Copyright © 2012 Carlson, O’Bryan & Joyner Worksheet #6: Vertex Form and Completing the Square Worksheet #7: The Quadratic Formula Worksheet #8: Imaginary

Copyright © 2012 Carlson, O’Bryan & Joyner

Worksheet #6: Vertex Form and Completing the Square

Worksheet #7: The Quadratic Formula

Worksheet #8: Imaginary Numbers

Worksheet #9: The Quadratic Formula, Complex Roots, and Conjugates

MODULE 4Quadratic Functions

Part 2: Worksheets #6-9

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1. Let f (x) = 2(x – 7)2 + 3. A classmate claims that the vertex of the function must be at (7, 3), and it must be the function’s minimum. His reasoning follows.

• When x = 7, (x – 7)2 = 0, so 2(x – 7)2 = 0, and so f (7) = 3.

• If x > 7, then (x – 7)2 > 0, so 2(x – 7)2 > 0, and so f (x) > 3.

• If x < 7, then (x – 7)2 > 0, so 2(x – 7)2 > 0, and so f (x) > 3.

• Therefore, the smallest output value is 3, and this occurs at x = 7. So (7, 3) must be the vertex, and it must be a minimum.

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VERTEX FORM AND COMPLETING THE SQUARE

a. Explore his argument. Is he correct?


Yes.

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b. What is the vertex for each of the following functions? Is the vertex a maximum or minimum in each case?


(5, –4)minimum

i) f (x) = (x – 5)2 – 4 ii) f (x) = 4(x – 1)2

iii) f (x) = 2(x + 6)2 + 5 iv) f (x) = (x + 2)2 – 8

(1, 0)minimum

(–6, 5)minimum

(–2, –8)minimum

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2. Let f (x) = –3(x – 2)2 + 5 with a vertex of (2, 5). Construct an argument to explain why the vertex must be a maximum.

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• When x = 2, (x – 2)2 = 0, so –3(x – 2)2 = 0, and so f (2) = 5.

• If x > 2, then (x – 2)2 > 0, so –3(x – 2)2 < 0, and so f (x) < 5.

• If x < 2, then (x – 2)2 > 0, so –3(x – 2)2 < 0, and so f (x) < 5.

• Therefore, the largest output value is 5, and this occurs at x = 2. So (2, 5) must be the vertex, and it must be a maximum.

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3. The form f (x) = a(x – h)2 + k is called the vertex form for a quadratic function. This is an appropriate name because the ordered pair (h, k) represents the vertex of the function. (Note: If a > 0, then the vertex is a minimum. If a < 0, the vertex is a maximum.)

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a. We have learned three forms for a quadratic function: standard form (f (x) = ax2 + bx + c), factored form (f (x) = a(x – u)(x – v)), and now vertex form (f (x) = a(x – h)2 + k). Each of these forms has an advantage over the other forms. What is the advantage of each form?

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The main advantage of standard form is that we can easily determine the vertical intercept, and perhaps that it is the easiest to identify as a quadratic function.

The main advantage of factored form is that it makes it easy to determine the function’s roots.

The main advantage of the vertex form is that it clearly shows the vertex of the function, and by extension it also tells us the axis of symmetry x = h.

a. We have learned three forms for a quadratic function: standard form (f (x) = ax2 + bx + c), factored form (f (x) = a(x – u)(x – v)), and now vertex form (f (x) = a(x – h)2 + k). Each of these forms has an advantage over the other forms. What is the advantage of each form?

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3. The form f (x) = a(x – h)2 + k is called the vertex form for a quadratic function. This is an appropriate name because the ordered pair (h, k) represents the vertex of the function. (Note: If a > 0, then the vertex is a minimum. If a < 0, the vertex is a maximum.)

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Answers will vary. Two examples are

f (x) = (x + 5)2 – 7and

g(x) = –2(x + 5)2 – 7.

b. Write two different functions that have their vertices at (–5, –7).

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4. Let f (x) = 2(x – 2)2 – 8.

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a. Write f in standard form. b. Write f in factored form.

c. Provide all of the details you can about the function f (such as vertex, vertical intercept, etc.), then sketch its graph.

vertical intercept: (0, 0), horizontal intercepts: (0, 0) and (4, 0), axis of symmetry: x = 2, vertex: (2, –8)

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4. Let f (x) = 2(x – 2)2 – 8.

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In general, it’s much easier to find the standard form for a quadratic function when given the vertex form than to find the vertex form when given the standard form. Working the other direction is possible, however.

First, it’s important to note that in the vertex form f (x) = a(x – h)2 + k we have an expression squared:

f (x) = a (x – h)2 + k.

Moreover, when we expand this expression we get (x – h)2 = x2 – 2hx + h2, a quadratic expression of the form f (x) = ax2 + bx + c where a = 1.

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5a. If h = 6, then (x – h)2 = (x – 6)2

= (x – 6)(x – 6) = x2 – 12x + c.

What is the value of c that completes this statement?

b. If h = –4, then (x – h)2 = (x + 4)2

= (x + 4)(x + 4) = x2 + bx + 16.

What is the value of b that completes this statement?

c = 36

b = 8

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If we substitute some different values for h, we get the following:

I. If h = 10, (x – h)2 = (x – 10)2 = (x – 10)(x – 10) = x2 – 20x + 100

II. If h = –1, (x – h)2 = (x – (–1))2 = (x + 1)2 = (x + 1)(x + 1) = x2 + 2x + 1

III. If h = 4, (x – h)2 = (x – 4)2 = (x – 4)(x – 4) = x2 – 8x + 16

IV. If h = –9, (x – h)2 = (x – (–9))2 = (x + 9)2 = (x + 9)(x + 9) = x2 + 18x + 81

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6. For each of the expressions in standard form (ax2 + bx + c with a = 1) just derived, show that (b/2)2 = c.

I. x2 – 20x + 100 II. x2 + 2x + 1

III. x2 – 8x + 16 IV. x2 + 18x + 81

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Therefore, if we have a quadratic expression of the form ax2 + bx + c where a = 1 and (b/2)2 = c, we can write it in the form (x – h)(x – h) or (x – h)2 [or (x + h)(x + h) or (x + h)2].

This realization unlocks the technique called completing the square. The name comes from the fact that we will provide the necessary value of c such that we can complete the expression x2 + bx + ____ with the appropriate value of c so that x2 + bx + c factors to the form (x – h)2 [or (x + h)2].

x2 + 12x + 36 = (x + 6)2

x2 – 30x + 225 = (x – 15)2

x2 + 3x + 2.25 = (x + 1.5)2

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a. Consider the expression x2 + 14x + ____. What must go in the blank so that we can factor the expression in the form (x – h)2 or (x + h)2?

7. Let f (x) = x2 + 14x – 5. The first step is to group the terms that contain variables: f (x) = (x2 + 14x) – 5.

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b. If we return to the function, we want to add this value in the blank space for f (x) = (x2 + 14x + ____ ) – 5. However, if we simply add something to the function then we change the output values of the function. We don’t want to do this. Therefore, anything we add to the function definition we must also subtract.

Fill in the blanks to write f in vertex form.


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c. What is the vertex of f ?


(–7, –54)

What follows are two more examples of using the completing the square process to help you understand it.

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8. Use the technique from Exercise #7 to write each of the following functions in vertex form, then state the

vertex. You are encouraged to graph each function on a calculator to verify your answers.a. f (x) = x2 + 2x + 7 b. g(x) = x2 – 10x + 3

c. f (x) = x2 + 6x + 30 d. h(x) = x2 – 32x + 27

e. f (x) = x2 + 9x +13 f. g(x) = x2 – 5x

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vertex. You are encouraged to graph each function on a calculator to verify your answers.a. f (x) = x2 + 2x + 7

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vertex. You are encouraged to graph each function on a calculator to verify your answers.b. g(x) = x2 – 10x + 3

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vertex. You are encouraged to graph each function on a calculator to verify your answers.c. f (x) = x2 + 6x + 30

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vertex. You are encouraged to graph each function on a calculator to verify your answers.d. h(x) = x2 – 32x + 27

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vertex. You are encouraged to graph each function on a calculator to verify your answers.e. f (x) = x2 + 9x +13

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vertex. You are encouraged to graph each function on a calculator to verify your answers.f. g(x) = x2 – 5x

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If a ≠ 1 for ax2 + bx + c, then completing the square becomes a little more difficult, but we just need to be a bit more careful.

Let f (x) = 2x2 + 16x + 1.

The first step is to again group the terms that contain a variable:

f (x) = (2x2 + 16x) + 1.

Next, we factor the coefficient on the x2 term out of the grouped portion:

f (x) = 2(x2 + 8x) + 1.

This is where we must be careful.

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We know that for x2 + 8x + ____, we need to replace the blank space with 16 so that we can write x2 + 8x + 16 as (x + 4)2.

However, when we examine the process in the context of the entire function, we see that we are adding 16 inside of a set of parentheses, and that we are multiplying 2 by everything inside of the parentheses.

This means that we aren’t actually adding 16 to the function definition – we are actually adding 2(16), or 32. Therefore we must subtract 2(16), or 32, not 16.

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The vertex form of f is f (x) = 2(x + 4)2 – 31.

The vertex is located at (–4, –31).

Here are two more examples of using the completing the square process for your reference.

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9. Write each of the following functions in vertex form. Then state the vertex. You are encouraged to graph each function on a calculator to verify your answers.

a. f (x) = 2x2 + 12x + 23 b. f (x) = 5x2 + 120x

c. h(x) = –3x2 – 18x + 12 d. j(x) = 4x2 – 22x + 19

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a. f (x) = 2x2 + 12x + 23

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b. f (x) = 5x2 + 120x

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c. h(x) = –3x2 – 18x + 12

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d. j(x) = 4x2 – 22x + 19

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10. A local school is planning a community carnival to raise money that they will use to implement a recycling program at the school. Based on data they have collected from similar events held around the country they believe that they can model the expected attendance by the function f (x) = –0.5x2 + 80x – 2350 where x is the forecasted high temperature for the day in degrees Fahrenheit.

a. Write the formula in vertex form using completing the square. The first step has been completed for you.

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b. Is the vertex a minimum or a maximum?

c. What is the vertex? What does this tell us about the situation?

a maximum

(80, 850); The carnival planners expect the greatest possible attendance to be 850 people, which will occur if the forecasted high temperature for the day is 80oF.

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THE QUADRATIC FORMULA


In Worksheet #5 we learned how to factor quadratic equations and use the zero-product property to find the zeros (or roots) of a quadratic function. However, we also saw that not every quadratic function is factorable over the integers. Therefore, we need an additional method that doesn’t rely on factoring.

The quadratic formula can be used to find the zeros of any quadratic function based on the coefficients of the function in standard form (that is, based on a, b, and c in f (x) = ax2 + bx + c). Deriving the formula is based on using the technique of completing the square to solve the equation ax2 + bx + c = 0 (to find the zeros of f (x) = ax2 + bx + c). The process is complicated, so we’ll break it down into steps and have you help us fill in some of the steps.

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1. Let f (x) = 3x2 + 13x + 7 and g(x) = ax2 + bx + c. We have started the process of finding the zeros of these functions using completing the square for these functions. (Note: We avoid simplifying the expressions that result from completing the square with function f to help you make connections with the process for the generic coefficients in function g.)

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a. In Step 3, why were and added inside of the parentheses?

To complete the square for , we needed to

divide by 2 and square the result.

To complete the square for , we needed to

divide by 2 and square the result.

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b. In Step 3, why were and subtracted

from the left sides of the equations instead of

and ?

213 / 3

2

When we add inside of the parentheses, and

everything inside of the parentheses is being multiplied by

3, then we’ve actually added to the left side of

the equation. Thus, to maintain the value of the left side of

the equation we need to subtract .

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b. In Step 3, why were and subtracted

from the left sides of the equations instead of

and ?

213 / 3

2

When we add inside of the parentheses, and

everything inside of the parentheses is being multiplied by

3, then we’ve actually added to the left side of

the equation. Thus, to maintain the value of the left side of

the equation we need to subtract .

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c. What did we do to move from Step 3 to Step 4?

We also rewrote the expression as (because

).

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c. What did we do to move from Step 3 to Step 4?

We rewrote the expression as (because

).

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d. What did we do to move from Step 4 to Step 5? (Note: there are two actions we took in each equation.)

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d. What did we do to move from Step 4 to Step 5? (Note: there are two actions we took in each equation.)

We wrote the perfect square trinomials and

in their factored forms and

.

We also applied the rule of exponents to rewrite

as and as .

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e. Fill in the blanks to complete Step 6.

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f. What was the goal of steps 9 and 10?

The goal of these steps is to create common denominators for subtracting the rational numbers on the right side of each equation.

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g. Subtract the fractions on the right sides of the equations to write Step 11.

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h. Fill in Step 14 that links Step 13 and Step 15.

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h. Fill in Step 14 that links Step 13 and Step 15.

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i. Why does the statement show

two unique values of x? What do these values of x represent?

The “±” tells us that there is a value of x that is some

amount larger than and a value of x some amount less

than .

Whatever these numbers turn out to be, they are the zeros of the original function f (x) = 3x2+ 13x + 7.

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j. Use a calculator to find the approximate values of x

represented by (round to two

decimal places).

x ≈ –0.63 and x ≈ –3.70

k. Graph function f using a calculator, then make a rough sketch of the graph below. Label the horizontal intercepts with their ordered pairs.

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l. Simplify to show the exact

values of the zeros of f. (Your final answer will contain a radical.)

[We are finding the exact values for the rounded decimals in part (j).]

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2. The quadratic formula is defined as .

When we substitute the values of a, b, and c from the coefficients of a given quadratic function in standard form, what does this formula give us?

The quadratic formula determines the zeros (roots) of the quadratic function. (And if they are real numbers, they tell us the location of the horizontal intercepts.)

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3. Let f (x) = 2x2 + 7x + 3.

a. Use the quadratic formula to find the roots of f.

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3. Let f (x) = 2x2 + 7x + 3.

b. Find the roots of f by factoring and using the zero- product property.

c. Explain how you can use the graph of f to check your answers in parts (a) and (b).

The graph of the function should cross the horizontal axis when x = –1/2 and when x = –3 if our solutions are correct.

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3. Let f (x) = 2x2 + 7x + 3.

d. What other method can you use to check your solutions to parts (a) and (b)?

We can evaluate f (–1/2) and f (–3) to make sure that f (–1/2) = 0 and f (–3) = 0.

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4. Use the quadratic formula to find the roots of these quadratic functions. (Recall that when finding the roots of a function, the output variable is set equal to 0, then the resulting equation is solved for x.) Give exact answers and the decimal equivalents. Round your answers to two decimals if necessary.

a. f (x) = 2x2 – 6x + 2.5 b. h(x) = x2 + 4x – 1

c. f (x) = 3x2 – 5x – 1 d. g(x) = –x2 + 7x – 2

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a. f (x) = 2x2 – 6x + 2.5

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b. h(x) = x2 + 4x – 1

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c. f (x) = 3x2 – 5x – 1

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d. g(x) = –x2 + 7x – 2

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5. We know from Exercise #1 that the solutions to ax2 + bx + c = 0 for any quadratic function are given by

and .

a. This tells us that the zeros (roots) of the function are

equidistant from . That is, the roots of the

function are units less than and greater

than . Using what we’ve learned in previous

worksheets, what must represent for a

quadratic function?

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a. This tells us that the zeros (roots) of the function are

equidistant from . That is, the roots of the

function are units less than and greater

than . Using what we’ve learned in previous

worksheets, what must represent for a

quadratic function?

is the axis of symmetry

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b. Using , find the axis of symmetry for each of

the following quadratic functions. i) f (x) = x2 – 6x + 10

ii) g(x) = –2x2 – 12x + 9

iii) h(x) = 10x2 + 5x – 1

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c. What are the coordinates for the vertex of each function in part (b)?

We evaluate each of the functions for the input value that corresponds with the axis of symmetry.

•f (3) = 1, so the vertex for function f is (3, 1)

•g(–3) = 27, so the vertex for function g is (–3, 27)

•h(–1/4) = 7/8, so the vertex for function h is [or (–0.25, 0.875)]

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6a. Use the quadratic formula to find the roots of f (x) = –3x2 + x – 4.

We ended up with a negative number inside of the radical. Since we can’t represent the square root of a negative number using real numbers, we are unable to evaluate the quadratic formula for this function. (At least at this point in the module.)

b. Describe what went “wrong” during the solution process.

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The graph of the function doesn’t cross the horizontal axis (the function has no horizontal intercepts).

c. The quadratic formula is supposed to find the zeros of the function. When the zeros are real numbers, they equate to the horizontal intercepts of the function’s graph. However, in this case the quadratic formula didn’t return any real number answers. Graph function f using a calculator and explain what this fact implies about the horizontal intercepts of the graph of f.

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y

x

f

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7a. Use the quadratic formula to find the roots of f (x) = 2x2 – 4x + 2.

The value of was 0, leaving us with only one

solution to the equation.

b. Describe what happened during the solution process.

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c. The quadratic formula is supposed to find the zeros of the function. When the zeros are real numbers, they equate to the horizontal intercepts of the function’s graph. However, in this case the quadratic formula only returned one value. Graph function f using a calculator and explain what this fact implies about the horizontal intercept(s) of the graph of f.

The graph of the function only intersects the horizontal axis at one point, so the horizontal intercept must also be the vertex of the function.

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y

x

f

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8. Perform the first several steps of using the quadratic formula to find the zeros of the following functions. Complete as many steps as necessary until you can predict the number of horizontal intercepts the function’s graph will have. (You do not have to find the intercepts – only how many will exist.) Use a graphing calculator to check your answers.

a. f (x) = x2 + 7x + 25

b. f (x) = x2 – 2x – 5

c. f (x) = 2x2 + 20x + 50

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a. f (x) = x2 + 7x + 25

no zeros

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b. f (x) = x2 – 2x – 5

two zeros

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c. f (x) = 2x2 + 20x + 50

one zero

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9. The expression b2 – 4ac from the formula is called the discriminant because it discriminates (makes a distinction, or highlights the difference) between functions with zero, one, or two horizontal intercepts.

a. Calculate the value of the discriminant (b2 – 4ac) for the functions from Exercise #8.

8a. b2 – 4ac = (7)2 – 4(1)(25) = –51

8b. b2 – 4ac = (–2)2 – 4(1)(–5) = 24

8c. b2 – 4ac = (20)2 – 4(2)(50) = 0

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b. Why does the discriminant alone give us enough information to predict the number of horizontal intercepts for the graph of a quadratic function?

When the discriminant is positive, then we get a real, non-zero number for .

So the function has zeros of and

.

When the discriminant is 0, then .

So , and the only zero is .

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b. Why does the discriminant alone give us enough information to predict the number of horizontal intercepts for the graph of a quadratic function?

When the discriminant is negative, then there is no real

number , and so does not

evaluate to real numbers.

Therefore, knowing the discriminant is enough to know the number of zeros for the function.

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c. Complete the following statements for any quadratic function in standard form f (x) = ax2 + bx + c.

i) If b2 – 4ac > 0, then …

ii) If b2 – 4ac = 0, then …

iii) If b2 – 4ac < 0, then …

i) If b2 – 4ac > 0, then there are two real number zeros (roots) for the quadratic function.

ii) If b2 – 4ac = 0, then …



ii) If b2 – 4ac = 0, then there is exactly one real number zero (root) for the quadratic function.



ii) If b2 – 4ac = 0, then there is exactly one real number zero (root) for the quadratic function.

iii) If b2 – 4ac < 0, then there are no real number zeros (roots) for the quadratic function.

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For each of the functions in Exercises #10-11, do the following.a. Determine the vertical intercept of the function.b.Find the axis of symmetry and the coordinates for the function’s vertex.c.Use the discriminant (b2 – 4ac) to predict the number of horizontal intercepts the function’s graph will have.d.Use the quadratic formula to find the real-number zeros of the function.e.Sketch a graph of each function and label the information you found in parts (a) through (d).

10. f (x) = x2 + 10x – 39

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10. f (x) = x2 + 10x – 39

a. vertical intercept: (0, –39)

b. axis of symmetry: x = –5 vertex: (–5, –64)

c. number of horizontal intercepts: 2

d. horizontal intercept(s): (–13, 0) and (3, 0)

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10. f (x) = x2 + 10x – 39

e.

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For each of the functions in Exercises #10-11, do the following.a. Determine the vertical intercept of the function.b.Find the axis of symmetry and the coordinates for the function’s vertex.c.Use the discriminant (b2 – 4ac) to predict the number of horizontal intercepts the function’s graph will have.d.Use the quadratic formula to find the real-number zeros of the function.e.Sketch a graph of each function and label the information you found in parts (a) through (d).

11. f (x) = –4x2 + 24x – 36

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11. f (x) = –4x2 + 24x – 36

a. vertical intercept: (0, –36)

b. axis of symmetry: x = 3 vertex: (3, 0)

c. number of horizontal intercepts: 1

d. horizontal intercept(s): (3, 0)

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11. f (x) = –4x2 + 24x – 36

e.

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12. In Worksheet #6 we examined the following context: A local school is planning a community carnival to raise money that they will use to implement a recycling program at the school. Based on data they have collected from similar events held around the country they believe that they can model the expected attendance by the function f (x) = –0.5x2 + 80x – 2350 where x is the forecasted high temperature for the day in degrees Fahrenheit.

a. Find the roots of f using the quadratic formula. Round your answers to the nearest whole number.

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a. Find the roots of f using the quadratic formula. Round your answers to the nearest whole number.

b. What do the roots of f tell us about the situation?

The carnival planning committee expects that no one will attend the carnival if the forecasted high temperature is 39oF or 121oF.

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IMAGINARY NUMBERSHave you ever tried to explain negative numbers to a child? For children, numbers are very concrete concepts – they use numbers to count items and (if they’re old enough) perhaps to measure attributes of an object such as the length of a string.

In such settings negative numbers don’t make much sense since they aren’t needed for counting and children don’t yet have a concept of a directional measurement. Believe it or not, mathematicians rejected the existence of negative numbers for centuries for basically the same reasons.


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Up until about 500 years ago, almost every mathematician grounded his or her work in geometry and geometrical representations for numbers, proofs, and theories. Since negative numbers aren’t naturally represented by a length, area, or other attribute of a geometric figure, most mathematicians ignored negative numbers, threw out negative solutions to equations as false, and generally thought such numbers were useless.

In this worksheet we’re going to explore another type of number that comes up when solving quadratic equations that is difficult to imagine and was thus originally ignored by mathematicians but now is used extensively in fields such as engineering and science.


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Let’s briefly look at square roots before continuing this discussion. The square root of a number is defined to be the number that, when multiplied by itself, gives you the original number.


1. Complete the following statements.

a. is 3 because ________________.

b. is 8 because _________________.

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For numbers that aren’t perfect squares, such as 18, the square root is typically left as or simplified as

This is because these square roots are irrational – they can’t be written exactly as a whole number or a non-repeating, terminating decimal.

We can approximate their value ( ), but in “exact” form the best we can do is say that the square root of 18 is the real number (or ).


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What about negative numbers? What is the square root of –9? It can’t be 3, since 3∙3 = 9, and it can’t be –3 since –3∙–3 = 9.

Therefore, it initially appears that there is no number that, when multiplied by itself, yields –9.

If there were such a number, however, it would be represented by . The number would be the number that, multiplied by itself, yields –9.


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2. Show that the only possible solutions to the equation x2 + 35 = 10 are and .

3. Show that the only possible solutions to the equation 2x2 + 14 = –6 are and .

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4. Given that f (x) = –3x2 + 15,

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4. Given that f (x) = –3x2 + 15,

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4. Given that f (x) = –3x2 + 15,

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4. Given that f (x) = –3x2 + 15,

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5. Since and , we know that when applying the rule of f to the input values and , they both result in an output value of 36.

Can you plot the point on the graph of f ?

No. When we graph f we are showing the real number ordered pairs (x, y) that satisfy y = –3x2 + 15. Since contains a non-real number, it doesn’t appear on the graph of f (at least not on the graph drawn on the x-y coordinate plane).

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Since mathematicians initially rejected negative numbers, just imagine what they thought about the square root of a negative number! Most mathematicians thought such “numbers” had no place in serious mathematics.

It was during this time that the square roots of negative numbers came to be called imaginary numbers. Like an unflattering nickname a child gets in elementary school that can stay with him or her for years afterwards, the unfortunate name stuck.

Even today we still call them “imaginary numbers” despite the fact that the square roots of negative numbers have applications in many branches of science and engineering.

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Over the course of a few hundred years, as mathematicians slowly became comfortable with negative numbers, they began to accept “imaginary numbers” as solutions to equations and began studying them as a serious topic and thereby unlocking the path to many scientific and mathematical breakthroughs.

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Recall that the zeros (or roots) of a function are the values of the input quantity such that the function’s output is 0. In other words, they are the values of x such that f (x) = 0.

6. Show that the zeros of function f are the imaginary numbers and if f (x) = x2 + 13.

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Recall that the zeros (or roots) of a function are the values of the input quantity such that the function’s output is 0. In other words, they are the values of x such that f (x) = 0.

6. Show that the zeros of function g are the imaginary numbers and if g(x) = –2x2 – 8.

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Recall that we can write a quadratic function in factored form f (x) = a(x – u)(x – v) if u and v are the function’s zeros.

8. Write function f from Exercise #6 in factored form.

9. Each of the following represents a factored form for a quadratic function with “imaginary” numbers for zeros. State the zeros of the function, then write the function

in standard form.

a.

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in standard form.

a.

the zeros are and

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in standard form.

b.

the zeros are and

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b.

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10. Rewrite each of the following expressions using i where .

a. b.

c. d.

e. f.

g. h.

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a. b.

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c. d.

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e. f.

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g. h.

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11. If , then what is the value of i2 ?

The square root of a number is the number that, when multiplied by itself, yields the original number.

So i is the number that, when multiplied by itself, yields –1.

i2 is product of this number and itself, so i2 = –1.

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12. Each of the following represents a factored form for a quadratic function with “imaginary” numbers for zeros. Write each function in standard form.

a. f (x) = (x – 2i)(x + 2i) b. f (x) = (x + 6i)(x – 6i)

c. d.

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a. f (x) = (x – 2i)(x + 2i)

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b. f (x) = (x + 6i)(x – 6i)

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c.

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d.

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Mathematicians have developed a shorthand for referring to “imaginary” numbers. They use the letter i to represent . That is, . With this notation, we can rewrite numbers such as , , , and as follows.

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13. If and i2 = –1, then what is the value of each of the following? (Simplify your answer if possible.)

a. i3 b. i4

c. i5 d. i6

e. i21 f. i46

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a. i3 b. i4

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c. i5 d. i6

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e. i21 f. i46

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A complex number is a number made up of both a real number portion and an imaginary number portion.

We’ve seen a couple of examples of complex numbers so far in this worksheet. For example, 3 + 2i is a complex number with 3 being the real number portion and 2i being the imaginary number portion, and 5 – 7i is a complex number (also written as 5 + (–7i)) with 5 being the real number portion and –7i the imaginary number portion.

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In general, a complex number has the form a + bi where a is the real number portion and bi the imaginary number portion.

One way to think of real numbers like 4 or 17 is to say they are complex numbers where b = 0 (i.e., 4 = 4 + 0i or 17 = 17 + 0i).

We can think of imaginary numbers such as 10i or –6i as complex numbers where a = 0 (i.e., 10i = 0 + 10i or –6i = 0 + (–6i)).

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Complex numbers have similar properties to expressions involving only real numbers. We can add them, subtract them, multiply them, etc. by mostly just following the processes we already know.

When the operations are complete, we make sure to collect the real numbers together and the imaginary numbers together so that the final answer is written as a single complex number. (Notice that when we perform operations with complex numbers we end up with complex numbers as the result.)

Here are a few examples of performing operations with complex numbers.

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14. Perform the indicated operations. Write your final answer as a complex number a + bi in simplest form.

a. 2i(9 + 4i) b. (–4 + i) + (19 – 6i)

c. 3(7 – 5i) – 8(6 + 3i) d. (1 + 4i)(5 – 2i)

e. (3 + 5i)2 f. (a + bi)(c + di)

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a. 2i(9 + 4i) b. (–4 + i) + (19 – 6i)

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c. 3(7 – 5i) – 8(6 + 3i)

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d. (1 + 4i)(5 – 2i)

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e. (3 + 5i)2

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f. (a + bi)(c + di)

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1. Use the quadratic formula to find the two real zeros for function f.

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THE QUADRATIC FORMULA, COMPLEXROOTS, AND CONJUGATES

Recall that in this module we learned that the quadratic

formula tells us the zeros of a

quadratic function (that is, the values of x such that f (x) = 0). Recall also that sometimes we had two real number zeros, such as with the function f (x) = 4x2 + 2x – 2.


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1. Use the quadratic formula to find the two real zeros for function f.

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2. Use the quadratic formula to find the one real number zero for function g.

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Sometimes we had one real number zero, such as with the function g(x) = 3x2 – 12x + 12.

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3. Let’s find the roots of h(x) = x2 + 6x + 10.

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However, there were also times when there were only imaginary zeros and we were unable to find any real number zeros

a. Complete the solution process and write the roots of h as complex numbers (in the form a + bi).

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b. Can the roots of h be represented on the graph of h?

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b. Can the roots of h be represented on the graph of h?

No. When we graph h we are showing the real number ordered pairs (x, y) that satisfy the formula y = x2 + 6x + 10.

Since these zeros are non-real numbers, they don’t appear on the graph of h (at least not on the graph drawn on the x-y coordinate plane).

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4. Rewrite the following imaginary numbers using i. (Recall that Mathematicians have developed a shorthand for referring to “imaginary” numbers. They use the letter i to represent and write .)

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5. We found that the roots of h(x) = x2 + 6x + 10 are

The zeros of the function are x = –3 + i and x = –3 – i. We thus expect h(–3 + i) = 0 and h(–3 – i) = 0.

We can show that –3 + i is a root of h by evaluating h(–3 + i).

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Prove that a –3 – i is a root of h by evaluating h(–3 – i) to show that h(–3 – i) = 0. [h(x) = x2 + 6x + 10 ]

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In Exercises #6-11, find the zeros (roots) of the following quadratic functions. For those with imaginary roots, make sure to write the roots as complex numbers (that is, in the form a + bi).

6. f (x) = x2 + x + 2.5 7. f (x) = x2 + 5x + 6

8. f (x) = x2 + 5x + 9 9. f (x) = 2x2 – 5x + 2

10. f (x) = 4x2 + 2x – 12 11. f (x) = –3x2 + 3x – 4

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6. f (x) = x2 + x + 2.5

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7. f (x) = x2 + 5x + 6

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8. f (x) = x2 + 5x + 9

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9. f (x) = 2x2 – 5x + 2

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10. f (x) = 4x2 + 2x – 12

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11. f (x) = –3x2 + 3x – 4

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In Exercises #12-14, prove that the given numbers are roots of each quadratic function.

12. x = 1 + i and x = 1 – i are roots for f (x) = x2 – 2x + 2

13. x = 2 + 5i and x = 2 – 5i are roots for g(x) = x2 – 4x + 29

14. x = –4 + 3i and x = –4 – 3i are roots for h(x) = x2 + 8x + 25

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12. x = 1 + i and x = 1 – i are roots for f (x) = x2 – 2x + 2

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13. x = 2 + 5i and x = 2 – 5i are roots for g(x) = x2 – 4x + 29

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14. x = –4 + 3i and x = –4 – 3i are roots for h(x) = x2 + 8x + 25

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15. Write each of the functions in Exercises #12-14 in factored form.

12. f (x) = x2 – 2x + 2

13. g(x) = x2 – 4x + 29

14. h(x) = x2 + 8x + 25

f (x) = (x – (1 + i))(x – (1 – i))

g(x) = (x – (2 + 5i))(x – (2 – 5i))

h(x) = (x – (–4 + 3i))(x – (–4 – 3i))

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You might have observed that when the discriminant is negative for a quadratic function with real-number coefficients (i.e., when b2 – 4ac < 0), we get two complex roots, and that they will always be in a form like 3 + 2i and 3 – 2i.

In other words, if we get one complex root for a quadratic function (call it a + bi), then a – bi is always a complex root of the same function as well.

Complex numbers whose real number portions are identical but whose imaginary number portions have opposite signs (such as a + bi and a – bi) are called conjugate pairs.

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a. 7 – 10i b. 11 + 2i

c. –2 – 2i d. –i

16. For each of the following complex numbers, write the other number in the conjugate pair.

7 + 10i 11 – 2i

–2 + 2i i

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There are two interesting facts about conjugate pairs. The first fact, as stated above, is that if you know one element of the pair is a zero of a quadratic function with real number coefficients, then the other element of the pair is automatically a zero of the function as well. [For example, if 6 + 2i is a root, then so is 6 – 2i. If –5 + i is a root, then so is –5 + i.] Let’s perform an exploration to determine the second interesting fact.

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For Exercises #17-22, find the product. Write your final answer as a complex number (in the form a + bi).

17. (1 + 2i)(1 – 2i) 18. (3 – 4i)(2 + 5i)

19. (7 – i)(6 + 2i) 20. (8 – i)(8 + i)

21. (4 + 3i)(5 – 3i) 22. (a + bi)(a – bi)

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17. (1 + 2i)(1 – 2i)

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18. (3 – 4i)(2 + 5i)

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19. (7 – i)(6 + 2i)

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20. (8 – i)(8 + i)

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21. (4 + 3i)(5 – 3i)

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22. (a + bi)(a – bi)

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23. What is true about conjugate pairs that isn’t generally true about all complex numbers?

When we find the product of a conjugate pair, the result is a real number.

[Specifically (a + bi)(a – bi) = a2 + b2 using the results from Exercise #22.]

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Copyright © 2012 Carlson, O’Bryan & Joyner Worksheet #6: Vertex Form and Completing the Square Worksheet #7: The Quadratic Formula Worksheet #8: Imaginary