Quadratic Functions Key Terms

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Quadratic Functions

Table of Contents Slide 2 / 200 Key Terms Identify Quadratic Functions Explain Characteristics of Quadratic Functions 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 The Discriminant Solving Non-Quadratics Solving Rational Equations Solving Radical Equations Quadratic & Rational Inequalities

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Key Terms

Return to Table of Contents

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Axis of symmetry: The vertical line that divides a parabola into two symmetrical halves

Completing the square: Adding a term to x2 + bx to form a trinomial that is a perfect square

Discriminant: b2 - 4ac in a quadratic in standard form

Slide 5 / 200 Maximum: The y-value of the vertex if a < 0 and the parabola opens downward

Minimum: The y-value of the vertex Max if a > 0 and the parabola opens upward Min Parabola: The curve result of graphing a quadratic equation

Slide 6 / 200 Quadratic Equation: An equation that can be written in the standard form ax2 + bx + c = 0. Where a, b and c are real numbers and a does not = 0. Quadratic Function: Any function that can be written in the form y = ax2 + bx + c. Where a, b and c are real numbers and a does not equal 0.

Vertex: The highest or lowest point on a parabola.

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

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Identifying

Quadratic Functions

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Any function that can be written in the form y = ax2 + bx + c Where a, b, and c are real numbers and a ≠ 0

Examples Question: Is 2x2 = x + 4 a quadratic equation? Answer: Yes

Question: Is 3x - 4 = x + 1 a quadratic equation? Answer: No

Slide 9 / 200

Explain

Characteristics of Quadratic Functions

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Slide 10 / 200 A quadratic equation is an equation of the form ax2 + bx + c = 0 , where a is not equal to 0.

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.

Slide 11 / 200 Practice writing quadratic equations in standard form: (Reduce if possible.) Write 2x2 = x + 4 in standard form:

2x2 - x - 4 = 0

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

x2 + 3x - 7 = 0

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

2 x - x - 2 = 0

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

Not a quadratic equation

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

3. The parabola from a polynomial function will open upward or downward.

Slide 17 / 200 4. 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.

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5. The domain of a quadratic function is all real numbers.

6. To determine the range of a quadratic function, ask yourself Slide 19 / 200 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 to

Slide 20 / 200 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 to 10

Slide 21 / 200 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

Slide 22 / 200 8. 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 function will have two, one or no real x-intercepts.

Slide 23 / 200 1 True or False: The vertex is the highest or lowest value on the parabola.

True

False

Slide 24 / 200 2 If a parabola opens upward then...

A a>0 B a<0

C a=0

Slide 25 / 200 3 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 26 / 200 Quadratic Equations

Finding Zeros (x- intercepts)

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Solve Quadratic Equations by

Graphing Return to Table of Contents

Slide 28 / 200 Vocabulary

Every quadratic function has a related quadratic equation.

A quadratic equation is used to find the zeroes of a quadratic function. When a function intersects the x-axis its y-value is zero.

When writing a quadratic function as its related quadratic equation, you replace y with 0. So y = 0. y = ax2 + bx + c 0 = ax2 + bx + c ax2 + bx + c = 0

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One way to solve a quadratic equation in standard form is find the zeros of the related function 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 30 / 200 How many zeros do the parabolas have? What are the values of the zeros?

No zeroes 2 zeroes; 1 zero; x = -1 and x=3 x=1

Slide 31 / 200 One way to solve a quadratic equation in standard form is to find the zeros or x-intercepts of the related function.

Solve a quadratic equation by graphing: Step 1 - Write the related function.

Step 2 - Graph the related function.

Step 3 - Find the zeros (or x intercepts) of the related function.

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Step 1 - Write the Related Function

2x2 - 18 = 0

2 2x - 18 = y

y = 2x2 + 0x - 18

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Step 2 - Graph the Function 2 y = 2x + 0x – 18 The vertex is found by substituting the Use the same six step process for graphing x-coordinate of the Axis of

Vertex Symmetry into the equation and solving for y. The axis of symmetry is x = 0. The vertex is (0,-18). Find the y intercept -- It is -18. Find two other points (2,-10) and (3,0) TwoPoints The point where the line passes through the y-axis. This occurs when the

x-value is 0. Y Intercept Y

Slide 34 / 200 Step 2 - Graph the Function y = 2x2 + 0x – 18

Graph the points and reflect them across the axis of symmetry.

x = 0 # # (3,0)

# # (2,-10)

# (0,-18)

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Step 3 - Find the zeros y = 2x2 + 0x – 18

Solve the equation by graphing the related function.

x = 0 # # The zeros appear (3,0) to be 3 and -3. # # (2,-10)

# (0,-18)

Slide 36 / 200 Step 3 - Find the zeros y = 2x2 + 0x – 18

Substitute 3 and -3 for x in the quadratic equation.

Check 2x2 – 18 = 0

2 2(3) – 18 = 0 2(-3)2 – 18 = 0 2(9) - 18 = 0 2(9) - 18 = 0 18 - 18 = 0 18 - 18 = 0 0 = 0 ü 0 = 0 ü

The zeros are 3 and -3.

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4 Solve the equation by graphing the related function.

-12x + 18 = -2x2

Step 1: What of these is the related function?

2 A y = -2x - 12x + 18 B y = 2x2 - 12x - 18

2 C y = -2x + 12x - 18

Slide 38 / 200 5 What is the axis of symmetry? y = -2x2 + 12x - 18

A -3 Formula: -b 2a B 3

C 4

D -5

Slide 39 / 200 2 6 y = -2x + 12x - 18

What is the vertex?

A (3,0) B (-3,0)

C (4,0)

D (-5,0)

Slide 40 / 200 2 7 y = -2x + 12x - 18

What is the y- intercept?

A (0,0) B (0, 18)

C (0, -18)

D (0, 12)

Slide 41 / 200 2 8 y = -2x + 12x - 18

If two other points are (5,-8) and (4,-2), what does the graph look like?

A B

Slide 42 / 200 9 If two other points are (5,-8) and (4,-2), what does the graph of y = -2x2 + 12x - 18 look like?

A B

C D

Slide 43 / 200 2 10 y = -2x + 12x - 18 What is(are) the zero(s)?

A -18

B 4 C 3 D -8

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Solve Quadratic Equations by Factoring

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Solving Quadratic Equations Slide 45 / 200 by Factoring 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

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

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When c is positive, its 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.

Slide 47 / 200 Remember the FOIL method for multiplying binomials

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

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

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

F O I L

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Zero Product Property

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

Slide 49 / 200 Example 1: Solve x2 + 4x - 12 = 0 Use "FUSE" !

(x + 6) (x - 2) = 0 Factor the trinomial using the FOIL method.

x + 6 = 0 or x - 2 = 0 Use the Zero - 6 - 6 + 2 +2 property x = -6 x = 2

-62 + 4(-6) - 12 = 0 Substitue found 2 -6 + (-24) - 12 = 0 value into original 36 - 24 - 12 = 0 equation 0 = 0 or Equal - problem 2 2 + 4(2) - 12 = 0 solved! The 4 + 8 - 12 = 0 solutions are -6 0 = 0 and 2.

Slide 50 / 200 2 Example 2: Solve x + 36 = 12x -12x -12x The equation has to be written in standard form x2 - 12x + 36 = 0 (ax2 + bx + c). So subtract 12x from both sides.

(x - 6)(x - 6) = 0 Factor the trinomial using the FOIL x - 6 = 0 method. +6 +6 x = 6 Use the Zero property 62 + 36 = 12(6) Substitue found value into original 36 + 36 = 72 equation

72 = 72 Equal - problem solved!

Slide 51 / 200 2 Example 3: Solve x - 16x + 64 = 0 (x - 8)(x - 8) = 0 Factor the trinomial

x - 8 = 0 using the FOIL +8 +8 method. x = 8 Use the Zero property

82 - 16(8) + 64 = 0 Substitue found value into original 64 - 128 + 64 = 0 equation

Equal - problem solved! -64 + 64 = 0

0 = 0

Slide 52 / 200 11 Solve

A -7 F 3

B -5 G 5 C -3 H 6 D -2 I 7 E 2 J 15

Slide 53 / 200 12 Solve

A -7 F 3

B -5 G 5 C -3 H 6 D -2 I 7 E 2 J 15

Slide 54 / 200 13 Solve

A -12 F 3

B -4 G 4 C -3 H 6 D -2 I 8 E 2 J 12

Slide 55 / 200 14 Solve

A -7 F 3

B -5 G 5 C -3 H 6 D -2 I 7 E 12 J 35

Slide 56 / 200 15 Solve

3 3 A - /4 F /4 1 1 B - /2 G /2 4 4 C - /3 H /3 D -2 I -3 E 2 J 3

Slide 57 / 200 16 The product of two consecutive even integers is 48. Find the smaller of the two integers. Hint: x(x+2) = 48

Slide 58 / 200 17 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.

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Solve Quadratic

Equations Using Square Roots Return to Table of Contents

Slide 60 / 200 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

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Solve for z: z² = 49 z = ± √49 z = ±7

The solution set is 7 and -7

Slide 62 / 200 A quadratic equation of the form x2 = c can be solved using the Square Root Property. Example: Solve 4x2 = 20

2 4x = 20 Divide both sides 4 4 by 4 to isolate x² x2 = 5

x = ±√5

The solution set is √5 and -√5

Slide 63 / 200 Solve 5x² = 20 using the square root method:

5x2 = 20 5 5 x2 = 4 x =√4 or x = -√4 x = ±2

Slide 64 / 200 Solve (2x - 1)² = 20 using the square root method.

Slide 65 / 200 18 When you take the square root of a real number, your answer will always be positive.

True

False

Slide 66 / 200 19 If x2 = 16, then x =

A 4 B 2 C -2 D 26

E -4

Slide 67 / 200 20 If y2 = 4, then y =

A 4 B 2 C -2 D 26

E -4

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Slide 70 / 200 2 23 If (3g - 9) + 7= 43, then g =

A B

D E

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Solving Quadratic Equations by Completing the Square Return to Table of Contents

Form a perfect square trinomial with lead coefficient of 1 Slide 72 / 200

2 b 2 x + bx +c where c = ( /2) Find the value that completes the square.

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b 2 24 Find ( /2) if b = 14

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b 2 25 Find ( /2) if b = -12

Slide 75 / 200 26 Complete the square to form a perfect square trinomial

x2 + 18x + ?

Slide 76 / 200 27 Complete the square to form a perfect square trinomial

x2 - 6x + ?

Slide 77 / 200 Solving quadratic equations by completing the square:

Step 1 - Write the equation in the form x2 + 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.

Let's look at an example to solve: x2 + 14x = 15 Slide 78 / 200 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

Another example to solve: x2 - 2x - 2 = 0 Slide 79 / 200 x2 - 2x - 2 = 0 Step 1 - Write as x2+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

Slide 80 / 200 Solve the following by completing the square : 28 x2 + 6x = -5

A -5 B -2 C -1 D 5 E 2

Slide 81 / 200 Solve the following by completing the square : 29 x2 - 8x = 20

A -10 B -2 C -1 D 10 E 2

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A more difficult example: Slide 83 / 200 Write as x2+bx=c

Find (b÷2)2

Add 25/9 to both sides

Factor and simplify

Take the square root of both sides

or Write and solve two equations

Slide 84 / 200 Solve the following by completing the square : 31

Another example to solve: x2 - 2x + 2 = 0 Slide 85 / 200 x2 - 2x + 2 = 0 Step 1 - Write as x2+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 = -1 Step 4 - Factor and simplify x - 1 = ± √-1 = +i Step 5 - Take the square root of both sides x - 1 = i or x - 1 = i Step 6 - Write and solve two equations x = 1 + i or x = 1 - i

Slide 86 / 200 Solve the following by completing the square : 32

Slide 87 / 200

Solve Quadratic Equations by Using

the Quadratic Formula

Return to Table of Contents

Slide 88 / 200 At this point you have learned how to solve quadratic equations by: · graphing · factoring · using square roots and · completing the square

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

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

Slide 89 / 200 The Quadratic Formula

The solutions of ax2 + bx + c = 0, where a ≠ 0, are: 2 x = -b ± √b - 4ac 2a

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

Example 1 Slide 90 / 200

2x2 + 3x - 5 = 0

2x2 + 3x + (-5) = 0 Identify values of a, b and c

2 x = -b ± √b -4ac Write the Quadratic Formula 2a

2 x = -3 ± √3 -4(2)(-5) Substitute the values of 2(2) a, b and c continued on next slide

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x = -3 ± √9 - (-40) Simplify 4

x = -3 ± √49 = -3 ± 7 4 4

x = -3 + 7 x = -3 - 7 or Write as two equations 4 4

x = 1 or x = -5 Solve each equation 2

Slide 92 / 200 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. 2 2x = x - 3 -2x -2x Use only addition for 0 = x2 + (-2x) + (-3) standard form

2 x + (-2x) + (-3) = 0 Flip the equation Now you are ready to use the Quadratic Formula Solution on next slide

Slide 93 / 200 2 x + (-2x) + (-3) = 0

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

2 x = -b ± √b -4ac Write the Quadratic Formula 2a x = -(-2) ± √(-2)2 -4(1)(-3) Substitute the values of 2(1) a, b and c

Continued on next slide

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x = 2 ± √4 - (-12) Simplify 2 x = 2 ± √16 = 2 ± 4 2 2

x = 2 + 4 or x = 2 - 4 Write as two equations 2 2

x = 3 or x = -1 Solve each equation

Slide 95 / 200 33 Solve the following equation using the quadratic formula:

A -5 F 1 B -4 G 2 C -3 H 3 I 4 D -2 E -1 J 5

Slide 96 / 200 34 Solve the following equation using the quadratic formula:

A -5 F 1 B -4 G 2 C -3 H 3 I 4 D -2 E -1 J 5

Slide 97 / 200 35 Solve the following equation using the quadratic formula:

A -5 F 1 B -4 G 2 C -3 H 3 I 4 D -2 E -1 J 5

Example 3 Slide 98 / 200

x2 - 2x - 4 = 0

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

2 x = -b ± √b -4ac Write the Quadratic Formula 2a

2 x = -(-2) ± √(-2) -4(1)(-4) Substitute the values of 2(1) a, b and c Continued on next slide

x = 2 ± √4 - (-16) Slide 99 / 200 Simplify 2 x = 2 ± √20 2 x = 2 + √20 x = 2 - √20 or 2 2 Write as two x = 2 +2 √5 or x = 2 - 2√5 equations 2 2

x = 1 + √5 or x = 1 - √5

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

Slide 100 / 200 36 Find the larger solution to

Slide 101 / 200 37 Find the smaller solution to

Slide 102 / 200

The Discriminant

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Slide 103 / 200

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

x = -b ± √b2 -4ac 2a

2 b - 4ac is the discriminant

Slide 104 / 200 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 105 / 200 Example

√4 = ± 2

(2) (2) = 4 and (-2)(-2) = 4

So BOTH 2 and -2 are solutions

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Slide 109 / 200 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.

If b2 - 4ac > 0, the equation has two real solutions If b2 - 4ac = 0, the equation has one real solution 2 If b - 4ac < 0, the equation has no real solutions

Slide 110 / 200 38 What is value of the discriminant of 2x2 - 2x + 3 = 0 ?

Slide 111 / 200

39 Find the number of solutions using the discriminant for 2x2 - 2x + 3 = 0

A 0

B 1

C 2

Slide 112 / 200 40 What is value of the discriminant of x2 - 8x + 4 = 0 ?

Slide 113 / 200

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

A 0

B 1

C 2

Slide 114 / 200

Solving Non-Quadratics

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Summary of Factoring Slide 115 / 200 Factor the Polynomial

Factor out GCF 2 Terms 4 Terms 3 Terms Difference Sum/Difference Group and Factor of Squares of Cubes out GCF. Look for a Perfect Square Factor the Common Binomial Trinomial Trinomial

a = 1 a = 1 Check each factor to see if it can be factored again. If a polynomial cannot be factored, then it is called prime.

Slide 116 / 200

Sum and

Difference of Cubes

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Slide 118 / 200 Examples:

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Slide 120 / 200 43 Factor A (3 + 2a)(9 + 4a + a2)

B (3 - 2a)(9 + 4a + a2) 2 C (3 + 2a)(9 - 4a + a ) D (3 - 2a)(9 - 4a + a2)

Slide 121 / 200 44 Factor completely A (a2 - 1)(a4 + 4a2 + 1)

B (a2 - 1)(a4 - 4a2 + 1) 4 2 C (a - 1)(a + 1)(a + 4a + 1) D (a + 1)(a - 1)(a4 - 4a2 + 1)

Slide 122 / 200

Factoring 4 Term Polynomials

Polynomials with four terms like ab - 4b + 6a - 24, can Slide 123 / 200 be factored by grouping terms of the polynomials.

Example 1: ab - 4b + 6a - 24

(ab - 4b) + (6a - 24) Group terms into binomials that can be factored using the distributive property b(a - 4) + 6(a - 4) Factor the GCF (a - 4) (b + 6) Notice that a - 4 is a common binomial factor and factor!

Slide 124 / 200 Example 2: 6xy + 8x - 21y - 28

(6xy + 8x) + (-21y - 28) Group 2x(3y + 4) + (-7)(3y + 4) Factor GCF (3y +4) (2x - 7) Factor common binomial

Slide 125 / 200 You must be able to recognize additive inverses!!! (3 - a and a - 3 are additive inverses because their sum is equal to zero.) Remember 3 - a = -1(a - 3).

Example 3: 15x - 3xy + 4y - 20 (15x - 3xy) + (4y - 20) Group 3x(5 - y) + 4(y - 5) Factor GCF 3x(-1)(-5 + y) + 4(y - 5) Notice additive inverses -3x(y - 5) + 4(y - 5) Simplify (y - 5) (-3x + 4) Factor common binomial

Remember to check each problem by using FOIL.

Slide 126 / 200 45 Factor 15ab - 3a + 10b - 2 A (5b - 1)(3a + 2)

B (5b + 1)(3a + 2) C (5b - 1)(3a - 2) D (5b + 1)(3a - 1)

Slide 127 / 200 2 46 Factor 10m n - 25mn + 6m - 15 A (2m-5)(5mn-3)

B (2m-5)(5mn+3) C (2m+5)(5mn-3) D (2m+5)(5mn+3)

Slide 128 / 200 47 Factor 20ab - 35b - 63 +36a A (4a - 7)(5b - 9)

B (4a - 7)(5b + 9) C (4a + 7)(5b - 9) D (4a + 7)(5b + 9)

Slide 129 / 200 2 48 Factor a - ab + 7b - 7a A (a - b)(a - 7)

B (a - b)(a + 7) C (a + b)(a - 7) D (a + b)(a + 7)

Slide 130 / 200

Mixed Factoring

Summary of Factoring Slide 131 / 200 Factor the Polynomial

Factor out GCF 2 Terms 4 Terms 3 Terms Difference Sum/Difference Group and Factor of Squares of Cubes out GCF. Look for a Perfect Square Factor the Common Binomial Trinomial Trinomial

a = 1 a = 1 Check each factor to see if it can be factored again. If a polynomial cannot be factored, then it is called prime.

Slide 132 / 200

Slide 133 / 200 49 Factor completely: A

B C D

Slide 134 / 200 50 Factor completely A

B C D prime polynomial

Slide 135 / 200 51 Factor A

B C D prime polynomial

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Slide 137 / 200 53 Factor A

B C D Prime Polynomial

Slide 138 / 200

Solving Equations by

Factoring

Slide 139 / 200 Given the following equation, what conclusion(s) can be drawn? ab = 0

Since the product is 0, one of the factors, a or b, must be 0. This is known as the Zero Product Property.

Slide 140 / 200 Given the following equation, what conclusion(s) can be drawn? (x - 4)(x + 3) = 0 Since the product is 0, one of the factors must be 0. Therefore, either x - 4 = 0 or x + 3 = 0. x - 4 = 0 or x + 3 = 0 + 4 + 4 - 3 - 3 x = 4 or x = -3

Therefore, our solution set is {-3, 4}. To verify the results, substitute each solution back into the original equation. To check x = -3: (x - 4)(x + 3) = 0 To check x = 4: (x - 4)(x + 3) = 0 (-3 - 4)(-3 + 3) = 0 (4 - 4)(4 + 3) = 0 (-7)(0) = 0 (0)(7) = 0 0 = 0 0 = 0

Slide 141 / 200 What if you were given the following equation?

How would you solve it?

We can use the Zero Product Property to solve it. How can we turn this polynomial into a multiplication problem? Factor it!

Factoring yields: x(x - 6)(x + 4) = 0 By the Zero Product Property: x = 0 x - 6 = 0 or x + 4 = 0 After solving each equation, we arrive at our solution: {0,-4, 6}

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Slide 143 / 200 Zero Product rule works only when the product of factors equals zero. If the equation equals some value other than zero subtract to make one side of the equation zero.

Example

Slide 144 / 200

Slide 145 / 200 54 Choose all of the solutions to: A

B C D

Slide 146 / 200 55 Choose all of the real solutions to: A

B C D

Slide 147 / 200 56 Choose all of the solutions to: A

B C D

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Slide 149 / 200 58 A ball is thrown with its height at any time given by

When does the ball hit the ground? A -1 seconds B 0 seconds

C 9 seconds D 10 seconds

Slide 150 / 200

Solving Rational Equations

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Steps to Solving a Rational Equation Slide 151 / 200 1) Find LCD 2) Multiply each term by LCD 3) Reduce 4) Solve 5) Verify answer works (Answer may make denominator = 0) Example:

Slide 152 / 200 Check:

Slide 153 / 200

Example: Slide 154 / 200

Check: x = 7 Check: x = -2 Slide 155 / 200

Slide 156 / 200 59 Solve the equation. Check to see it works.

Slide 157 / 200 60 Solve the equation. Check to see it works.

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Slide 160 / 200

Solving Radical Equations

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Slide 161 / 200 To solve a radical equation: · isolate the radical on one side of the equation

· use the index to determine the power to eliminate the radical · solve the equation

· check to see if solution is extraneous

Slide 162 / 200 Example

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Slide 164 / 200 64 Find the solution to

Slide 165 / 200 65 Find the solution to

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Slide 167 / 200 67 Find the solution to

If an equation has multiple roots, move them to opposite Slide 168 / 200 sides of the equal sign and then solve.

Slide 169 / 200 68 Solve the following:

Slide 170 / 200 69 Solve the following:

Slide 171 / 200 70 If the distance between (3,5) and (x,9) is 7, find x.

Slide 172 / 200

Quadratic & Rational

Inequalities

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Slide 175 / 200 Graph Step 1: Graph Points on the Bounds

X Y -5 -7 -4 -10 -3 -11 -2 -10 -1 -7 0 -2

Slide 176 / 200 Graph Step 2: Solid or Dotted?

Slide 177 / 200 Graph Step 3: Shade

Slide 178 / 200 Graph Step 1: Graph Points on the Bounds

X Y

-3

-2

-1

Slide 179 / 200 Graph Step 2: Solid or Dotted?

Slide 180 / 200 Graph Step 3: Shade

Slide 181 / 200 Graph

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Slide 184 / 200 73 Which equation is graphed? A f(x) > -4x2 + 2x + 5

B f(x) > -4x2 + 2x + 5 2 C f(x) < -4x + 2x + 5 D f(x) < -4x2 + 2x + 5

Slide 185 / 200 74 Which equation is graphed? A f(x) > -4x2 + 2x + 5

B f(x) > -4x2 + 2x + 5 2 C f(x) < -4x + 2x + 5 D f(x) < -4x2 + 2x + 5

Slide 186 / 200 Solving

Method 1: Graphically Graph the related function The solution is where the shaded region intersects the x-axis.

-3 2 -3 2

Note: It is possible to have a solution of a point or the empty set. Slide 187 / 200

Find the solution set given the graph of the related function. Slide 188 / 200

-2 3 -3 -.5

1 2 -.5 1

Slide 189 / 200 75 Solve the following inequality: A 1 < x< 4

B 1 < x < 4 C x < 1 or x > 4 1 4 D x < 1 or x > 4

Slide 190 / 200 76 Solve the following inequality: A 1 < x< 4

B 1 < x < 4 C x < 1 or x > 4 1 4 D x < 1 or x > 4

Slide 191 / 200 77 Solve the following inequality: A -4 < x< 2

B -4 < x < 2 C x < -4 or x > 2 D x < -4 or x > 2

Slide 192 / 200 78 Solve the following inequality: A 2 < x< 5

B 2 < x < 5 C x < 2 or x > 5 D x < 2 or x > 5

Slide 193 / 200 Steps to Solving Quadratic Inequalities Algebraically

1) Get inequality so that it is compared to zero ie. ax2 + bx + c > 0

2) Factor

3) Set each factor equal to zero and solve

4) Create a number line with the solution as the points 5) Test points in each region to see if they satisfy the inequality 6) Write the solution

Slide 194 / 200 Solve 1) Rewrite inequality <0 2) Factor 3) Solve 4) Create number line x -4 -2 Test points: -10, -3, 0 5) Test points

F T F x -4 -2

6) Write the solution

Slide 195 / 200 Solve 1) Rewrite inequality <0 2) Factor 3) Solve 4) Create number line x -1.5 1 Test points: -2, 0, 2 5) Test points