Transcript Document

MODULE 4
Quadratic Functions
Part 2: Worksheets #6-9
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
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VERTEX FORM AND COMPLETING THE SQUARE
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.
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?
i) f (x) = (x – 5)2 – 4
ii) f (x) = 4(x – 1)2
(5, –4)
minimum
(1, 0)
minimum
iii) f (x) = 2(x + 6)2 + 5
iv) f (x) = (x + 2)2 – 8
(–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.
•
•
•
•
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.)
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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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?
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.
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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.)
b. Write two different functions that have their vertices at
(–5, –7).
Answers will vary. Two examples are
f (x) = (x + 5)2 – 7
and
g(x) = –2(x + 5)2 – 7.
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4. Let f (x) = 2(x – 2)2 – 8.
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?
c = 36
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?
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].
x2 + 12x + 36 = (x + 6)2
x2 – 30x + 225 = (x – 15)2
x2 + 3x + 2.25 = (x + 1.5)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].
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7. Let f (x) = x2 + 14x – 5. The first step is to group the
terms that contain variables: f (x) = (x2 + 14x) – 5.
a. Consider the expression x2 + 14x + ____.
49 What must go
in the blank so that we can factor the expression in the
form (x – h)2 or (x + h)2?
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7. Let f (x) = x2 + 14x – 5. The first step is to group the
terms that contain variables: f (x) = (x2 + 14x) – 5.
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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7. Let f (x) = x2 + 14x – 5. The first step is to group the
terms that contain variables: f (x) = (x2 + 14x) – 5.
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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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
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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.
b. g(x) = x2 – 10x + 3
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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.
c. f (x) = x2 + 6x + 30
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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.
d. h(x) = x2 – 32x + 27
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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.
e. f (x) = x2 + 9x +13
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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.
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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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
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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.
b. f (x) = 5x2 + 120x
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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.
c. h(x) = –3x2 – 18x + 12
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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.
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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a. Write the formula in vertex form using completing the
square. The first step has been completed for you.
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a. Write the formula in vertex form using completing the
square. The first step has been completed for you.
b. Is the vertex a minimum or a maximum?
a maximum
c. What is the vertex? What does this tell us about the
situation?
(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
the parentheses?
and
To complete the square for
divide
, we needed to
by 2 and square the result.
To complete the square for
divide
added inside of
, we needed to
by 2 and square the result.
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 13 / 3 


 2 
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b. In Step 3, why were
and
subtracted
from the left sides of the equations instead of
and
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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 13 / 3 


 2 
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b. In Step 3, why were
and
subtracted
from the left sides of the equations instead of
and
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?
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c. What did we do to move from Step 3 to Step 4?
We rewrote the expression
as
(because
).
We also 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
in their factored forms
and
and
.
We also applied the rule of exponents
as
and
as
to rewrite
.
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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
than
and a value of x some amount less
.
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 zeroproduct 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
function are
than
. That is, the roots of the
units less than and greater
. 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
function are
than
. That is, the roots of the
units less than and greater
. 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.
b. Describe what went “wrong” during the solution
process.
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.)
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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
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.
The graph of the function doesn’t cross the horizontal axis
(the function has no horizontal intercepts).
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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.
b. Describe what happened during the solution process.
The value of
was 0, leaving us with only one
solution to the equation.
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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
f
x
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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, nonzero 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,
0, then
then …
there are two real number
zeros (roots) for the quadratic function.
ii) If b2 – 4ac =
= 0,
0, then
then there
… is exactly one real
number zero (root) for the quadratic function.
iii) If b2 – 4ac <
< 0,
0, then
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 NUMBERS
Have 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.
b.
is 3 because ________________.
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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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.
the zeros are
and
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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.
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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10. Rewrite each of the following expressions using i
where
.
a.
b.
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10. Rewrite each of the following expressions using i
where
.
c.
d.
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10. Rewrite each of the following expressions using i
where
.
e.
f.
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10. Rewrite each of the following expressions using i
where
.
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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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)
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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.
b. f (x) = (x + 6i)(x – 6i)
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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.
c.
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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.
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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13. If
and i2 = –1, then what is the value of each of
the following? (Simplify your answer if possible.)
a. i3
b. i4
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13. If
and i2 = –1, then what is the value of each of
the following? (Simplify your answer if possible.)
c. i5
d. i6
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13. If
and i2 = –1, then what is the value of each of
the following? (Simplify your answer if possible.)
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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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)
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14. Perform the indicated operations. Write your final
answer as a complex number a + bi in simplest form.
c. 3(7 – 5i) – 8(6 + 3i)
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14. Perform the indicated operations. Write your final
answer as a complex number a + bi in simplest form.
d. (1 + 4i)(5 – 2i)
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14. Perform the indicated operations. Write your final
answer as a complex number a + bi in simplest form.
e. (3 + 5i)2
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14. Perform the indicated operations. Write your final
answer as a complex number a + bi in simplest form.
f. (a + bi)(c + di)
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THE QUADRATIC FORMULA, COMPLEX
ROOTS, 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.
1. Use the quadratic formula to find the two real zeros for
function f.
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1. Use the quadratic formula to find the two real zeros for
function f.
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Sometimes we had one real number zero, such as with the
function g(x) = 3x2 – 12x + 12.
2. Use the quadratic formula to find the one real number
zero for function g.
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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
3. Let’s find the roots of h(x) = x2 + 6x + 10.
a. Complete the solution process and write the roots of h as
complex numbers (in the form a + bi).
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3. Let’s find the roots of h(x) = x2 + 6x + 10.
a. Complete the solution process and write the roots of h as
complex numbers (in the form a + bi).
b. Can the roots of h be represented on the graph of h?
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3. Let’s find the roots of h(x) = x2 + 6x + 10.
a. Complete the solution process and write the roots of h as
complex numbers (in the form a + bi).
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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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
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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).
7. f (x) = x2 + 5x + 6
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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).
8. f (x) = x2 + 5x + 9
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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).
9. f (x) = 2x2 – 5x + 2
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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).
10. f (x) = 4x2 + 2x – 12
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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).
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
f (x) = (x – (1 + i))(x – (1 – i))
13. g(x) = x2 – 4x + 29
g(x) = (x – (2 + 5i))(x – (2 – 5i))
14. h(x) = x2 + 8x + 25
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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16. For each of the following complex numbers, write the
other number in the conjugate pair.
a. 7 – 10i
b. 11 + 2i
11 – 2i
7 + 10i
c. –2 – 2i
d. –i
–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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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)
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For Exercises #17-22, find the product. Write your final
answer as a complex number (in the form a + bi).
18. (3 – 4i)(2 + 5i)
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For Exercises #17-22, find the product. Write your final
answer as a complex number (in the form a + bi).
19. (7 – i)(6 + 2i)
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For Exercises #17-22, find the product. Write your final
answer as a complex number (in the form a + bi).
20. (8 – i)(8 + i)
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For Exercises #17-22, find the product. Write your final
answer as a complex number (in the form a + bi).
21. (4 + 3i)(5 – 3i)
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For Exercises #17-22, find the product. Write your final
answer as a complex number (in the form a + bi).
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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