Download Finding Complex Solutions of Quadratic Equations

LESSON
3.3
Name
Finding Complex
Solutions of Quadratic
Equations
3.3
A
MP.2 Reasoning
B
Work with a partner or small group to determine whether solutions to
quadratic equations are real or not real and justify reasoning.
2x 2 + 4x + 2 = 0
2x + 4x = -2
f(x) = 2x + 4x
g(x) = -2
2x 2 + 4x + 3 = 0
2x + 4x = -3
f(x) = 2x + 4x
g(x) = -3
2
2
2
2
The graph of ƒ(x) = 2x 2 + 4x is shown. Graph each g(x). Complete the table.
-4
© Houghton Mifflin Harcourt Publishing Company
g(x) = -1
2x + 4x = -1
x
0
-2
2
4
Equation
Number of Real
Solutions
2x 2 + 4x + 1 = 0
2
2x + 4x + 2 = 0
1
2x 2 + 4x + 3 = 0
0
2
Repeat Steps A and B when ƒ(x) = -2x 2 + 4x.
ax 2 + bx + c = 0
ax 2 + bx = -c
f(x) = ax 2 + bx
g(x) = -c
-2x 2 + 4x - 1 = 0
-2x 2 + 4x = 1
f(x) = -2x 2 + 4x
g(x) = 1
-2x 2 + 4x - 2 = 0
-2x 2 + 4x = 2
f(x) = -2x 2 + 4x
g(x) = 2
-2x + 4x - 3 = 0
-2x + 4x = 3
f(x) = -2x + 4x
g(x) = 3
2
2
2
y
Equation
Number of Real
Solutions
-2x 2 + 4x - 1 = 0
2
-2x 2 + 4x - 2 = 0
1
-2x 2 + 4x - 3 = 0
0
2
-4
Module 3
-2
0
-2
x
4
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Lesson 3
139
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Finding Co ic Equations
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A2_MNLESE385894_U2M03L3.indd 139
find the
can you
equations
quadratic
N-CN.C.7 Solve A-REI.B.4b
.2,
Also N-CN.C
COMMON
CORE
s of
l Solution
g Rea
ax2 + bx
=0
2
+3
2x + 4x
. Graph each
is shown
2
2x + 4x
-2

A
Repeat Steps
and
Publishin
x2 + 4x.
(x) = -2
B when ƒ
Harcour t
2
-1
-2x + 4x
2
-2x + 4x
© Houghto
n Mifflin
2
-2x +
ax2 + bx
2=0
0
g(x) = -c
+ bx
2
-2x + 4x
g(x) = 1
g(x) = 2
g(x) = 3
of Real
Number ions
Solut
2
-1
-2x + 4x
4
2
x + 4x
f(x) = -2
x2 + 4x
f(x) = -2
x2 + 4x
f(x) = -2
Equation
x
0
-2
1
2
=3
2
-2x + 4x
0
4x - 3 =
2
-2
2
f(x) = ax
= -c
=1
2
-2x + 4x
=2
2
-2x + 4x
=0
y
-4
of Real
Number ions
Solut
=0
2
+3=0
2x + 4x
+c=0
ax2 + bx
g(x) = -3
=0
2
+2
2x + 4x
x
4
2
Turn to these pages to
find this lesson in the
hardcover student
edition.
g(x) = -2
+ 4x
lete the table.
g(x). Comp
2
+1
2x + 4x
0
g(x) = -c
g(x) = -1
+ 4x
2
f(x) = 2x
Equation
y
2
-4
2
f(x) = 2x
= -3
2x + 4x
2
=0
of ƒ(x) =
ns
ic Equatio
Quadrat
2 + bx
f(x) = ax
2
+ 4x
f(x) = 2x
= -c
= -1
2
2x + 4x
= -2
2
2x + 4x
=0
2
+2
2x + 4x
The graph
ax2 + bx
+c=0
2
+1
2x + 4x
HARDCOVER PAGES 99108
Resource
Locker
ion?
atic equat
of any quadr
ns.
ex solutio
have compl
ients that
the table.
Complete

solutions
complex
with real coeffic
Investigatin
Explore

y
g Compan
View the Engage section online. Discuss the photo
and how to solve a quadratic equation to determine
how high a baseball will go after it is hit. Then
preview the Lesson Performance Task.
g(x) = -c
2x + 4x + 1 = 0
y
C
f(x) = ax 2 + bx
f(x) = 2x 2 + 4x
2
2
ENGAGE
PREVIEW: LESSON
PERFORMANCE TASK
ax 2 + bx = -c
2
Mathematical Practices
Possible answer: You can factor, if possible, to find
real solutions; approximate from a graph; find a
square root (which may be part of completing the
square); complete the square; or apply the
quadratic formula. For the general equation
ax 2 + bx + c = 0, you must either complete the
square or use the quadratic formula to find the
complex solutions of the equation.
Investigating Real Solutions of Quadratic Equations
ax 2 + bx + c = 0
Solve quadratic equations with real coefficients that have complex
solutions. Also N-CN.C.2, A-REI.B.4b
Language Objective
=0
2=0
2
-2x + 4x
3=0
2
1
0
Lesson 3
139
Module 3
Lesson 3.3
L3.indd
4_U2M03
SE38589
A2_MNLE
139
Resource
Locker
Complete the table.
N-CN.C.7
Essential Question: How can you
find the complex solutions of any
quadratic equation?
Finding Complex Solutions
of Quadratic Equations
Explore
The student is expected to:
COMMON
CORE
Date
Essential Question: How can you find the complex solutions of any quadratic equation?
Common Core Math Standards
COMMON
CORE
Class
139
19/03/14
12:05 PM
19/03/14 12:00 PM
Reflect
1.
2.
3.
Look back at Steps A and B. Notice
that the minimum value of f(x) in Steps
A and B is -2. Complete the table by
identifying how many real solutions
the equation ƒ(x) = g(x) has for the
given values of g(x).
Look back at Step C. Notice that the
maximum value of ƒ(x) in Step C is 2.
Complete the table by identifying how
many real solutions the equation ƒ
(x) = g(x) has for the given values
of g(x).
EXPLORE
Value of g(x)
Number of Real Solutions
of
f(x) = g(x)
g(x) = -2
1
g(x) > -2
2
g(x) < -2
0
Value of g(x)
Number of Real Solutions
of
f(x) = g(x)
g(x) = 2
1
g(x) > 2
0
g(x) < 2
2
Investigating Real Solutions of
Quadratic Equations
INTEGRATE TECHNOLOGY
Students can use a graphing calculator to graph f(x)
and each function g(x) to verify the number of real
solutions to each equation.
QUESTIONING STRATEGIES
If an equation is written in vertex form, what
information can you use to find out if it has
real solutions? The sign of a determines the
direction of the opening and the maximum or
minimum value tells you whether there are real
solutions.
You can generalize Reflect 1: For ƒ(x) = ax 2 + bx where a > 0, ƒ(x) = g(x) where g(x) = -c has real
solutions when g(x) is greater than or equal to the minimum value of ƒ(x). The minimum value of ƒ(x) is
2
b
b2 - _
b2 - _
b2 - _
b2 = _
b2 = _
2b 2 = -_
b + b -_
b =a _
b2 .
= a -_
ƒ4a
4a
4a
2a
2a
4a
2a
2a
2a
4a 2
b2 .
So, ƒ(x) = g(x) has real solutions when g(x) ≥ -_
4a
b2
Substitute -c for g(x). -c ≥ -_
4a
b2 - c ≥ 0
b2
to both sides. _
Add __
4a
4a
Multiply both sides by 4a, which is positive. b 2 - 4ac ≥ 0
( _) ( ) ( ) ( )
How do you determine where the graph of a
quadratic function crosses the x-axis? You
can find the x-intercepts of the graph of a quadratic
function in standard form by factoring the function
to get its intercept form. If the function is not
factorable, the x-intercepts can be found by using
the quadratic formula to find the zeros of the
function.
In other words, the equation ax + bx + c = 0 where a > 0 has real solutions when b - 4ac ≥ 0.
2
2
For f(x) = ax 2 + bx where a < 0, f(x) = g(x) where g(x) = -c has real solutions when g(x) is
( 2a )
b
b
less than or equal to the maximum value of f(x). The maximum value of f(x) is f -___
= -___
.
4a
b
So, f(x) = g(x) has real solutions when g(x) ≤ -___
.
4a
b2
__
Substitute -c for g(x). -c ≤ - 4a
2
2
b2
b2
Add ___
to both sides. __
-c≤0
4a
4a
Multiply both sides by 4a, which is negative. b 2 - 4ac ≥ 0
Whether a > 0 or a < 0, b 2 - 4ac ≥ 0 tells when ax 2 + bx + c = 0 has real solutions.
Module 3
140
© Houghton Mifflin Harcourt Publishing Company
Generalize the results of Reflect 2 in a similar way. What do you notice?
Lesson 3
PROFESSIONAL DEVELOPMENT
A2_MNLESE385894_U2M03L3 140
5/22/14 10:47 AM
Math Background
In Algebra 1, students used the quadratic formula to find real solutions to a
quadratic equation. Students now revisit the formula to extend its use to complex
solutions.
The sign of the expression b 2 - 4ac determines whether the quadratic
equation has two real solutions, one real solution, or two nonreal solutions. For
cubic equations of the form ax 3 + bx 2 + cx + d = 0, the sign of the discriminant
b 2c 2 - 4ac 3 - 4b 3d - 27a 2d 2 determines whether the equation has three real
solutions, two real solutions, or one real solution.
Finding Complex Solutions of Quadratic Equations
140
Finding Complex Solutions by Completing
the Square
Explain 1
EXPLAIN 1
Finding Complex Solutions by
Completing the Square
2
()
2
(
)
2
trinomial x 2 + bx + __b2 , which you can factor as x + __b2 . Don’t forget that when x 2 + bx appears on one side of an
()
()
2
2
equation, adding __b2 to it requires adding __b2 to the other side as well.
Example 1
QUESTIONING STRATEGIES
How do you convert quadratic functions to
vertex form? Explain. You can convert
quadratic functions from standard form to vertex
2
form f(x) = a(x - h) + k by completing the square
on ax 2 + bx. You have to add and subtract the same
constant to keep the function value the same.

Solve the equation by completing the square. State whether the solutions
are real or non-real.
3x 2 + 9x - 6 = 0
1. Write the equation in the form x 2 + bx = c.
4. Solve for x.
(x + _23 ) = 2 + _49
(x + _23 ) = _174
2
3x 2 + 9x - 6 = 0
3x 2 + 9x = 6
2
――
x 2 + 3x = 2
(2)
17
3 = ±_
x+_
4
2
_
√ 17
3 = ±_
x+_
2
2
_
√ 17
3 ±_
x = -_
2
2
_
-3 ± √17
x=_
2
_
-3 + √17
There are two_real solutions: _
2
-3 - √17
and _.
2
2
2. Identify b and __b .
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Technology
MP.5 Discuss with students how to use the
b=3
() ()
2
2
9
b = _
3 =_
_
4
2
2
()
2
3. Add __b2 to both sides of the equation.
9 =2+_
9
x 2 + 3x + _
4
4

x 2 - 2x + 7 = 0
1. Write the equation in the form x 2 + bx = c.
© Houghton Mifflin Harcourt Publishing Company
graphing calculator to find a maximum or minimum
value of a quadratic function. Students can solve
problems algebraically and then use their graphing
calculators to check their solutions.
()
Recall that completing the square for the expression x 2 + bx requires adding __b2 to it, resulting in the perfect square
4. Solve for x.
x - 2x = -7
2
()
x 2 + 2x 1 = -7 + 1
(x - 1 ) = -6
2
2
2. Identify b and __b2 .
()
b = -2
( )
2
-2
b = _ = 1
_
2
2
()
――

x- 1 =±
2
-6
――

x=1±
-6
2
3. Add __b2 to both sides.
―
There are two real/non-real solutions:
√
and 1 - i 6 .
x 2 - 2x + 1 = -7 + 1
Module 3
141
―
1 + i √6
Lesson 3
COLLABORATIVE LEARNING
A2_MNLESE385894_U2M03L3.indd 141
Peer-to-Peer Activity
Have students work in pairs. Provide each pair with several quadratic equations
written in various forms. Have one student verbally instruct the partner in how to
find the nonreal solutions to the equation. Then have partners switch roles,
repeating the activity for a different quadratic equation. Have students discuss
how their steps for solving the equation were similar or different.
141
Lesson 3.3
19/03/14 12:00 PM
Reflect
EXPLAIN 2
How many complex solutions do the equations in Parts A and B have? Explain.
Each equation has two complex solutions, because the set of complex numbers includes
4.
Identifying Whether Solutions are
Real or Non-real
all real numbers as well as all non-real numbers.
Your Turn
Solve the equation by completing the square. State whether the solutions are real or non-real.
x 2 + 8x + 17 = 0
5.
6.
x + 8x = - 17
QUESTIONING STRATEGIES
x 2 + 10x - 7 = 0
x 2 + 10x = 7
2
x 2 + 8x +16 = - 17 + 16
Does the discriminant give the solution of a
quadratic equation? Explain. No, it gives the
number of solutions and type of solution, but it
does not give the actual solution.
x 2 + 10x + 25 = 7 + 25
(x + 4) 2 = -1
(x + 5) 2 = 32
x + 4 = ± √――
-1
_
x + 5 = ±√32
x = -4 ± i
_
x = -5 ± 4√2
There are two non-real solutions:
-4 + i and -4 - i.
There are_two non-real solutions:
_
-5 + 4√2 and -5 - 4√ 2 .
AVOID COMMON ERRORS
Identifying Whether Solutions Are Real or Non-real
Explain 2
Remind students that they must write the quadratic
equation in standard form before applying the
quadratic formula.
By completing the square for the general quadratic equation ax 2 + bx + c = 0, you can obtain the quadratic
__
√
-b ± b - 4ac
, which gives the solutions of the general quadratic equation. In the quadratic formula, the
formula, x = ___________
2a
expression under the radical sign, b 2 - 4ac, is called the discriminant, and its value determines whether the solutions
of the quadratic equation are real or non-real.
2
Number and Type of Solutions
b - 4ac > 0
Two real solutions
b 2 - 4ac = 0
One real solution
b 2 - 4ac < 0
Two non-real solutions
Example 2

Answer the question by writing an equation and determining whether the
solutions of the equation are real or non-real.
A ball is thrown in the air with an initial vertical velocity of
14 m/s from an initial height of 2 m. The ball’s height h (in meters)
at time t (in seconds) can be modeled by the quadratic function
h(t) = -4.9t 2 + 14t + 2. Does the ball reach a height of 12 m?
Set h(t) equal to 12.
Subtract 12 from both sides.
Find the value of the discriminant.
CONNECT VOCABULARY
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Fred
Fokkelman/Shutterstock
Value of Discriminant
2
-4.9t 2 + 14t + 2 = 12
-4.9t 2 + 14t + 10 = 0
Review vocabulary related to quadratic functions,
such as discriminant and real numbers, by having
students label the parts of a quadratic function
written in various forms.
14 2 - 4(-4.9)(-10) = 196 - 196 = 0
Because the discriminant is zero, the equation has one real solution, so the ball
does reach a height of 12 m.
Module 3
142
Lesson 3
DIFFERENTIATE INSTRUCTION
A2_MNLESE385894_U2M03L3 142
6/8/15 1:22 PM
Cognitive Strategies
Some students have trouble completing the square because there are so many
steps. Show them how to break the process into three parts:
(1) Get the equation into the form needed for completing the square.
(2) Complete the square.
(3) Finish the solution by taking square roots of both sides and simplifying the results.
When students make errors, analyze their work carefully to see what part of
the process is giving them trouble, and give them extra practice on that part of the
process.
Finding Complex Solutions of Quadratic Equations
142

INTEGRATE MATHEMATICAL
PRACTICES
Focus on Reasoning
MP.2 The discriminant can be used to distinguish
A person wants to create a vegetable garden and keep the rabbits
out by enclosing it with 100 feet of fencing. The area of the garden
is given by the function A(w) = w(50 - w) where w is the width
(in feet) of the garden. Can the garden have an area of 700 ft 2?
Set A(w) equal to 700.
Multiply on the left side.
between rational and irrational solutions. Give
students several quadratic equations for which
b 2 - 4ac is positive, some with rational solutions, and
some with irrational solutions. Ask them to make a
conjecture about how the value of the discriminant is
related to whether the solutions are rational or
irrational. Students should be able to explain why the
solutions will be rational when the value of the
discriminant is a perfect square.
Subtract 700 from both sides.
w(50 - w) = 700
50w - w 2 = 700
-w 2 + 50w - 700 = 0
Find the value of the discriminant. 50 2 - 4(-1)(-700) = 2500 -2800 = -300
Because the discriminant is [positive/zero/negative], the equation has [two real/one real/two non-real]
solutions, so the garden [can/cannot] have an area of 700 ft 2.
Your Turn
Answer the question by writing an equation and determining if the solutions are
real or non-real.
7.
EXPLAIN 3
A hobbyist is making a toy sailboat. For the triangular sail, she wants the height h (in inches)
to be twice the length of the base b (in inches). Can the area of the sail be 10 in 2?
1 b(2b) = b 2
Write the area A of the sail as a function of b.
A=_
2
Substitute 10 for A.
10 = b 2
0 = b 2 - 10
Subtract 10 from both sides.
QUESTIONING STRATEGIES
Why are there always two solutions to a
quadratic equation that has nonreal solutions?
How are they related? Since √b 2 - 4ac is not zero,
its value will be both added to and subtracted from
-b in the numerator, resulting in two solutions;
they are complex conjugates.
―――
What is the general solution of a quadratic
b
equation with only one solution? x = - ___
2a
Find the discriminant.
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©David
Burton/Alamy
Finding Complex Solutions Using the
Quadratic Formula
0 2 - 4(1)(-10) = 0 + 40 = 40
Because the discriminant is positive, the equation has two real solutions, so the area of the
sail can be 10 in 2.
Explain 3
Finding Complex Solutions
Using the Quadratic Formula
When using the quadratic formula to solve a quadratic equation, be sure the equation is in the
form ax 2 + bx + c = 0.
Example 3

Solve the equation using the quadratic formula.
Check a solution by substitution.
-5x 2 - 2x - 8 = 0
Write the quadratic formula.
―――
-b ± b 2 - 4ac
x = __
2a
―――――――
Substitute values.
2
-(-2) ± (-2) - 4(-5)(-8)
= ___
(
2 -5)
Simplify.
1 ± i39
2 ± -156
= __ = _
-5
-10
Module 3
――
143
―
Lesson 3
LANGUAGE SUPPORT
A2_MNLESE385894_U2M03L3 143
Communicate Math
Students play “How do you know?” Give students several cards containing
quadratic equations; some have real number solutions, others nonreal or complex
solutions. In small groups, students draw a card and state whether the solution is
real or not real. They then answer the question “How do you know?” Players take
turns and sort cards into piles according to the kind of solution. By the end of the
game, all players in a group must agree on card placement.
143
Lesson 3.3
5/22/14 10:47 AM
_
_
i√39
i√39
1 -_
1 +_
So, the two solutions are -_
and -_
.
5
5
5
5
AVOID COMMON ERRORS
Check by substituting one of the values.
Substitute.
Square.
―
―
) (
)
2i ―
39 _
39
i ―
1 +_
1 -_
- 39 ) - 2(-_
-8≟0
-5(_
5 )
5
25
25
25
(
Distribute.
√
√
―
―
2i √39 _
2i √39
2 +_
1 -_
-_
+ 39 + _
-8≟0
5
5
5
5
5
40 - 8 ≟ 0
_
5
0=0
Simplify.
B
Students may have difficulty remembering the
quadratic formula. Encourage students to copy the
formula and have it on hand when they are working.
Caution them to write the equation in standard form
before identifying the values of a, b, and c to be used
in the formula.
2
i √39
i √39
1 -_
1 -_
- 2 -_
-8≟0
-5 -_
5
5
5
5
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Communication
MP.3 You may wish to point out that quadratic
7x 2 + 2x + 3 = -1
7x 2 + 2x + 4 = 0
Write the equation with 0 on one side.
Write the quadratic formula.
x=
―――
-b ± √b - 4ac
__
2
2a
―――――――――
√ ( ) ( )( )
equations always have two roots. However, when the
value of the discriminant is 0, the two roots happen
to be the same. In this case, the quadratic is said to
have a double root.
2
4
2
7
-4
- 2 ±
= ____
7
2
( )
Substitute values.
―――
- 2 ±- 108
= __
14
Simplify.
――
――
√
√
_ _
Check by substituting one of the values.
Square.
Distribute.
(
3i √―
3
_ _
)
7 -1 +
7
Substitute.
(_
7
2
7
_― _)
6i √3
27
1
49
49
49
2
_ _
(
3i √―
3
_ _
)
+ 2 -1 +
7
(
2
7
3i √―
3
_ _
)
+ 2 -1 +
7
7
+4≟0
6i √―
6i √―
3 _
3
27 _
2
_1 - _
- +_+4≟0
7
7
7
Simplify.
A2_MNLESE385894_U2M03L3.indd 144
+4≟0
2
7
7
-
Module 3
© Houghton Mifflin Harcourt Publishing Company
3
3
- 2 ± 6 i
- 1 ± 3 i
= __ = __
7
14
3i ―
3i ―
3
3
-1 -1 +
7
7
7
7
So, the two solutions are
and
.
144
28
_
+4≟0
7
0=0
Lesson 3
19/03/14 11:59 AM
Finding Complex Solutions of Quadratic Equations
144
Your Turn
ELABORATE
Solve the equation using the quadratic formula. Check a solution by substitution.
8. 6x 2 - 5x - 4 = 0
9. x 2 + 8x + 12 = 2x
―――
-b ± √b - 4ac
__
2a
――――――
-(-5) ± √(-5) - 4(6)(-4)
___
=
2(6)
5 ± √――
121
_
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Critical Thinking
MP.3 Emphasize that choosing which method to
x 2 + 6x + 12 = 0
2
x=
―――
-b ± √b - 4ac
__
2a
――――――
-(-6) ± √(6) - 4(1)(12)
___
=
2(1)
-6 ± √――
-12
__
=
2
12
=
5 ± 11
=_
use to solve a quadratic equation is as important as
being able to use each method. Have students discuss
when each method might be preferred.
_ _
(_) (_)
AVOID COMMON ERRORS
―
32 _
20
_
-4≟ 0
Students may sometimes make a mistake in sign
when calculating the discriminant, particularly when
the quantity 4ac is less than 0. Remind them that
subtracting a negative number is the same as adding
the opposite, or positive, number. If a and c are
opposite signs, the discriminant will always be
positive.
3
3
0=0
Elaborate
__―
―
―
2
-6 ± 2i √3
=
2
= -3 ± i √3
So, the solutions are = -3 + i √3
and = -3 - i √3 .
Check
2
(-3 + i √3 ) + 8(-3 + i √3 ) + 12 ≟ 2(-3 + i √3 )
6 - 6i √3 - 24 + 8i √3 + 12 ≟ -6 + 2i √3
-6 + 2i √3 = -6 + 2i √3
12
5 + 11
4
So, the solutions are
=
12
3
5 - 11
1.
and
= -_
2
12
Check
2
4
4
6
-5
-4≟ 0
3
3
_
2
x=
2
―
―
―
―
―
―
―
―
10. Discussion Suppose that the quadratic equation ax 2 + bx + c = 0 has p + qi where q ≠ 0 as one of its
solutions. What must the other solution be? How do you know?
The other solution must be p − qi. The radical √b 2 - 4ac in the quadratic formula
―――
―――
produces imaginary numbers when b - 4ac < 0. Since √b 2 - 4ac is both added to and
2
subtracted from –b in the numerator of the quadratic formula, one solution will have the
SUMMARIZE THE LESSON
When does a quadratic equation have nonreal
solutions, and how do you find them? When
the value of the discriminant is negative, the
quadratic equation will have two nonreal solutions.
You find the solutions by using the quadratic
formula to solve the equation, and then writing the
solutions as a pair of complex conjugates of the
form a±bi.
© Houghton Mifflin Harcourt Publishing Company
form p + qi, and the other will have the form p − qi.
11. Discussion You know that the graph of the quadratic function ƒ(x) = ax 2 + bx + c has the vertical line
b
x = -__
as its axis of symmetry. If the graph of ƒ(x) crosses the x-axis, where do the x-intercepts occur
2a
relative to the axis of symmetry? Explain.
―――
-b ± √b 2 - 4ac
by the
The x-intercepts are the solutions of f(x) = 0, which are x = ___________
2a
b
quadratic formula. Writing the x-intercepts as x = -__
±
2a
the x-intercepts are the same distance,
√―――
b - 4ac
________
2
2a
2
2a
, away from the axis of symmetry,
b
with one x-intercept on each side of the line: x = -__
2a
b
x = -__
+
2a
√―――
b - 4ac
________
shows that
√―――
b - 4ac
________
on the other side.
2
√―――
b - 4ac
________
on one side and
2
2a
2a
12. Essential Question Check-In Why is using the quadratic formula to solve a quadratic equation easier
than completing the square?
The quadratic formula is the result of completing the square on the general
quadratic equation ax + bx + c = 0. As long as any particular equation is in the form
2
ax + bx + c = 0, you can simply substitute the values of a, b, and c into the quadratic
2
formula and obtain the solutions of the equation.
Module 3
A2_MNLESE385894_U2M03L3 145
145
Lesson 3.3
145
Lesson 3
10/17/14 4:53 PM
EVALUATE
Evaluate: Homework and Practice
1.
The graph of ƒ(x) = x 2 + 6x is shown. Use the
graph to determine how many real solutions
the following equations have: x 2 + 6x + 6 = 0,
x 2 + 6x + 9 = 0, and x 2 + 6x + 12 = 0. Explain.
-12 -8
For each equation, subtract the constant
from both sides to obtain these equations:
x 2 + 6x = -6, x 2 + 6x = -9, and x 2 + 6x = -12.
x
0
4
-4
ASSIGNMENT GUIDE
-12
The graph of g(x) = -6 intersects the graph
of f(x) twice, so the equation x 2 + 6x + 6 = 0
has two real solutions. The graph of g(x) = -9
intersects the graph of f(x) once, so the equation
x 2 + 6x + 9 = 0 has one real solution. The graph
of g(x) = -12 doesn’t intersect the graph of
f(x) , so the equation x 2 + 6x + 12 = 0 has no
real solutions.
2.
-4
• Online Homework
• Hints and Help
• Extra Practice
y
4
The graph of ƒ(x) = -_12 x 2 + 3x is shown. Use the graph to determine how
many real solutions the following equations have: -_21x 2 + 3x - 3 = 0,
-_12 x 2 + 3x - _29 = 0, and -_12x 2 + 3x - 6 = 0. Explain.
6
5
4
3
2
1
For each equation, subtract the constant from both sides to
9
1 2
1 2
x + 3x = 3, -_
x + 3x = _
, and
obtain these equations: -_
2
2
2
1 2
-_
x + 3x = 6. The graph of g(x) = 3 intersects the graph of
2
0
y
x
1 2 3 4 5 6 7
1 2
f(x) twice, so the equation -_
x + 3x - 3 = 0 has two real
2
9
solutions. The graph of g(x) = _
intersects the graph of f(x)
2
9
1 2
once, so the equation -_
x + 3x - _
= 0 has one real solution.
2
2
1 2
equation -_
x + 3x - 6 = 0 has no real solutions.
2
Solve the equation by completing the square. State whether the
solutions are real or non-real.
3.
x 2 + 4x + 1 = 0
4.
x 2 + 4x = -1
x 2 + 2x = -8
x 2 + 4x + 4 = -1 + 4
(x + 2)2 = 3
―
x + 2 = ±√3
x 2 + 2x + 8 = 0
x + 2x + 1 = -8 + 1
2
(x + 1)2 = -7
――
x + 1 = ± i √―
7
x + 1 = ± √-7
―
x = -2 ± √3
―
―
two real solutions:
-2 + √3 and -2 - √3 .
Module 3
Exercise
Explore
Investigating Real Solutions of
Quadratic Equations
Exercises 1–2
Example 1
Finding Complex Solutions by
Completing the Square
Exercises 3–8
Example 2
Identifying Whether Solutions are
Real or Non-real
Exercises 9–16
Example 3
Finding Complex Solutions Using
the Quadratic Formula
Exercises 17–20
What information does the value of the discriminant
give about a quadratic equation? The value of the
discriminant indicates the number and types of
roots.
―
x = -1 ± i √7
―
―
two non-real solutions:
-1 + i √7 and -1 - i √7 .
Lesson 3
146
A2_MNLESE385894_U2M03L3 146
Practice
CONNECT VOCABULARY
© Houghton Mifflin Harcourt Publishing Company
The graph of g(x) = 6 doesn’t intersect the graph of f(x), so the
Concepts and Skills
Depth of Knowledge (D.O.K.)
COMMON
CORE
Mathematical Practices
1–8
1 Recall of Information
MP.2 Reasoning
9–16
1 Recall of Information
MP.3 Logic
17–20
2 Skills/Concepts
MP.2 Reasoning
21
1 Recall of Information
MP.3 Logic
22
3 Strategic Thinking
MP.3 Logic
23–24
3 Strategic Thinking
MP.4 Modeling
5/22/14 10:47 AM
Finding Complex Solutions of Quadratic Equations
146
VISUAL CUES
5.
If students have difficulty evaluating the
discriminant, have them organize the variables in a
table. Ask students to create a table for each of the
variables (a, b, c, b 2, 4ac, b 2 - 4ac) and have them
predict the number and type of solutions based on
the variables they list in the table.
x 2 - 5x = -20
6.
25
25
x 2 - 5x + __
= -20 + __
4
4
x 2 - 1.2x = 1.6
2
55
(x - _52) = -__
4
――
5
55
x-_
=
±
-__
√
2
4
―
i √55
5
____
x-_
=
±
2
2
i √―
55
5
x=_
± ____
2
x 2 - 1.2x + 0.36 = 1.6 + 0.36
(x - 0.6)2 = 1.96
x - 0.6 = ±1.4
x = 0.6 ± 1.4
2
two real solutions:
2 and -0.8.
i √―
55
i √―
55
5
_5 + ____
and _
- ____.
7.
2
2
2
7x + 13x = 5
2
x
2
8.
13
5
+ __
x=_
7
-x 2 - 6x - 11 = 0
x 2 + 6x + 11 = 0
7
x 2 + 6x + 9 = -11 + 9
13
169
5
169
x + ___
=_
+ ___
x 2 + __
7
7
196
196
(x + 3)2 = -2
――
x + 3 = ± i √―
2
2
13
309
) = ___
(x + __
x + 3 = ± √-2
――
13
309
= ± √___
x + __
14
196
14
――
x - 0.6 = ± √1.96
two non-real solutions:
2
5x 2 - 6x = 8
196
√――
309
13
= ± _____
x + __
√――
309
13
± _____
x = -__
14
14
14
two real solutions:
――
14
―
―
x = -3 ± i √2
―
two non-real solutions:
-3 + i√2 and -3 - i√2 .
――
-13
+ √309
-13 - √309
_________
and _________
.
14
14
Without solving the equation, state the number of solutions and whether
they are real or non-real.
9. -16x 2 + 4x + 13 = 0
10. 7x 2 - 11x + 10 = 0
Find the discriminant.
Find the discriminant.
© Houghton Mifflin Harcourt Publishing Company
4 2 - 4 (-16)(13) = 16 + 832 = 848
Because the discriminant is positive,
the equation has two real solutions.
2x = 1
11. -x 2 - _
5
2
x-1=0
-x 2 - _
5
Find the discriminant.
4
96
- 4 = -__
(-_25) - 4(-1)(-1) = __
25
25
2
Because the discriminant is negative,
the equation has two non-real solutions.
Module 3
A2_MNLESE385894_U2M03L3.indd 147
147
Lesson 3.3
(-11)2 - 4(7)(10) = 121 - 280 = -159
Because the discriminant is negative,
the equation has two non-real solutions.
12. 4x 2 + 9 = 12x
4x 2 - 12x + 9 = 0
Find the discriminant.
(-12)2 - 4(4)(9) = 144 - 144 = 0
Because the discriminant is zero, the equation
has one real solution.
147
Lesson 3
19/03/14 11:59 AM
Answer the question by writing an equation and determining whether the solutions of
the equation are real or non-real.
13. A gardener has 140 feet of fencing to put around a rectangular vegetable garden. The function
A(w) = 70w - w 2 gives the garden’s area A (in square feet) for any width w (in feet). Does the gardener
have enough fencing for the area of the garden to be 1300 ft 2?
Write an equation by setting A(w) equal to 1300. Then rewrite the equation
with 0 on one side. 70w - w 2 = 1300
-w 2 + 70w − 1300 = 0
Find the discriminant. 70 2 − 4(−1)(−1300) = 4900 − 5200 = − 300
Because the discriminant is negative, the equation has two non-real
solutions, so the gardener does not have enough fencing.
14. A golf ball is hit with an initial vertical velocity of 64 ft/s. The
function h(t) = -16t 2 + 64t models the height h (in feet) of the golf
ball at time t (in seconds). Does the golf ball reach a height of 60 ft?
Write an equation by setting h(t) equal to 60. Then rewrite the
equation with 0 on one side. -16t 2 + 64t = 60
-16t 2 + 64t - 60 = 0
4t 2 - 16t + 15 = 0
Find the discriminant. (-16) − 4(4)(15) = 256 − 240 = 16
2
Because the discriminant is positive, the equation has two real
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Aflo Foto
Agency/Alamy
solutions, so the golf ball does reach a height of 60 ft.
15. As a decoration for a school dance, the student council creates a
parabolic arch with balloons attached to it for students to walk
through as they enter the dance. The shape of the arch is modeled by
the equation y = x(5 - x), where x and y are measured in feet and
where the origin is at one end of the arch. Can a student who is 6 feet
6 inches tall walk through the arch without ducking?
Write an equation by setting y equal to 6.5. Then rewrite the equation with 0 on
one side. x(5 - x) = 6.5
5x - x 2 = 6.5
-x 2 + 5x - 6.5 = 0
Find the discriminant. 5 2 -4(-1)(-6.5) = 25 - 26 = -1
Because the discriminant is negative, the equation has two non-real solutions, so a
student who is 6 feet 6 inches tall cannot walk through the arch without ducking.
Module 3
A2_MNLESE385894_U2M03L3 148
148
Lesson 3
10/17/14 4:56 PM
Finding Complex Solutions of Quadratic Equations
148
16. A small theater company currently has 200 subscribers who each pay $120 for a season ticket. The
revenue from season-ticket subscriptions is $24,000. Market research indicates that for each $10
increase in the cost of a season ticket, the theater company will lose 10 subscribers. A model for the
projected revenue R (in dollars) from season-ticket subscriptions is R(p) = (120 + 10p)(200 - 10p),
where p is the number of $10 price increases. According to this model, is it possible for the theater
company to generate $25,600 in revenue by increasing the price of a season ticket?
PEER-TO-PEER DISCUSSION
Ask students to discuss with a partner how the graphs
of the following three parabolas would look: a
parabola with two real solutions, a parabola with one
real solution, and a parabola with two nonreal
solutions. Students should say that a parabola with
two solutions will have two x-intercepts, and the
parabola will open from the vertex toward the x-axis;
that a parabola with one solution will have one
x-intercept with the vertex on the x-axis; and that a
parabola with two nonreal solutions will open from
the vertex away from the axis and have no
x-intercept.
Write an equation by setting R(p) equal to 25,600. Then rewrite the equation with 0 on
one side. (120 + 10p) (200 - 10p) = 25,600
-100p 2 + 800p + 24, 000 = 25,600
-100p 2 + 800p - 1600 = 0
p 2 - 8p + 16 = 0
Find the discriminant. (-8) - 4(1)(16) = 64 - 64 = 0
2
Because the discriminant is zero, the equation has one real solution, so it is possible to
generate $25,600 in revenue by increasing the price of a season ticket.
Solve the equation using the quadratic formula. Check a solution by substitution.
17. x 2 - 8x + 27 = 0
―――
-b ± √b - 4ac
x = __
2a
――――――
-(-8) ± √(-8) - 4(1)(27)
= ___
2
18. x 2 - 30x+ 50 = 0
―――
-b ± √b - 4ac
__
2a
―――――――
-(-30) ± √(-30) - 4(1)(50)
= ___
2
2
=
=
8 ± √――
-44
_
2
8 ± 2i √―
11
_
2(1)
=
=
2
© Houghton Mifflin Harcourt Publishing Company
11
= 4 ± i √―
4 + i √―
11 and 4 - i √―
11 .
Check
(4 + i √―
11 ) - 8(4 + i √―
11 ) + 27 ≟ 0
5 + 8i √―
11 - 8(4 + i √―
11 ) + 27 ≟ 0
5 + 8i √―
11 - 32 - 8i √―
11 + 27 ≟ 0
2
5 - 32 + 27 ≟ 0
0=0
A2_MNLESE385894_U2M03L3 149
149
Lesson 3.3
――
30 ± √700
__
2
30 ± 10 √―
7
__
2
2(1)
―
= 15 ± 5 √7
So, the solutions are
Module 3
2
x=
So, the solutions are
―
―
15 + 5 √7 and 15 - 5 √7 .
―
―
―
―
―
7 + 50 ≟ 0
400 + 150 √7 - 450 - 150 √―
Check
2
(15 + 5 √7 ) - 30(15 + 5 √7 ) + 50 ≟ 0
400 + 150 √7 - 30(15 + 5 √7 ) + 50 ≟ 0
400 - 150 + 50 ≟ 0
0=0
149
Lesson 3
5/22/14 10:47 AM
19. x + 3 = x 2
20. 2x 2 + 7 = 4x
Rewrite the equation with 0 on one side.
2x 2 - 4x + 7 = 0
x 2 - x -3 = 0 Use the quadratic formula.
x=
―――
-b ± √b - 4ac
__
2a ――――――
-(-1)± √(-1) - 4(1)(-3)
= ___
2(1)
1 ± √―
13
_
2
x=
―――
-b ± √b - 4ac
__
2a
――――――
-(-4) ± √(-4) - 4(2)(7)
___
=
2(2)
4 ± √――
-40
__
=
4
4 ± 2i √―
2 ± i √―
10
10
__
_
=
2
So, the two solutions are
2
and
1
- √―
13
______
2
Students need to be careful to avoid making sign
errors when completing the square. Point out that
when the rule representing vertex form is simplified,
the result should be the original rule written in
standard form. Students can use this fact to perform a
quick check of the reasonableness of their results, and
in order to catch any sign errors they may have made.
2
2
2
1
+ √―
13
______
AVOID COMMON ERRORS
=
=
4
2
So, the two solutions are
.
―
―
)
) (_
(_
―
―
__ (_)
― _
__
( ―)
Check
2
1 + √13
1 + √13
-3≟0
2
2
14 + 2 √13
1 + √13
-3≟0
4
2
14 + 2 √13
2 + 2 √13
-3≟0
4
4
12
-3≟0
4
0=0
_
―
―
1 + ____
and 1 - ____
.
2
2
i √10
i √10
Check.
i √―
10
i √―
10
_
- 4(1 + _) + 7 ≟ 0
2 )
2
i √―
10
3
2(-_ + i √―
10 ) - 4(1 + _) + 7 ≟ 0
2
2
-3 + 2i √―
10 - 4 - 2i √―
10 + 7 ≟ 0
(
2
2 1+
-3 - 4 + 7 ≟ 0
0=0
21. Place an X in the appropriate column of the table to classify each equation by the
number and type of its solutions.
Equation
x 2 - 3x + 1 = 0
Two Real
Solutions
One Real
Solution
Two Non-Real
Solutions
X
x - 2x + 1 = 0
2
x - x +1 = 0
X
x2 + 1 = 0
X
x +x+1=0
X
2
x + 2x + 1 = 0
X
2
x + 3x + 1 = 0
2
Module 3
A2_MNLESE385894_U2M03L3.indd 150
© Houghton Mifflin Harcourt Publishing Company
X
2
X
150
Lesson 3
19/03/14 11:59 AM
Finding Complex Solutions of Quadratic Equations
150
JOURNAL
H.O.T. Focus on Higher Order Thinking
Have students summarize how to use the
discriminant to help solve any quadratic equation.
Have them include examples of quadratic equations
with one or two real solutions and with two nonreal
solutions.
22. Explain the Error A student used the method of completing the square to solve the
equation -x 2 + 2x - 3 = 0. Describe and correct the error.
-x 2 + 2x - 3 = 0
-x 2 + 2x = 3
-x 2 + 2x + 1 = 3 + 1
(x + 1) 2 = 4
_
x + 1 = ±√4
x + 1 = ±2
x = -1 ± 2
So, the two solutions are -1 + 2 = 1 and -1 - 2 = -3.
The student did not divide both sides by –1 first to make the coefficient of the x 2-term be 1.
The correct solution is as follows.
x 2 - 2x + 3 = 0
x 2 - 2x = -3
x - 2x + 1 = -3 + 1
2
(x -1)2 = -2
――
x - 1 = ± i √―
2
x = 1 ± i √―
2
x - 1 = ± √-2
―
―
So, the two solutions are 1 + i √2 and 1 - i √2 .
23. Make a Conjecture Describe the values of c for which the equation
x 2 + 8x + c = 0 has two real solutions, one real solution, and two non-real solutions.
Find the value of the discriminant.
© Houghton Mifflin Harcourt Publishing Company
b 2 - 4ac = 8 2 - 4(1) c = 64 - 4c
The equation has two real solutions when the discriminant is positive, so solving
64 - 4c > 0 for c gives c < 16. The equation has one real solution when the discriminant
is zero, so solving 64 - 4c = 0 for c gives c = 16. The equation has two non-real solutions
when the discriminant is negative, so solving 64 - 4c < 0 for c gives c > 16.
24. Analyze Relationships When you rewrite y = ax 2 + bx + c in vertex form by
(
)
b
b
, c - __
.
completing the square, you obtain these coordinates for the vertex: -__
4a
2a
Suppose the vertex of the graph of y = ax 2 + bx + c is located on the x-axis. Explain
2
how the coordinates of the vertex and the quadratic formula are in agreement in this
situation.
When the vertex is on the x-axis, the y-coordinate of the vertex must be 0,
b
= 0, which can be rewritten as b 2 - 4ac = 0. When you set y equal
so c - __
4a
2
to 0 in y = ax 2 + bx + c and solve for x, you get one real solution, namely,
b
x = -__
, which is the x-coordinate of the vertex.
2a
Module 3
A2_MNLESE385894_U2M03L3.indd 151
151
Lesson 3.3
151
Lesson 3
19/03/14 11:59 AM
Lesson Performance Task
AVOID COMMON ERRORS
Students may sometimes make a mistake in sign
when calculating the discriminant, particularly when
the quantity 4ac is less than 0. Remind them that
subtracting a negative number is the same as adding
the opposite, or positive, number. If a and c have
opposite signs, the discriminant will always be
positive.
Matt and his friends are enjoying an afternoon at a baseball game. A batter hits a towering
home run, and Matt shouts, “Wow, that must have been 110 feet high!” The ball was 4 feet off
the ground when the batter hit it, and the ball came off the bat traveling vertically at 80 feet per
second.
a. Model the ball’s height h (in feet) at time t (in seconds) using the projectile motion
model h(t) = -16t 2 + v 0t + h 0 where v 0 is the projectile’s initial vertical velocity
(in feet per second) and h 0 is the projectile’s initial height (in feet). Use the model
to write an equation based on Matt’s claim, and then determine whether Matt’s
claim is correct.
b. Did the ball reach a height of 100 feet? Explain.
INTEGRATE TECHNOLOGY
c. Let h max be the ball’s maximum height. By setting the projectile motion model
equal to h max, show how you can find h max using the discriminant of the quadratic
formula.
Students can use a graphing utility to graph a
parabola and find the maximum value.
d. Find the time at which the ball reached its maximum height.
a. The ball’s height h at time t is given by h(t) = -16t 2 + 80t + 4. Matt’s claim
is that h(t) = 110 at some time t. Applying the discriminant of the quadratic
formula to the equation -16t 2 + 80t + 4 = 110, or -16t 2 + 80t - 106 = 0,
gives b 2 - 4ac = 80 2 - 4(-16)(-106) = 6400 - 6784 = -384. Since the discriminant is
negative, there are no real values of t that solve the equation, so Matt’s claim is incorrect.
QUESTIONING STRATEGIES
How can the symmetry of a parabola help you
to find the maximum or minimum if you
know two different points on the graph with the same
y-coordinate? The x-coordinate of the maximum or
minimum will be halfway between the
x-coordinates of the two points on the graph.
b. For the ball to reach of height of 100 feet, h(t) must equal 100. Applying the discriminant
of the quadratic formula to the equation -16t 2 + 80t + 4 = 100, or
-16t 2 + 80t - 96 = 0, gives b 2 - 4ac = 80 2 - 4(-16)(-96) = 6400 - 6144 = 256.
Since the discriminant is positive, there are two real values of t that solve the equation,
so the ball did reach a height of 100 feet at two different times (once before reaching its
maximum height and once after).
b 2 - 4ac = 0
80 - 4(-16)(4 - h max) = 0
2
6400 + 64(4 - h max) = 0
64(4 - h max )= -6400
4 - h max = -100
-h max = -104
h max = 104
© Houghton Mifflin Harcourt Publishing Company
c. Setting h(t) equal to h max gives -16t 2 + 80t + 4 = h max = 0, or -16t 2 + 80t + 4 - h max = 0.
Since the maximum height occurs for a single real value of t, the discriminant of the
quadratic equation must equal 0.
So, the ball reached a maximum height of 104 feet.
d. Solve the equation -16t 2 + 80t + 4 = 104, or -16t 2 + 80t - 100 = 0, using the quadratic
formula. You already know that the discriminant is 0 when the ball reached its maximum
―
-80
= 2.5. So, the ball reached its maximum height 2.5 seconds
height, so t = ________ = ___
-32
2(-16)
after it was hit.
-80 ± √0
Module 3
152
EXTENSION ACTIVITY
A2_MNLESE385894_U2M03L3 152
_
Lesson 3
On the moon, the force of gravity is 1 Earth’s gravity, so the equation for simple
6
projectile motion for a ball hit at a height of 4 feet above the ground is
16
y = - t 2 + vt + 4. Have students determine if a baseball hit upward traveling
6
at an initial vertical velocity of v = 80 ft/s on the moon reaches a height of 200
feet. They should find the real solutions. t = 2.7 and 27.3 s, so the ball does
reach a height of 200 feet.
6/8/15 1:23 PM
_
Scoring Rubric
2 points: Student correctly solves the problem and explains his/her reasoning.
1 point: Student shows good understanding of the problem but does not fully
solve or explain his/her reasoning.
0 points: Student does not demonstrate understanding of the problem.
Finding Complex Solutions of Quadratic Equations
152