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Chapter 3
Quadratic Functions
Section 3.1 Introduction to Quadratic Functions
Section 3.2 Quadratic Expressions
Section 3.3 Converting to Factored and Vertex Form
Section 3.4 Quadratic Equations
Section 3.5 Factoring Hidden Quadratics
Section 3.6 Complex Numbers
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
3.1 Introduction to
Quadratic Functions
Section 3.1
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 1
Let ℎ(t) be the height of the ball thrown.
Show that Figure 3.1 is shaped like the
graph of h(t) = −16t2 + 32t + 128, t ≥ 0.
Solution
Figure 3.1: Height of a ball
thrown from the top of a building
Table 3.1 shows values of ℎ, which fit the points marked in Figure 3.1. The ball
rises from 128 ft when t = 0 to a height of 144 ft when t = 1, returns to 128 ft
when t = 2, and reaches the ground height 0 ft, when t = 4.
Table 3.1 Height of a ball
t (seconds)
h(t) (ft.)
0
1
2
3
4
128
144
128
80
0
Figure 3.1 shows that quadratic functions have graphs that bend, and this
shape is consistent with the values in Table 3.1.
Section 3.1
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 2
In Example 1, the height ℎ(t), in feet of the ball t seconds after it is
thrown off the top of a building is given by
h(t) = −16t2 + 32t + 128.
(a) What is the practical meaning of the value 128? How can you see this
value in the graph of the function in Figure 3.1?
(b) What effect does the sign of the coefficient −16 have on the graph of the
function?
Solution
(a) The constant term 128 gives the height of the ball when t = 0. This means
the top of the building (where the ball is at time t = 0) is 128 feet above the
ground. We see this since Figure 3.1 has a vertical intercept at 128.
(b) Because the coefficient −16 is negative, the term −16t2 counteracts the
positive terms 32t and 128. As a result of the negative coefficient −16, the
graph in Figure 3.1 goes up at first, reaches a maximum height, and then
goes down as the ball falls back to the ground.
Section 3.1
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Interpreting Quadratic Functions Expressed
in Standard Form
A quadratic function in x is one that can be written in standard form
y = f(x) = ax2 + bx + c, a, b, c constants, a ≠ 0.
• The term ax2 is called the quadratic term or leading term, and its
coefficient a is the leading coefficient.
· If a > 0, the graph of f is a parabola that opens upward.
· If a < 0, the graph of f is a parabola that opens downward.
• The term bx is called the linear term.
• The term c is called the constant term. It is the vertical intercept of the
graph of f.
The shape of the graph of a quadratic function is called a parabola. In Example 2,
the coefficient of the quadratic term is negative, and as a result the parabola opens
downward. If the coefficient of the quadratic term is positive then the parabola
opens upward.
Section 3.1
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Creating Computer Graphics
Curved graphs are used in computer graphics. Early computer graphics
software stored curves as large sets of points in the coordinate plane.
Modern software uses graphs of functions to draw curves because it
takes less memory to store an expression for a function than to store all
the points. For example, Figures 3.2 and 3.3 display a smile and frown
drawn using quadratic functions.
Figure 3.2: A smile drawn
using two quadratic functions
Section 3.1
Figure 3.3: A frown drawn
using two quadratic functions
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Interpreting Quadratic Functions to Fit Graphics
Example 3a
(a) Write the quadratic function in standard form and identify the
coefficients a, b, and c.
(i) f(x) = 2x(x − 1)
(ii) g(x) = x(x − 1)
Solution
(a) (i) Distributing 2x, we have
f(x) = 2x(x − 1) = 2x2 − 2x,
which is in standard form with a = 2, b = −2, and c = 0.
(ii) Distributing x, we have
g(x) = x(x − 1) = x2 − x,
which is in standard form with a = 1, b = −1, and c = 0.
Section 3.1
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Interpreting Quadratic Functions to Fit Graphics
Example 3b
(b) Evaluate f(0), f(0.5), and f(1), and identify which one of the graphs (I)–
(IV) in Figures 3.2 and 3.3 corresponds to the graph of f.
f(x) = 2x(x − 1) = 2x2 − 2x.
Solution
(b) First notice since a > 0, the
graph of f opens upward. So the
graph is either graph (I) or (II) in
Figure 3.2. Evaluating, we have
f(0) = 2 ⋅ 0 ⋅ (0 − 1) = 0.
f(0.5) = 2 ⋅ 0.5 ⋅ (0.5 − 1) = −0.5.
f(1) = 2 ⋅ 1 ⋅ (1 − 1) = 0.
The graph of f is therefore graph
(II) in Figure 3.2.
Section 3.1
Figure 3.2: A smile drawn
using two quadratic functions
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Interpreting Quadratic Functions to Fit Graphics
Example 3c
(c) Evaluate g(0), g(0.5), and g(1), and identify which one of the graphs (I)–
(IV) in Figures 3.2 and 3.3 corresponds to the graph of g
g(x) = x(x − 1) = x2 − x.
Solution
(c Since a > 0, the graph of g also
opens upward. We evaluate
g(0), g(0.5), and g(1) to check if
the graph of g could be graph (I)
in Figure 3.2.We have
g(0) = 0 ⋅ (0 − 1) = 0.
g(0.5) = 0.5 ⋅ (0.5 − 1) = −0.25.
g(1) = 1 ⋅ (1 − 1) = 0.
Thus the graph of g is indeed
graph (I) in Figure 3.2.
Section 3.1
Figure 3.2: A smile drawn
using two quadratic functions
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
3.1 QUADRATIC FUNCTIONS
Key Points
• The standard form shows the vertical intercept
• Curves in computer graphics
• Expressing quadratic functions in different forms reveals
different properties of the function
Section 3.1
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
3.2 Quadratic Expressions
Section 3.2
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Interpreting Quadratic Functions Expressed in Factored Form
Example 1
The function h(t) = −16t2 + 32t + 128 in Example 1 on page 100 can be
expressed in the form
h(t) = −16(t − 4)(t + 2), t ≥ 0.
What is the practical interpretation of the factors (t − 4) and (t + 2)?
Solution
When t = 4, the factor (t − 4) has the value 4 − 4 = 0. So
ℎ(4) = −16(0)(4 + 2) = 0.
In practical terms, this means that the ball hits the ground 4 seconds
after it is thrown. When t = −2, the factor (t + 2) has the value −2 + 2 = 0.
However, there is no
practical interpretation
for this, since the domain
of ℎ is t ≥ 0. See Figure 3.4.
Figure 3.4: Interpretation of the
factored form ℎ(t) = −16(t − 4)(t + 2)
Section 3.2
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Interpreting Quadratic Functions Expressed in Factored Form
Values of the independent variable where a function has the value zero,
such as the t = 4 and t = −20 in the previous example, are called zeros of
the function. In general, we have the following definition.
A quadratic function in x is expressed in factored form if it is
written as
y = f(x) = a(x − r)(x − s),
where a, r, and s are
constants and a ≠ 0.
• The constants r and s are zeros of the function
f(x) = a(x − r)(x − s).
• The constant a is the leading coefficient, the same as the
constant a in the standard form.
Section 3.2
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 2abc
For the functions y = (x − r)(x − s) graphed in Figure 3.5, give possible
values for the constants r and s, and write formulas for the functions in
factored form (if possible).
Figure 3.5
Solution
(a) Since the function has zeros at x = 1 and x = 3, we have r = 1 and
s = 3. The function can be written as y = (x − 1)(x − 3).
(b) Since the function has zeros at x = −2 and x = 4, we have r = −2 and
s = 4. The function can be written as y = (x + 2)(x − 4).
(c) Since the function has zeros at x = −5 and x = 0, we have r = −5 and
s = 0. The function can be written as y = (x + 5)(x − 0), which is
equivalent to y = x(x + 5).
Section 3.2
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 2de
For the functions y = (x − r)(x − s) graphed in Figure 3.5, give possible
values for the constants r and s, and write formulas for the functions in
factored form (if possible).
Figure 3.5
Solution
(d) The function has only one zero at x = 2, so we have r = s = 2. The
function can be written as y = (x − 2)(x − 2), which is equivalent to
y = (x − 2)2.
(e) Since the function has no zeros, it must not have any factors, and thus
the function cannot be written in factored form. Every quadratic
function can be written in standard form, but if a quadratic function
does not have any zeros, it cannot be expressed in factored form.
Section 3.2
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 3
A college bookstore finds that if it charges p dollars for a T-shirt, it sells 1000 −
20p T-shirts. Its revenue is the product of the price and the number of T-shirts
it sells.
(a) Express its revenue R(p) as a quadratic function of the price p in
factored form.
(b) For what prices is the revenue equal to zero?
Solution
(a) The revenue R(p) at price p is given by
Revenue = R(p) = (Price)(Number sold) = p(1000 − 20p).
Factoring −20 from each term in 1000 − 20p, we have
R(p) = −20p(−50 + p) = −20p(p − 50).
(b) The revenue has factors p and p−50. From the first factor, we see the
revenue is zero when p = 0. This makes sense since if the bookstore
does not charge any money for the T-shirt, it does not earn any revenue.
From the second factor, we see the revenue is also zero when p = 50. If
the price of a T-shirt is p = 50, then the number of T-shirts the bookstore
sells is 1000−20(50) = 0. The T-shirts are so expensive that nobody is
willing to buy them.
Section 3.2
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Interpreting Quadratic Functions Expressed in Vertex Form
Example 4
The function h(t) = −16t2 + 32t + 128 in Example 1 can be expressed in the
form
h(t) = −16(t − 1)2 + 144.
Use this form to show that the ball reaches its maximum height h = 144
when t = 1.
Solution
Looking at the right-hand side, we see that the term −16(t − 1)2 is a
negative number times a square, so it is always negative or zero, and it
is zero when t = 1. Therefore h(t) is always less than or equal to 144 and
is equal to 144 when t = 1. This means the maximum height the ball
reaches is 144 feet, and it reaches that height after 1 second. See Table
3.2 and Figure 3.6.
(See next slide for Table 3.2 and Figure 3.6.)
Section 3.2
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Interpreting Quadratic Functions Expressed in Vertex Form (continued)
Example 6 (continued)
(See previous slide.)
Table 3.2 Values of h(t) = −16(t − 1)2 + 144
are less than or equal to 144
−16(t − 1)2
h(t)
0
-16
128
1
0
144
2
-16
128
3
-64
80
4
-144
0
t (seconds)
Section 3.2
Figure 3.6: Ball reaches its greatest
height of 144 ft at t = 1
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Interpreting Quadratic Functions Expressed in Vertex Form (continued)
The point (1, 144) in Figure 3.6 is called the vertex of the graph. For
quadratic functions the vertex shows where the function reaches either its
largest value, called the maximum, or its smallest value, called the minimum.
In Example 4 the function reaches its maximum value at the vertex because
the coefficient is negative in the term −16(t − 1)2.
A quadratic function in x is expressed in vertex form if it is written as
y = f(x) = a(x − h)2 + k,
where a, h, and k are constants and a ≠ 0.
For the function f(x) = a(x − h)2 + k,
• f(h) = k, and the point (h, k) is the vertex of the graph.
• The coefficient a is the leading coefficient, the same a as in the standard
form.
・ If a > 0 then k is the minimum value of the function, and the graph
opens upward.
・ If a < 0 then k is the maximum value of the function, and the graph
opens downward
Section 3.2
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 5a
For each function, find the maximum or minimum and sketch the graph,
indicating the vertex.
(a) g(x) = (x − 3)2 + 2
Solution
(a) The expression for g is in vertex form. We have
g(x) = (x − 3)2 + 2 = Positive number (or zero) + 2.
Thus g(x) ≥ 2 for all values of x except x = 3, where it equals 2. The
minimum value is 2, and the vertex is at (3, 2) where the graph reaches
its lowest point.
See Figure 3.7(a).
Figure3.7: Interpretation
of the vertex form of a
quadratic expression
Section 3.2
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 5b
For each function, find the maximum or minimum and sketch the graph,
indicating the vertex.
(b) A(x) = 5 − (x + 2)2
Solution
(b) The expression for A is also in vertex form. We have
A(x) = 5 − (x + 2)2 = 5 − Positive number (or zero).
Thus A(x) ≤ 5 for all x except x = −2, where it equals 5. The maximum
value is 5, and the vertex is at (−2, 5), where the graph reaches its
highest point.
See Figure 3.7(b).
Figure 3.7: Interpretation
of the vertex form of a
quadratic expression
Section 3.2
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 7c
For each function, find the maximum or minimum and sketch the graph,
indicating the vertex.
(c) h(x) = x2 − 4x + 4
Solution
(c) The expression for h is not in vertex form. However, recognizing that
it is a perfect square, we can write it as
h(x) = x2 − 4x + 4 = (x − 2)2,
which is in vertex form with k = 0. So h(2) = 0 and h(x) is positive for all
other values of x. Thus the minimum value is 0, and it occurs at x = 2.
The vertex is at (2, 0).
See Figure 3.7(c).
Figure 3.7: Interpretation of
the vertex form of a quadratic
expression
Section 3.2
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Constructing Quadratic Expressions
The standard form tells us the vertical intercept, the factored
form tells us the horizontal intercepts, and the vertex form
tells us the maximum or minimum value of the function.
We can also go the other way and use this information to construct an
expression for a given quadratic function.
Section 3.2
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 6a
Find a possible formula for the
quadratic function whose graph is given
in Figure 3.8 (a).
Solution
Figure 3.8:
(a) Since we know the zeros, we start with a function in factored form:
f(x) = a(x − r)(x − s).
The graph has x-intercepts at x = 1 and x = 4, so the function has zeros at
those values, so we choose r = 1 and s = 4, which gives
f(x) = a(x − 1)(x − 4).
Since the y-intercept is −12, we know that y = −12 when x = 0. So
−12 = a(0 − 1)(0 − 4)
−12 = 4a
−3 = a.
So the function f(x) = −3(x − 1)(x − 4) has the right graph. Notice that the
value of a is negative, which we expect because the graph opens downward.
Section 3.2
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 6b
Find a possible formula for the
quadratic function whose graph is given
in Figure 3.8 (b).
Figure 3.8:
Solution
(b) We are given the vertex of the parabola, so we try to write its equation
using the vertex form y = f(x) = a(x − h)2 + k.
Since the coordinates of the vertex are (3,−13), we let h = 3 and k = −13. This
gives
f(x) = a(x − 3)2 − 13.
The y-intercept is (0, 5), so we know that y = 5 when x = 0. Substituting, we
get
5 = a(0 − 3)2 − 13
5 = 9a − 13
18 = 9a
2 = a.
Therefore, the function f(x) = 2(x − 3)2 − 13 has the correct graph.
Notice that the value of a is positive, which we expect because the graph
opens upward.
Section 3.2
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
3.2 WORKING WITH QUADRATIC EXPRESSIONS
Key Points
• The factored form shows where the function is equal to
zero
• The vertex form shows where the function reaches its
maximum or minimum
• Constructing quadratic expressions
Section 3.2
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
3.3 Converting to Factored
and Vertex Form
Section 3.3
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Converting Quadratic Expressions to
Standard and Factored Form
In the previous section we saw three forms for a function
giving the height of a ball:
h(t) = −16t2 + 32t + 128
(standard form)
= −16(t − 4)(t + 2)
(factored form)
= −16(t − 1)2 + 144
(vertex form).
One way to see that these forms are equivalent is to convert
them all to standard form.
Section 3.3
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Converting to Factored Form
Example 1
If possible, express the quadratic function in factored form.
(a) f(x) = x2 − 2x − 15
(b) g(t) = 2t2 − 16t + 14
Solution
(a) To factor x2 −2x−15, we write −2x as a sum of two terms whose
coefficients multiply to −15. Writing −2x = −5x + 3x works since −5 ⋅ 3 =
−15. This gives
f(x) = x2 − 2x − 15 = x2 − 5x + 3x − 15
= (x2 − 5x) + (3x − 15)
= x(x − 5) + 3(x − 5)
= (x + 3)(x − 5).
(b) First we take out the common factor of 2 from 2t2 − 16t + 14. This gives
2t2 − 16t + 14 = 2(t2 − 8t + 7).
To factor t2−8t+7, we write −8t as a sum of two terms whose coefficients
multiply to 7.Writing −8t = −7t − t works since −7 ⋅ (−1) = 7. This gives
g(t) = 2(t2 − 8t + 7) = 2(t2 − 7t − t + 7)
= 2 ((t2 − 7t) + (−t + 7)) = 2(t(t − 7) − (t − 7))
= 2(t − 7)(t − 1).
Section 3.3
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
How Do We Put an Expression in Vertex Form?
The key to putting an expression in vertex form is recognizing a
quadratic expression that is a perfect square—that is, in the following
form:
(x + p)2 = x2 + 2px + p2.
For example, in looking at the function y = x2 + 6x + 9 we recognize that
x2 + 6x + 9 = x2 + 2 ⋅ 3x + 32,
so p = 3 and
y = x2 + 6x + 9 = (x + 3)2.
When we have an expression that is not a perfect square, such as x2 +
6x + 8 , we can put it in vertex form by first transforming it into a perfect
square. Starting with
y = x2 + 6x + 8
we recognize that adding 1 to both sides leads to
y + 1 = x2 + 6x + 9 = (x + 3)2.
Finally, subtracting 1 from both sides gives the vertex form
y = (x + 3)2 − 1.
Because we added a constant that made the right-hand side into a
perfect square, this process is called completing the square.
Section 3.3
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
How Do We Put an Expression in Vertex Form? (continued)
Example 3
Express the function in vertex form by completing the square:
y = 4z2 − 40z + 16.
Solution
We first divide both sides by 4, the value of the leading coefficient of the
expression on the right-hand side. This gives
y/4 = (4z2 − 40z + 16)/4
= z2 − 10z + 4.
For the z-terms to match we must have 2p = −10, so p = −5, and p2 = 25.
To get a 25 on the right-hand side we add 21 to both sides:
y/4 + 21 = z2 − 10z + 4 + 21 = z2 − 10z + 25 = (z − 5)2.
Finally, to write the function in vertex form we solve for y:
y/4 + 21 = (z − 5)2
y/4 = (z − 5)2 − 21
y = 4((z − 5)2 − 21)
y = 4(z − 5)2 − 84.
Section 3.3
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
To find the maximum or minimum value of a quadratic function, we can use
the method of completing the square to express the function in vertex form.
Example 7
A bookstore finds that if it charges $p for a T-shirt then its revenue from T-shirt
sales is given by
R = f(p) = p(1000 − 20p)
What price should it charge in order to maximize the revenue?
Solution
We first expand the revenue function into standard form:
R = f(p) = p(1000 − 20p) = 1000p − 20p2.
Since the coefficient of p2 is −20, we know the graph of the quadratic opens
downward. So the vertex form of the equation gives us its maximum value.
R = 1000p − 20p2
−R/20 = p2 − 50p
divide by −20
−R/20 + 252 = p2 − 50p + 252
add 252 to both sides
−R/20 + 252 = (p − 25)2
(Solution continued on next slide.)
Section 3.3
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 7 (continued)
A bookstore finds that if it charges $p for a T-shirt then its revenue from T-shirt
sales is given by
R = f(p) = p(1000 − 20p)
What price should it charge in order to maximize the revenue?
Solution (continued)
−R/20 + 252 = (p − 25)2
R = −20(p − 25)2 + 20 · 252
R = −20(p − 25)2 + 12,500.
So the vertex is (25, 12,500),
indicating that the price that
maximizes the revenue is p = $25.
Figure 3.9 shows the graph of f. The
graph reveals that revenue initially
rises as the price increases, but
eventually starts to fall again when
the high price begins to deter
customers.
Section 3.3
Figure 3.9: Revenue from the sale
of T-shirts
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Visualizing The Process of Completing The Square
We can visualize how to find the constant that needs to be added to x2 + bx in
order to obtain a perfect square by thinking of x2 + bx as the area of a rectangle.
For example, the rectangle in Figure 3.10 has area x(x + 8) = x2 + 8x.
Figure 3.10: Rectangle with
sides x and x + 8
Now imagine cutting the rectangle into pieces as in Figure 3.11 and trying to
rearrange them to make a square, as in Figure 3.12. The corner piece, whose
area is 42 = 16, is missing. By adding this piece to our expression, we “complete”
the square: x2 + 8x + 16 = (x + 4)2.
Figure 3.11: Cutting off a
strip of width 4
Section 3.3
Figure 3.12: Rearranging the piece to make
a square with a missing corner
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
3.3 WORKING WITH QUADRATIC EXPRESSIONS
Key Points
• Converting to factored form to show zeros.
• Converting to vertex form to show where the function
reaches its maximum or minimum
• Completing the square procedure
Section 3.3
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
3.4 Quadratic Equations
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
When we find the zeros of a quadratic function
f(x) = ax2 + bx + c,
we are solving the equation
ax2 + bx + c = 0.
A quadratic equation in x is one which can be put into the
standard form
ax2 + bx + c = 0,
where a, b, c are constants,
with a ≠ 0.
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Solving Quadratic Equations by Factoring
The zero-factor principle states that
If A · B = 0 then either A = 0 or B = 0 (or both).
The factors A and B can be any numbers, including those represented by
algebraic expressions.
In Example 3 on page 105 we considered the quadratic function which
gives the revenue of selling T-shirts at price p dollars,
R(p) = p(1000 − 20p).
Since this function is already given as a product of linear factors, we can
find its zeros by finding where the factors are zero.
According to the zero-factor principle, the only possibilities for
R(p) = p(1000 − 20p) = 0
are for
p = 0 or
1000 – 20p = 0
This gives the solutions p = 0 and p = 50.
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 1a
Solve the quadratic equations using the zero-factor principle and factoring.
(a) x2 − 4x + 3 = 0
Solution
(a) We have
f(x) = x2 − 4x + 3 = (x − 1)(x − 3),
so x = 1 and x = 3 are zeros. To see that they are the only zeros, we let
A = x − 1 and B = x − 3, and we apply the zero-factor principle. If x is a
zero, then
( x  1)( x  3)  0
A
B
so either A = 0, which implies
x−1=0
x = 1,
or B = 0, which implies
x−3=0
x = 3.
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 1b
Solve the quadratic equations using the zero-factor principle and factoring.
(b) 3x2 + 3x = 6
Solution
(b) In order to use the zero-factor principle, we must first write the
quadratic equation in the standard form by subtracting 6 from both
sides. This gives
3x2 + 3x − 6 = 0
x2 + x − 2 = 0
divide both sides by 3
(x + 2)(x − 1) = 0.
This implies x = −2 and x = 1 are the solutions.
In general, the zero-factor principle tells us that for a quadratic equation in
the form
a(x − r)(x − s) = 0, where a, r, s are constants, a ≠ 0,
the only solutions are x = r and x = s.
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Solving Quadratic Equations with Perfect Squares
Example 2ab
Solve the quadratic equation by taking square roots of both sides (if
possible).
(a) x2 − 4 = 0
(b) x2 − 5 = 0
Solution
(a) Rewriting the equation as
we see the solutions are
x2 = 4,
x   4  2.
(b) Similarly, the solutions to x2 − 5 = 0 are
x   5  2.236.
We can use the result of Example 2b to solve (x − 2)2 = 5.
Since the equation x2 = 5 has solutions x   5., the equation
(x − 2)2 = 5 has solutions
Thus the solutions to (x − 2)2 = 5 are x  2  5 and x  2  5.
which can be combined as x  2   5.
Using a calculator, these solutions are approximately
x = 2 + 2.236 = 4.236
and
x = 2 − 2.236 = −0.236.
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
In general, if we can put a quadratic equation in the form
(x − h)2 = Constant,
then we can solve it by taking square roots of both sides.
Example 2cd
Solve the quadratic equation by taking square roots of both sides (if
possible).
(a) 2(y + 1)2 = 0
(b) 2(y − 3)2 + 4 = 0.
Solution
(a) Since 2(y + 1)2 = 0, dividing by 2 gives (y + 1)2 = 0 so y + 1 = 0.
Thus the only solution is y = −1. It is possible for a quadratic equation
to have only one solution.
(b) Since 2(y − 3)2 + 4 = 0, we have 2(y − 3)2 = −4,
so dividing by 2 gives
(y − 3)2 = −2.
But since no number squared is −2, this equation has no real number
solutions.
Example 2 illustrates that it is possible for a quadratic equation to have two
solutions, just one solution, or no solution.
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 3
The height of a ball t seconds after it is thrown is
h(t) = 16t2 +32t + 128 = −16(t − 1)2 + 144
Find the times where the ball reaches a height of 135 feet.
Solution
We want to find the values of t such that h(t) = 135, so we want to solve
the equation with the perfect square −16(t − 1)2 + 144 = 135.
Isolating the (t − 1)2 term we get
135  144 9
(t  1)2 
16
3
t 1  
4
3
t  1 .
4

16
Therefore, the solutions are t = 0.25 and t = 1.75. So the ball reaches a
height of 135 ft on its way up very soon after being thrown and again on
its way down about 2 seconds after being thrown.
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Solving Quadratic Equations by Completing the Square
Example 4a
Solve the quadratic equation by completing the square: (a) x2 + 6x + 8 = 1.
Solution
First, we move the constant to the right by adding −8 to both sides:
x2 + 6x + 8 − 8 = 1 − 8
add −8 to each side
x2 + 6x = −7
x2 + 6x + 9 = −7 + 9
complete the square by adding
(6/2)2 = 9 to both sides
(x + 3)2 = 2
taking the square root of both sides and solving for x
x3 2
x  3  2.
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Solving Quadratic Equations by Completing the Square
If the coefficient of x2 is not 1 we can divide through by it before completing
the square.
Example 4b
Solve the quadratic equation by completing the square: (b) 3x2 + 6x − 2 = 0
.
Solution
First, divide both sides of the equation by 3:
x2 + 2x − 2/3 = 0.
Next, move the constant to the right by adding 2/3 to both sides:
x2 + 2x = 2/3.
Now complete the square. The coefficient of x is 2, and half this is 1, so
we add 12 = 1 to each side to complete the square:
x2 + 2x + 1 = 2/3 + 1
or
(x + 1)2 = 5/3
5
x1 
3
x  1 
Section 3.4
5
.
3
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
The Quadratic Formula
Completing the square on the equation ax2 + bx + c = 0 gives a formula
for the solution of any quadratic equation. First we divide by a, getting
b
c
x2  x   0.
a
a
Now subtract the constant c/a from both sides:
b
c
x2  x   .
a
a
We complete the square by adding a constant to both sides of the
equation. The coefficient of x is b/a, and half this is
1 b b
  .
2 a 2a
(Continued on next slide.)
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
The Quadratic Formula (continued)
We square this and add the result, (b/2a)2, to each side:
2
b
c  b 
 b 
x2  x        
a
a  2a 
 2a 
2
2
2
b
b
c
b
x2  x  2    2
a
a 4a
4a
complete the square
expand parentheses
c 4a b2
   2
a 4a 4a
find a common denominator
b2  4ac

4a2
simplify right-hand side.
(Continued on next slide.)
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
The Quadratic Formula (continued)
We now rewrite the left-hand side as a perfect square:
2
b  b 2  4ac

 x  2a   4a2 .


Taking square roots gives
b
b2  4ac
x 
2a
2a
or
b  b2  4ac
x 
2a
2a
or
b
b2  4ac
x

2a
2a
or
b  b2  4ac
x
2a
So
b
b2  4ac
x

2a
2a
which gives the solutions
b  b2  4ac
x
2a
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
The Quadratic Formula (continued)
The quadratic formula combines both solutions of
ax2 + bx + c = 0:
b  b2  4ac
x
.
2a
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 5a
Use the quadratic formula to solve for x:
(a) 2x2 − 2x − 7 = 0
Solution
(a) We have a = 2, b = −2, c = −7, so
( 2)  ( 2)2  4  2( 7) 2  4  56 2  60
x


.
22
4
4
If we write 60  4  15  2 15, we get
2  2 15 1
15
x
 
.
4
2
2
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 5b
Use the quadratic formula to solve for x:
(b) 3x2 + 3x = 10
Solution
(b) We first put the equation in standard form by subtracting 10
from both sides:
3x2 + 3x − 10 = 0.
Thus a = 3, b = 3, and c = −10, so
3  32  4  3( 10) 3  9  120 3  129
1
129
x


 
.
23
6
6
2
6
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 6
The distance it takes a driver to stop in an emergency is the sum of the
reaction distance (the distance traveled while the driver is reacting to the
emergency) and the braking distance (the distance traveled once the driver
has put on the brakes). For a car traveling at v km/hr, the reaction distance
is 0.42v meters and the stopping distance is 0.0085v2 meters. What is the
speed of a car that stops in 100 meters?
Solution
The total stopping distance is 0.42v + 0.0085v2 meters, so we want to
solve the equation
0.42v + 0.0085v2 = 100, or, in standard form, 0.0085v2 + 0.42v − 100 = 0.
Using the quadratic formula, we get
0.42  0.42 2  4  0.0085( 100)
v
 86.537 or  135.950.
2  0.0085
The positive root is the only one that makes sense in this context, so the
car was going about 87 km/hr.
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
The Discriminant
We can tell how many solutions a quadratic equation has without actually
solving the equation. Look at the expression, b2 − 4ac, under the square
root sign in the quadratic formula.
For the equation ax2 + bx + c = 0, we call D = b2 − 4ac the
discriminant
• If D = b2 − 4ac is positive, then  b  4ac has two different
values, so the quadratic equation has two distinct solutions
(roots).
2
• If D = b2 − 4ac is negative, then b2  4ac is the square root of
a negative number, so the quadratic equation has no real
solutions.
• If D = b2 − 4ac = 0, then b  4ac  0, so the quadratic equation
has only one solution. This is sometimes referred to as a
repeated root.
2
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 7
How many solutions does each equation have?
(a) 4x2 − 10x + 7 = 0 (b) 0.3w2 + 1.5w + 1.8 = 0
(c) 3t2 − 18t + 27 = 0
Solution
(a) We use the discriminant. In this case, a = 4, b = −10, and c = 7. The
value of the discriminant is thus
b2 − 4ac = (−10)2 − 4 · 4 · 7 = 100 − 112 = −12.
Since the discriminant is negative, we know that the quadratic equation
has no real solutions.
(b) We again use the discriminant. In this case, a = 0.3, b = 1.5, and c = 1.8.
The value of the discriminant is thus
b2 − 4ac = (1.5)2 − 4 · 0.3 · 1.8 = 0.09.
Since the discriminant is positive, we know that the quadratic equation
has two distinct solutions.
(c) Checking the discriminant, we find
b2 − 4ac = (−18)2 − 4 · 3 · 27 = 0.
Since the discriminant is zero, we know that the quadratic equation has
only one solution.
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
3.4 SOLVING QUADRATIC EQUATIONS
Key Points
• Solving quadratic equations by factoring
• Solving equations by taking square roots
• Solving equations by completing the square
• Using the quadratic formula to solve quadratic equations
• The discriminant
Section 3.4
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
3.5 Factoring Hidden Quadratics
Section 3.5
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 1abc
The revenue to a bookstore from selling 1000 − 20p T-shirts at p dollars each
is R(p) = p(1000 − 20p). Suppose that each T-shirt costs the bookstore $3 to
make.
(a) Write an expression for the cost of making the T-shirts.
(b) Write an expression for the profit, which is the revenue minus the cost.
(c) For what values of p is the profit positive?
Solution
(a) Since each T-shirt costs $3 to make, we have
Cost = 3(Number of T-shirts sold) = 3(1000 − 2p).
(b) We have
Profit = Revenue − Cost = p(1000 − 20p) − 3(1000 − 20p).
(c) Factored form is the most useful for answering this question. We do not need
to expand the expression into standard form before trying to factor since we
see there is a common factor of 1000 − 20p. Taking out this factor, we get
Profit = p(1000 − 20p) − 3(1000 − 20p) = (p − 3)(1000 − 20p) = 20(p − 3)(50 − p).
(Continued on next slide.)
Section 3.5
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 1cd
(continued)
The revenue to a bookstore from selling 1000 − 20p T-shirts at p dollars each is
R(p) = p(1000 − 20p). Suppose that each T-shirt costs the bookstore $3 to make.
(c) For what values of p is the profit positive?
(d) Interpret your answer in practical terms.
Solution
(c) Profit = p(1000 − 20p) − 3(1000 − 20p) = (p − 3)(1000 − 20p) = 20(p − 3)(50 − p).
This first factor is p − 3, which is positive when p > 3 and negative when p < 3.
So p > 50 ∶
g(p) = 20(p − 3)(50 − p) = Positive × Negative = Negative.
3 < p < 50 ∶ g(p) = 20(p − 3)(50 − p) = Positive × Positive = Positive.
p<3∶
g(p) = 20(p − 3)(50 − p) = Negative × Positive = Negative.
So the profit is positive only if the price is greater than $3 but less than $50.
(d) The bookstore has to sell the T-shirts for more than it costs to make them, so that
is why the price has to be greater than $3. But if it charges too much then nobody
will buy the T-shirts. Since the number it can sell at price p is 1000−20p, it prices
itself out of the market when 1000−20p = 0, which means p = 50. That is why the
price has to be less than $50.
Section 3.5
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 2b
Solve the equation using the zero-factor principle and factoring:
(b) 2x3 − 5x2 = 3x
Solution
(b) In order to use the zero-factor principle, we first subtract 3x from both
sides. Then we factor the expression on the left-hand side:
2x3 − 5x2 − 3x = 0
x(2x2 − 5x − 3) = 0
factor out an x
x(2x + 1)(x − 3) = 0.
factor the quadratic
If the product of three numbers is 0, then at least one of them must be 0.
Thus
x = 0 or 2x + 1 = 0 or x − 3 = 0,
so
x = 0 or x = −1/2 or x = 3.
We call 2x3 − 5x2 − 3x = 0 a cubic equation, because the highest power of
x is 3.
Section 3.5
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 3
Solve y4 − 10y2 + 9 = 0.
Solution
Since y4 = (y2)2, the equation can be written as
y4 − 10y2 + 9 = (y2)2 − 10y2 + 9 = 0.
We can think of this as a quadratic equation in y2. Letting z = y2, we get
z2 − 10z + 9 = 0
replacing y2 with z
(z − 1)(z − 9) = 0
factoring the left side
(y2 − 1)(y2 − 9) = 0
replacing z with y2.
Thus, the solutions are given by
y2 − 1 = 0 or y2 − 9 = 0.
Solving for y gives
y = ±1 or y = ±3
The equation y4 − 10y2 + 9 = 0 is called a quartic equation because the
highest power of y is 4.
Section 3.5
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Sometimes an equation that does not look like a quadratic equation can be
transformed into one. We must be careful to check our answers after making a
transformation, because sometimes the transformed equation has solutions that
are not solutions to the original equation.
Example 4
Solve
Solution
a
4
 .
a 1 a
a
4
 a( a  1)   a( a  1).
Multiplying both sides by a(a − 1) gives
a 1
a
Canceling (a − 1) on the left and a on the right, we get
a2 = 4(a − 1),
giving the quadratic equation
a2 − 4a + 4 = 0.
Factoring gives
(a − 2)2 = 0,
which has solution a = 2. Since we multiplied both sides by a factor that
2
4
could be zero, we check the solution:
  2.
2 1 2
Section 3.5
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 5
Solve
Solution
x2  1
8

.
x3 x3
Notice that the two equal fractions have the same denominator. Therefore,
it will suffice to find the values of x that make the numerators equal,
provided that such values do not make the denominator zero.
x2 − 1 = 8
Rewriting this equation as
x2 = 9
and taking the square root of both sides, we see that
x = 3 and
x = −3.
The denominator of the original equation, x + 3, is not zero when x = 3.
Therefore, it is a solution to the original equation. However, the
denominator in the original equation is zero when x = −3. This means that
x = −3 is not a solution to the original equation. So x = −3 is an extraneous
solution that was introduced during the solving procedure.
Section 3.5
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
3.5 FACTORING HIDDEN QUADRATICS
Key Points
• Solving and interpreting equations by factoring
• Solving other equations using quadratic equations
Section 3.5
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
3.6 Complex Numbers
Section 3.6
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Until now, we have been regarding expressions like
4 as
undefined, because there is no real number whose square is
−4. In this section, we expand our idea of number to include
complex numbers. In the system of complex numbers, there
is a number whose square is −4.
Section 3.6
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Using Complex Numbers to Solve Equations
The general solution of cubic equations was discovered during the sixteenth
century. The solution introduced square roots of negative numbers, called
imaginary numbers. This discovery led mathematicians to find the solutions of
quadratic equations, such as
x2 − 2x + 10 = 0,
which is not satisfied by any real number x. Applying the quadratic formula gives
2  4  40
36
 1
.
2
2
The number −36 does not have a square root which is a real number. To
x
overcome this problem, we define the imaginary number i  1.
Then
i2 = −1.
Using i, we see that (6i)2 = 36i2 = −36, so
(6i )2
36
6i
x  1
 1
 1   1  3i.
2
2
2
There are two solutions for this quadratic equation just as there were two
solutions in the case of real numbers. The numbers 1 + 3i and 1 − 3i are
examples of complex numbers.
Section 3.6
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Using Complex Numbers to Solve Equations
A complex number is defined as any number that can be
written in the form
z = a + bi,
where a and b are real numbers and i  1. The real part
of z is the number a; the imaginary part is the number bi.
Calling the number i imaginary makes it sound as though i does not exist
in the same way as real numbers exist. In practice, if we measure mass
or position, we want our answers to be real. However, there are realworld phenomena, such as electro-magnetic waves, which are described
using complex numbers.
Section 3.6
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 1
Solve x2 − 6x + 15 = 4.
Solution
To solve this equation, we put it in the form ax2 + bx + c = 0.
x2 − 6x + 15 = 4
x2 − 6x + 11 = 0.
Applying the quadratic formula gives
6  36  44
x
2
6  8 6  1  8


2
2
6i 8

2
6 2i 2
 
2
2
 3  i 2.
Section 3.6
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Algebra of Complex Numbers
We can perform operations on complex numbers much like on real
numbers.
Two complex numbers are equal if and only if their real parts are equal
and their imaginary parts are equal.
That is, a + bi = c + di means that a = c and b = d.
In particular, the equality a + bi = 0 is equivalent to a = 0, b = 0.
A complex number with an imaginary part equal to zero is a real number.
Two complex numbers are called conjugates if their real parts are equal
and if their imaginary parts differ only in sign.
The complex conjugate of the complex number z = a + bi is denoted z .
So
z  a  bi
(Note that z is real if and only if z  z . ).
Section 3,6
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 2
What is the conjugate of 5 − 7i?
Solution
To find the conjugate, we simply change the subtraction sign to an
addition sign. The conjugate of 5 − 7i is 5 + 7i.
Example 3
Find the real numbers a and b that will make the following equation true:
2 − 6i = 2a + 3bi.
Solution
For the equation to be true, we must have 2a = 2 and 3bi = −6i.
Thus, a = 1 and 3b = −6, which gives us b = −2.
Section 3.6
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Addition and Subtraction of Complex Numbers
To add two complex numbers, we add the real and
imaginary parts separately:
(a + bi) + (c + di) = (a + c) + (b + d)i.
Subtracting one complex number from another is similar:
(a + bi) − (c + di) = (a − c) + (b − d)i.
Section 3.6
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 4
Compute the sum (2 + 7i) + (3 − 5i).
Solution
Adding real and imaginary parts gives
(2 + 7i) + (3 − 5i) = (2 + 3) + (7i − 5i) = 5 + 2i.
Example 5
Compute the difference (5 − 4i) − (8 − 3i).
Solution
Subtracting real and imaginary parts gives
(5 − 4i) − (8 − 3i)
= (5 − 8) + (−4i − (−3i))
= (5 − 8) + (−4i + 3i) = −3 − i.
Section 3.6
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Multiplication of Complex Numbers
Multiplication of complex numbers follows the distributive
law. We use the identity i2 = −1 and separate the real and
imaginary parts to expand the product (a + bi)(c + di):
(a + bi)(c + di) = a(c + di) + bi(c + di)
= ac + adi + bci + bdi2
= ac + bd(−1) + adi + bci
= ac − bd + adi + bci
= (ac − bd) + (ad + bc)i.
Section 3.6
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 6
Compute the product (4 + 6i)(5 + 2i).
Solution
Multiplying out gives (4 + 6i)(5 + 2i) = 20 + 8i + 30i + 12i2
= 20 + 38i − 12
= 8 + 38i.
Example 7
Simplify (2 + 4i)(4 − i) + 6 + 10i.
Solution
We wish to obtain a single complex number in the form a + bi.
Multiplying out and adding real and imaginary parts separately:
(2 + 4i)(4 − i) + 6 + 10i = 8 − 2i + 16i − 4i2 + 6 + 10i
= 14 + 24i + 4
= 18 + 24i.
Section 3.6
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Multiplication of Complex Numbers (continued)
Multiplying a number by its complex conjugate gives a real, non-negative
number. This property allows us to divide complex numbers easily.
Example 8
(a) Compute the product of −5 + 4i and its conjugate.
(b) Compute z  z , where z = a + bi and a, b are real.
Solution
(a) We have
(−5 + 4i)(−5 − 4i) = 25 + 20i − 20i − 16i2
= 25 − 16(−1) = 25 + 16 = 41.
(b) We have
z  z  ( a  bi)( a  bi)
= a2 + abi − abi − b2i2
= a2 − b2(−1)
= a 2 + b2
Section 3.6
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Multiplication of Complex Numbers
A special case of multiplication is the multiplication of i by itself, that is,
powers of i. We know that i2 = −1; then, i3 = i · i2 = −i, and i4 = (i2)2 =
(−1)2 = 1. Then i5 = i · i4 = i, and so on.
That is, for a nonnegative integer n, in takes on only four values. Thus
we have
 i
1

n
i 
 i
 1
Section 3.6
for n  1,5,9,13,
for n  2,6,10,14,
for n  3,7,11,15,
for n  4,8,12,16,
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Example 9
Simplify each of the following.
(a) i34
(b) 3i8 + 2i21 − 3i43 − 4i26
Solution
Since i4 = 1, we use this fact to help us simplify.
(a) i34 = i2 · i32 = i2(i4)8 = (−1)(1) = −1
(b) 3i8 + 2i21 − 3i43 − 4i26 = 3(i8) + 2i · i20 − 3i3 · i40 − 4i2 · i24
= 3(i4)2 + 2i(i4)5 − 3i3(i4)10 − 4i2(i4)6
= 3(1) + 2i(1) − 3i3(1) − 4i2(1)
= 3 + 2i − 3i3 − 4i2
= 3 + 2i − 3(−i) − 4(−1)
= 3 + 2i + 3i + 4
= 7 + 5i.
Section 3.6
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Division of Complex Numbers
How can we divide two complex numbers?
Even a very simple case such as (2 + i)/(1 − i) does not
have an obvious solution.
However, suppose that we divide 2 + i by a real number,
such as 5. Then we have
2i 2 i 2 1
    i
5
5 5 5 5
In order to divide any two complex numbers, we use the
complex conjugate to create a division by a real number,
since the product of a number and its complex conjugate is
always real.
Section 3.6
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
Division of Complex Numbers (continued)
Example 10
Compute
2  3i
.
3  2i
Solution
The conjugate of the denominator is 3 − 2i, so we multiply by
(3 − 2i)/(3 − 2i).
2  3i 2  3i 3  2i


3  2i 3  2i 3  2i
6  4i  9i  6i 2

32  2 2
12  5i 12 5

  i.
13
13 13
Section 3.6
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.
3.6 COMPLEX NUMBERS
Key Points
• Definition of imaginary and complex numbers
• Quadratic equations with complex solutions
• Algebra of complex numbers
Section 3.6
ALGEBRA: FORM AND FUNCTION 2nd edition
by McCallum, Connally, Hughes-Hallett, et
al.,Copyright 2015, John Wiley & Sons, Inc.