FACTOR

Number Reasoning

Slide 148 / 175

PLEASE KEEP THIS IN MIND

When solving applied problems that lead to quadratic equations, you might get a solution that does not satisfy the physical constraints of the problem.

For example, if x represents a width and the two solutions of the quadratic equations are –9 and 1, the value –9 is rejected since a width must be a positive number.

Application Problems

Slide 149 / 175

61 The product of two consecutive even integers is 48. Find the smaller of the two integers.

Hint: x(x+2) = 48Click to reveal hint

Slide 150 / 175

The product of two consecutive integers is 272.

What are the numbers?

TRY THIS:

Application Problems

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Slide 151 / 175

The product of two consecutive even integers is 528. What is the smaller number?

62

Slide 152 / 175

The product of two consecutive odd integers is 1 less than four times their sum. Find the two integers.

Let n = 1st number n + 2 = 2nd number

More of a challenge...

n(n + 2) = 4[n + (n + 2)] – 1n2 + 2n = 4[2n + 2] – 1n2 + 2n = 8n + 8 – 1n2 + 2n = 8n + 7n2 – 6n - 7 = 0(n – 7)(n + 1) = 0

n = 7 and n = –1

Which one do you use?Or do you use both?

Slide 153 / 175

More of a challenge...

If n = 7 then n + 2 = 9

7 x 9 = 4[7 + (7 + 2)] – 163 = 4(16) – 163 = 64 – 163 = 63

If n = –1 then n + 2 = –1 + 2 = 1

(–1) x 1 = 4[–1 + (–1 + 2)] – 1–1 = 4[–1 + 1] – 1–1 = 4(0) – 1–1 = –1

We get two sets of answers.

Slide 154 / 175

63 The product of a number and a number 3 more than the original is 418. What is the smallest value the original number can be?

Slide 155 / 175

64 Find three consecutive positive even integers such that the product of the second and third integers is twenty more than ten times the first integer. Enter the value of the smaller even integer.

From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

Slide 156 / 175

65 When 36 is subtracted from the square of a number, the result is five times the number. What is the positive solution?

A 9B 6C 3D 4

From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

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66 Tamara has two sisters. One of the sisters is 7 years older than Tamara.The other sister is 3 years younger than Tamara. The product of Tamara’s sisters’ ages is 24. How old is Tamara?

From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

Slide 158 / 175

Two cars left an intersection at the same time, one heading north and one heading west. Some time later, they were exactly 100 miles apart. The car that headed north had gone 20 miles farther than the car headed west. How far had each car traveled?

Step 1 - Read the problem carefully.

Step 2 - Illustrate or draw your information.x+20

100

x

Example

Step 3 - Assign a variableLet x = the distance traveled by the car heading westThen (x + 20) = the distance traveled by the car heading northStep 4 - Write an equationDoes your drawing remind you of the Pythagorean Theorem? a2 + b2 = c2 Continued on next slide

Distance Problems

Slide 159 / 175

Step 5 - Solve a2 + b2 = c2

x2 + (x+20)2 = 1002

x2 + x2 + 40x + 400 = 10,000

2x2 + 40x – 9600 = 0

2(x2 +20x – 4800) = 0

x2 + 20x – 4800 = 0

100x+20

x

Square the binomial

Standard form

Factor

Divide each side by 2

Think about your options for solving the rest of this equation. Completing the square? Quadratic Formula?

Continued on next slide

Distance Problems

Slide 160 / 175

x = –20 ±√400 – 4(1)(–4800) 2

Distance ProblemsDid you try the quadratic formula?

x = –20 ±√19,600 2

x = 60 or x = -80

Since the distance cannot be negative, discard the negative solution. The distances are 60 miles and 60 + 20 = 80 miles.

Step 6 - Check your answers.

Slide 161 / 175

67 Two cars left an intersection at the same time,one heading north and the other heading east. Some time later they were 200 miles apart. If the car heading east traveled 40 miles farther, how far did the car traveling north go?

Slide 162 / 175

x

x + 6

Geometry Applications

Area ProblemThe length of a rectangle is 6 inches more than its width. The area of the rectangle is 91 square inches. Find the dimensions of the rectangle.

Step 1 - Draw the picture of the rectangle. Let the width = x and the length = x + 6

Step 2 - Write the equation using theformula Area = length x width

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Slide 163 / 175

Step 3 - Solve the equation

x( x + 6) = 91

x2 + 6x = 91

x2 + 6x – 91 = 0

(x – 7)(x + 13) = 0

x = 7 or x = –13

Since a length cannot be negative...

The width is 7 and the length is 13.

Geometry Applications

Slide 164 / 175

68 The width of a rectangular swimming pool is 10 feet less than its length. The surface area of the pool is 600 square feet. What is the pool's width?

Hint: (L)(L – 10) = 600.Click to reveal hint

Slide 165 / 175

69 A square's length is increased by 4 units and its width is increased by 6 units. The result of this transformation is a rectangle with an area that 195 square units. Find the area of the original square.

Slide 166 / 175

length

x

x

70 The rectangular picture frame below is the same width all the way around. The photo it surrounds measures 17" by 11". The area of the frame and photo combined is 315 sq. in. What is the length of the outer frame?

Slide 167 / 175

71 The area of the rectangular playground enclosure at South School is 500 square meters. The length of the playground is 5 meters longer than the width. Find the dimensions of the playground, in meters.[Only an algebraic solution will be accepted.]

From the New York State Education Department. Office of Assessment Policy,

Development and Administration. Internet. Available fromwww.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

Slide 168 / 175

72 Jack is building a rectangular dog pen that he wishes to enclose. The width of the pen is 2 yards less than the length. If the area of the dog pen is 15 square yards, how many yards of fencing would he need to completely enclose the pen?

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Slide 169 / 175

Free Falling Objects Problems

Slide 170 / 175

73 A person walking across a bridge accidentally drops an orange in the river below from a height of 40 ft. The function h = –16t2 + 40 gives the orange's approximate height h above the water, in feet, after t seconds. In how many t seconds will the orange hit the water? (Round to the nearest tenth.)

Hint: when it hits the water it is at 0.

Slide 171 / 175

74 Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determined by the equation d = 144 – 16t2 where t is the number of seconds it takes the car to travel down to each point on the ride. How many seconds will it take Greg to reach the ground?

Slide 172 / 175

75 The height of a golf ball hit into the air is modeled by the equation h = –16 t2 + 48t, where h represents the height, in feet, and t represents the number of seconds that have passed since the ball was hit. What is the height of the ball after 2 seconds?

From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from www.nysedregents.org/IntegratedAlgebra; accessed 17,

June, 2011.

A 16 ft

B 32 ft

C 64 ft

D 80 ft

Slide 173 / 175

Height of Projectiles Problems

Slide 174 / 175

76 A skyrocket is shot into the air. It's altitude in feet, h, after t seconds is given by the functionh = –16 t2 + 128 t.

What is the rocket's maximum altitude?

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Slide 175 / 175

77 A rocket is launched from the ground and follows a parabolic path represented by the equation y = –x 2 + 10x. At the same time, a flare is launched from a height of 10 feet and follows a straight path represented by the equation y = –x + 10. Using the accompanying set of axes, graph the equations that represent the paths of the rocket and the flare, and find the coordinates of the point or points where the paths intersect.

From the New York State Education Department. Office of Assessment Policy,

Development and Administration. Internet. Available fromwww.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.