Key Terms Solve Quadratic Equations by Factoring Solve Quadratic - - PDF document

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Algebra I

Quadratics

2015-11-04 www.njctl.org

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· 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

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Key Terms

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• f Contents

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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

• f graphing a

quadratic equation

(+ a)

Max Min

(- a)

Parabolas

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

Quadratics Slide 8 / 175

Characteristics

• f Quadratic

Equations

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• f Contents

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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?

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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

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

• A. 2x + 2 = 0
• B. -2x - 2 = 0
• C. not a quadratic equation

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← 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

• r 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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• 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 = –b 2a x = –(–8) 2(2) = 2 y = 2x2 – 8x + 2

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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=2 x = –b 2a x = –(–8) 2(2) = 2 y = 2x2 – 8x + 2 a b c

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The x-intercepts are the points at which a parabola intersects the x-axis. These points are also known as 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

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6 The equation y = x2 + 3x − 18 is graphed on the set of axes below. −3 and 6 0 and −18 3 and −6 3 and −18 A B C 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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7 The equation y = − x 2 − 2x + 8 is graphed 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.

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8 What is an equation of the axis of symmetry of the parabola represented by y = −x2 + 6x − 4? A x = 3 B y = 3 C x = 6 D 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.

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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 = –5 C x = 5 D 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.

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Transforming Quadratic Equations

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x x2

• 3

9

• 2

4

• 1

1 1 1 2 4 3 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 = – – x2 2 3

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

x 2

• 3

18

• 2

8

• 1

2 1 2 2 8 3 18

Quadratic Parent Equation

y = 2x2 y = x2

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x 0.5

• 3

4.5

• 2

2

• 1

0.5 1 0.5 2 2 3 4.5

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

Quadratic Parent Equation

y = x2 y = – x2 1 2

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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

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What does "a" also do in y =ax

2 + bx +c ?

How does your conclusion about "a" change as "a" changes?

What Does "A" Do?

y = x2 y = – x2 1 2 y = 3x2 y = –1x2 y = –3x2 y = – – x2 1 2

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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 ?

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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

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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

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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 = – – x2 2 3

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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 = – – x2 7 5

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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

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"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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16 Without graphing, what is the y- intercept of the the given parabola? y = x2 + 17

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17 Without graphing, what is the y- intercept of the the given parabola? y = –x2 –6

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18 Without graphing, what is the y- intercept of the the given parabola? y = –3x2 + 13x – 9

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19 Without graphing, what is the y- intercept of the the given parabola? y = 2x2 + 5x

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20 Choose all that apply to the following quadratic:

• pens up
• pens 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

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21 Choose all that apply to the following quadratic: A

• pens up

B

• pens 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 = 2 I y-intercept of y = 4 J y-intercept of y = 6 f(x) = – – x2 –6x 4 3

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Graphing Quadratic Equations

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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

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Axis of Symmetry

Axis of Symmetry

Step 1 - Find Axis of Symmetry What is the Axis of Symmetry?

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Slide 55 / 175 Step 3 - Find y intercept

What is the y-intercept? y- intercept

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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

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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 + 1 y = 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.

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(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)?

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24 What is the vertex for y = x2 + 2x - 3 (Step 2)? A (-1, -4) B (1, -4) C (-1, 4)

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

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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)

Graph

y= x2 + 2x – 3

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y = 2x2 – 6x + 4

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y = –x2 – 4x + 5

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y = 3x2 – 7

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Solve Quadratic Equations by Graphing

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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.

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No zeroes (doesn't cross the "x" axis) 2 zeroes; x = -1 and x=3 1 zero; x=1

Find the Zeros

How 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 the 6 step process we already learned.

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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 –3 B 3 C 4 D –5

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28 y = –2x2 + 12x – 18 What is the vertex? A (3,0) B (–3,0) C (4,0) D (–5,0)

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29 What is the y- intercept? A (0, 0) B (0, 18) C (0, –18) D (0, 12) y = –2x2 + 12x – 18

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30

A B C D

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

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31 y = –2x2 + 12x – 18

What is(are) the zero(s)?

A –18 B 4 C 3 D –8

click for graph of answer

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Solve Quadratic Equations by Factoring

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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 Equations by Factoring

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

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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 Equations by 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 Equations by Factoring

(x + 3)(x + 2) = x

2 + 2x + 3x + 6 = x 2 + 5x + 6

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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 x 2 + 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

• r

22 + 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

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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

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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

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33 Solve m

2 + 10m + 25 = 0

A –7 B –5 C –3 D –2 E 2 F 3 G 5 H 6 I 7 J 15

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34 Solve h

2 – h = 12

A –12 B –4 C –3 D –2 E 2 F 3 G 4 H 6 I 8 J 12

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35 Solve d

2 – 35d = 2d

A –7 B –5 C –3 D 35 E 12 F G 5 H 6 I 7 J 37

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36 Solve 8y

2 + 2y = 3

A –3/4 B –1/2 C –4/3 D –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 = 0 B x2 − 2x − 15 = 0 C x2 + 2x + 15 = 0 D x2 − 2x + 15 = 0

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Solve Quadratic Equations Using Square Roots

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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

√ √ √

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Solve for z: z² = 49 z = ± 49 z = ±7 The solution set is 7 and –7

Square Root Method

√

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The solution set is and – A quadratic equation of the form x

2 = c can be solved

using the Square Root Property. Example: Solve 4x

2 = 20

x = ±

Square Root Method

4x2 = 20 4 4 x2 = 5 Divide both sides by 4 to isolate x² 5 √ 5 √ 5 √

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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 = 20 2x – 1 = (4)(5) 2x – 1 = 2 5 2x = 1 + 2 5 1 + 2 5 x = 2 Solve (2x – 1)² = 20 using the square root method.

• r

Square Root Method

2x – 1 = – 20 2x – 1 = – (4)(5) 2x – 1 = –2 5 2x = 1 – 2 5 1 – 2 5 x = 2 solution: x = 1 ± 2 5 2 √ √ √ √ √ √ √ √ √ √ √

click click click

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

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39 If x2 = 16, then x = A 4 B 2 C –2 D 26 E –4

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40 If y2 = 4, then y = A 4 B 2 C –2 D 26 E –4

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41 If 8j2 = 96, then j = A – 3 2 B – 2 3 C 2 3 D 3 2 E ±12

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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

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Solving Quadratic Equations by Completing the Square

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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

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45 Find (b/2)2 if b = –12

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46 Complete the square to form a perfect square trinomial x2 + 18x + ?

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47 Complete the square to form a perfect square trinomial x2 – 6x + ?

Slide 110 / 175 Solving Quadratic Equations by 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

• f 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.

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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 Equations by 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 +2 x2 – 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 equations x = 1 + 3 or x = 1 – 3

Solving Quadratic Equations by Completing the Square

√ √ √ √ √

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48 Solve the following by completing the square : x2 + 6x = –5 A –5 B –2 C –1 D 5 E 2

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49 Solve the following by completing the square: x2 – 8x = 20 A –10 B –2 C –1 D 10 E 2

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50 Solve the following by completing the square : –36x = 3x2 + 108 A –6 B 6 C D 6 E – 6

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x – = ± 5 4 3 3 10x 3 x2 – = –1 Write as x 2+bx=c Find (b÷2)2 Add 25/9 to both sides Factor and simplify A more difficult example:

Solve

3x2 – 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 16 3 9

) (

2

√

Take the square root of both sides

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– 2 5 4 51 Solve the following by completing the square: 4x2 – 7x – 2 = 0 – 1 4 A B C D E 1 4 – 2 5 4 – 2

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The Discriminant

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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

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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 solutions If b2 – 4ac = 0, the equation has one real solution If b2 – 4ac < 0, the equation has no real solutions

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.

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Example √4 = ± 2 (2) (2) = 4 and (–2)(–2) = 4 So 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

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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

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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

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

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53 Find the number of solutions using the discriminant for 2x2 – 3x + 5 = 0 A B 1 C 2

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

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55 Find the number of solutions using the discriminant for x2 – 8x + 4 = 0 A B 1 C 2

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Solve Quadratic Equations by Using the Quadratic Formula

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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

• f 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

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

• r

The Quadratic Formula

Simplify Write as two equations Solve each equation x = 1 or x = –5 2

Slide 135 / 175

Solution on next slide

The Quadratic Formula

Example 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 Formula

x2 + (–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 ± √16 2 = 2 ± 4 2 x = 2 ± √4 – (–12) 2a Simplify x = 2 ± 4 2

• r

x = 2 - 4 2 x = 3

• r

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

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 4 J 5 x2 = x + 20

Slide 140 / 175

–3 2 58 Solve the following equation using the quadratic f

• rmula:

A –5 B –4 C D –2 E –1 F 1 G 2 H I 4 J 5 2x2 + 12 = 11x 3 2

Slide 141 / 175

x = -b ± √b2 -4ac 2a Continued on next slide

The Quadratic Formula

Example 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 ± √20 2 x = 2 ± √4 – (–16) 2 Simplify x = 2 ± 2√5 2 Write as two equations

• r

x = 2 - 2√5 2

• r

x = 2 ± √20 2 x = 2 - √20 2 x = 1 + √5

• r

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

Slide 145 / 175

Application Problems

Return to Table

• f Contents

Slide 146 / 175

A sampling of applied problems that lend themselves to being solved by quadratic equations: Number Reasoning Free Falling Objects Distances Geometry: 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 direction ask 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

• f 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) = 48

Click to reveal hint

Slide 150 / 175

The product of two consecutive integers is 272. What are the numbers? TRY THIS:

Application Problems

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)] – 1 n2 + 2n = 4[2n + 2] – 1 n2 + 2n = 8n + 8 – 1 n2 + 2n = 8n + 7 n2 – 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)] – 1 63 = 4(16) – 1 63 = 64 – 1 63 = 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

• riginal 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 th e 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 9 B 6 C 3 D 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.

Slide 157 / 175

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 variable Let x = the distance traveled by the car heading west Then (x + 20) = the distance traveled by the car heading north Step 4 - Write an equation Does 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 = 0 100 x+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 Problems

Did 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 Problem The 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 the formula Area = length x width

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 from www.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?

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 – 16 t2 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 = –16t2 + 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 function h = –16t2 + 128t. What is the rocket's maximum altitude?

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

• r points where the paths intersect.

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.