Download Module 2, Section 2 Solving Equations

Principles of Mathematics 11
Section 2, Introduction
103
Module 2, Section 2
Solving Equations
Introduction
In this section, you will learn to solve quadratic equations
graphically, by factoring, and by applying the quadratic
formula. You will learn to draw connections between the zeros
of a quadratic function, the x-intercepts of the graph of the
quadratic function, and the roots of the quadratic equation.
Furthermore, you will become familiar with the nature of the
roots of a quadratic equation by using the discriminant of a
quadratic equation. Other types of equations that you will learn
to solve are radical equations, rational equations, and absolute
value equations.
Section 2 — Outline
Lesson 1
Solving Quadratic Equations Graphically
Lesson 2
Solving Quadratic Equations by Factoring
Lesson 3
Solving Quadratic Equations by Completing the
Square and by the Quadratic Formula
Lesson 4
The Nature of Roots
Lesson 5
Radical Equations
Lesson 6
Rational and Absolute Value Equations
Review
Module 2
104
Notes
Module 2
Section 2, Introduction
Principles of Mathematics 11
Principles of Mathematics 11
Section 2, Lesson 1
105
Lesson 1
Solving Quadratic Equations Graphically
Outcomes
When you complete this lesson, you will be able to:
• solve quadratic equations graphically and draw
connections between the zeros of a quadratic function, the
x-intercepts of the graph of a quadratic function, and the
roots of the quadratic equation
• solve quadratic equations of the form ax2 + bx + c = 0
• solve quadratic equations where the solution is found in
the set of imaginary numbers
The Quadratic Equation
Definition
A Quadratic Equation is one which can be expressed in the form
ax2 + bx + c = 0 where a ¹ 0 and a, b, and c are Real numbers.
ax2 + bx + c = 0 is called the general form of the equation.
The roots or solutions for a quadratic equation are simply the
values of the variable x which make the equation a true
mathematical statement.
For example, 1 is a root of the quadratic equation x2 + 5x – 6 = 0
because (1)2 + 5(1) – 6 = 1 + 5 – 6 = 0.
A given Quadratic Equation might have as many as two roots
(solutions) or it may have no roots (solutions) at all!
The Corresponding Quadratic Function and Its Graph
For any Quadratic Equation ax2 + bx + c = 0, there is a
corresponding Quadratic Function (which we analyzed in detail
in Module 2, Section 1) f(x) = ax2 + bx + c. The graph of this
function was a parabola.
The roots of the quadratic equation are the zeros of the
corresponding quadratic function. These, in turn, correspond to
the x-intercepts of the parabola which is its graph.
In this lesson we will find the roots (solutions) of some
quadratic equations by graphing its corresponding function and
inspecting its x-intercepts.
Module 2
106
Section 2, Lesson 1
Principles of Mathematics 11
Example 1
Determine the roots of the equation 3x2 = 12.
Solution
Our first step is to put the equation into general form,
ax2 + bx + c = 0.
3x2 – 12 = 0.
In this case a = 3, b = 0, and c = -12.
In order to graph the corresponding function y = 3x2 – 12 we first
need the vertex. In Module 2, Section 1 we studied two methods of
determining the vertex. We can either
• complete the square, placing the function in standard form y
= a(x – h)2 + k and read the vertex (h,k) directly from the
equation, or
• we can use the formula for the vertex developed in Lesson 7.
In these examples we will use the latter method.
 –0 4 (3 )(–12 ) – 02 
 –b 4ac – b2 
=
V ,
,
 =V 
 2a
 2 (3 )

4a 
4 (3 )



Because the function y = 3x2 – 12 represents a vertical stretch
(by a factor of 3) of the standard parabola y = x2, the formation is
as follows:
y
z
12
12
z
z
3
3
x
}
}
1 1
2
Module 2
Now we sketch
the graph. Five
points are
usually
sufficient for the
sketch of a
parabola.
z
2
Principles of Mathematics 11
Section 2, Lesson 1
107
y
3
x
z
z
–2 –1
1 2
–5
By inspection of the curve, we
determine that the xintercepts are -2 and 2.
Therefore the roots (solutions)
for the original equation are 2 and 2. It's always a good
idea to check these values in
the original equation...just to
be sure.
z
z
–10
z
V(0,–12)
LHS
RHS
3x2
12
3(-2)2
3(4)
12
LHS
RHS
3x2
12
3(2)2
3(4)
12
Since both of the solutions check, we can state that the
Solution Set = { -2, 2 }.
Module 2
108
Section 2, Lesson 1
Principles of Mathematics 11
Example 2
Solve the equation x2 = -2x.
Solution
Our first step is to put the equation into general form,
ax2 + bx + c = 0 re-arrange it to x2 + 2x = 0. In this case a = 1,
b = 2, and c = 0.
In order to graph the corresponding function y = x2 + 2x we
need the vertex.
 –2 4 (1 )(0 ) – 22 
 –b 4ac – b2 
 = (-1,-1 )
=V 
V
,
,

 2a

 2 (1 )

4
a
4
1
(
)




The graph is standard in size (because the leading coefficient is
1), so we start with formation.
y
Then using the vertex (–1,–1),
the graph is shown below:
5
4
3
2
1
1 2 3 4 x
4 3 2 1 0
1
y
5
z
z
–3
z
–1
z
z 1
3
x
–3
The x-intercepts of the corresponding quadratic function are -2,
and 0. Therefore the Solution Set = { -2, 0 }. As an exercise you
can check these values in the original equation.
Module 2
Principles of Mathematics 11
Section 2, Lesson 1
109
Example 3
-2 x 2 − 2
Solve the equation
= −4.
x
Solution
First rearrange the original equation so that it's in general form.
-2x2 – 2 = -4x
-2x2 + 4x – 2 = 0
Now define y = -2x2 + 4x – 2. In this case a = -2, b = 4, and c = -2.
Determine the Vertex:
 –4 4 (−2 )(−2 ) – 42 
 –b 4ac – b2 
 = (1,0 )
V ,
,
 =V 
 2a
 2 (−2 )

−
4a 
4
2
(
)



The graph of this function is obtained by flipping (reflecting) the
standard y = x2 over (through) the x axis, and then stretching it
vertically by a factor of 2. So the formation will be as shown
below.
}
}
y
2
2
1
z
1
x
2
2
z
z
8
8
z
z
Q: Why does this
equation use
negative signs with
two different
lengths?
A: On the T1-83
there are two keys
for –
- a short one in
parenthesis (-) to
indicate negative as
in -2 or -x2 and a
longer one – to
indicate subtraction
as in 5 – 3 or x2 – 6x.
The short one is only
used before anumber
or variable and
never to operate on
two terms. The
longer one must be
used between two
terms.
When writing
equations by hand,
you may ignore the
difference
completely! When
using a graphing
calculator, you have
to follow that rule or
the calculator will
report an error.
The graph is shown on the next page:
Module 2
110
Section 2, Lesson 1
Principles of Mathematics 11
y
1
–4 –3 –2 –1
2 3 4
–1
z
z
x
There is only one xintercept in this case,
so the quadratic
equation has only one
root (solution),
namely 1. Solution
Set is singleton set
{ 1 }.
Of course, not all
Quadratic Equations
z
z –8
have integer
solutions. If one of its
solutions is nonintegral (in other
words, it's not an integer) then manually graphing the
corresponding function will not give you a very accurate answer.
The Graphing Calculator, however, can give you a very close
approximation. If you have a TI-83 (or the Langara Graph
Explorer) you can follow along with our final example.
Example 4
Solve the equation 2x(x + 1) = 5.
Solution:
2x(x + 1) = 5 ⇒ 2x2 + 2x = 5 ⇒ 2x2 + 2x – 5 = 0
Define Y1 = 2x2 + 2x – 5
Press Y= . Then type 2 X,T, θ,n x 2 + 2 X,T, θ,n – 5
Press GRAPH . You can see that the x-intercepts ar e not integers.
Press 2nd TRACE to access the "CALCULATE" menu.
Press 2 to locate your zeros (roots, x-intercep ts).
Module 2
Principles of Mathematics 11
Other graphing
calculators have
similar methods of
determining the
value of a noninteger root. Using
the Langara Graph
Explorer, you have
to “zoom in” to the xintercept to find it’s
approximate value.
Section 2, Lesson 1
111
You will see your graph and be prompted for a “Left Bound.”
Using the left arrow key locate the cursor just to the left of the
point of intersection of the parabola and the x-axis.
(x = -2.553191 is a good place). Press ENTER to confirm.
Now you will see a prompt for a “Right Bound.” This time use
the right arrow key to locate the cursor just to the right of the
same point of intersection. (x = -1.702128 is a good place). Press
ENTER to confirm.
You are now prompted for a “Guess”. Again using the left arrow
key, locate the cursor as close as possible to being on the actual
point of intersection and press ENTER to confirm. Voilà! At the
bottom of your screen your zero will appear, x = -2.158312
which we will round to -2.16.
In a similar manner you will determine that the second zero is
1.1583124 which we will round to 1.16.
So the Solution Set for the given equation is { -2.16, 1.16 }.
Now it's your turn to try a few. In the assignment which
follows, #2 g) will require a graphing calculator solution. If you
have one you can use it for that solution and you can also use it
to check your other hand-drawn graphs. If you do not have one
you may omit #2 g).
Self-Marking Activity
1. Rearrange each of the Quadratic Equations so that they are
in “general” form.
a) −2x 2 + 10 = −6
b
g
b) 3x x + 2 = 8
2
c) 5x − 7 + 3x = 7
d)
e)
bx + 3g
bx + 3g
2
= −4
2
−4 =5
x2 − 4
=x
3
g) 2x − 3 x + 2 = 0
f)
b
gb
g
Module 2
112
Section 2, Lesson 1
2. Solve each of the following by manually graphing the
corresponding quadratic function. You should use your
graphing calculator for (g), if you have one. Use the blank
grids that follow.
a) x 2 + 2 x − 8 = 0
b) x 2 + 4 x + 3 = 0
c) x 2 + 8 x = −15
d) 9 − x 2 = 0
e) 2x 2 − 12x + 10 = 0
f) -10 = x 2 − 12 x + 22
2
g ) -1 = (x − 2 ) − 3
y
a)
8
6
4
2
4
2
2
2
4
Module 2
4
6
8
x
Principles of Mathematics 11
Principles of Mathematics 11
Section 2, Lesson 1
113
y
b)
8
6
4
2
4
2
2
4
6
8
2
4
6
8
x
2
4
y
c)
8
6
4
2
4
2
x
2
4
Module 2
114
Section 2, Lesson 1
Principles of Mathematics 11
y
d)
8
6
4
2
4
2
2
4
6
8
2
4
6
8
x
2
4
y
e)
8
6
4
2
4
2
2
4
Module 2
x
Principles of Mathematics 11
Section 2, Lesson 1
115
y
f)
8
6
4
2
4
2
2
4
6
8
2
4
6
8
x
2
4
y
g)
8
6
4
2
4
2
x
2
4
Check your answers in the Module 2 Answer Key.
Module 2
116
Module 2
Section 2, Lesson 1
Principles of Mathematics 11
Principles of Mathematics 11
Section 2, Lesson 2
117
Lesson 2
Solving Quadratic Equations by Factoring
Outcome
When you complete this lesson, you will be able to
• solve quadratic equations by factoring and by drawing
connections among the zeros of functions, x-intercepts of the
graph, and roots of the quadratic equation.
Overview
You have solved quadratic equations by graphing. In some cases
only approximate roots can be determined. However, it is
possible to find the roots of some quadratics by factoring. This
solution method is based on the principle of zero products from
Module 2, Section 1. When using this property to solve
quadratic equations, make sure the equation is in general form
(with one side equal to zero).
In Lesson 1 you examined a quadratic equation of the form
ax2 + bx + c = 0, where b = 0. You solved this equation
graphically. In this lesson we will solve the same type of
equation by factoring. Recall from Principles of Mathematics 10
(and the Review in Module 1, Section 2) that the “difference of
squares” A2 – B2 factors as (A – B)(A + B).
Example 1
2
Solve 4x – 9 = 0 by factoring.
Solution
(2x − 3 )(2x + 3 ) = 0
(A difference of two squares.)
If the product of two expressions is equal to zero, then one of the
two expressions must be zero.
2x − 3 = 0 or 2 x + 3 = 0
2x = 3
or 2 x = −3
3
−3
x=
x=
2
2
You will use factoring to solve some equations of the type
2
2
ax + bx = 0 or ax + bx + c = 0.
Module 2
118
Section 2, Lesson 2
Example 2
Solve by factoring:
2
a) 3x = 7x
Solution
3 x 2 − 7x = 0
Set up equation in general form
x 3x − 7 = 0
Factor out the common factor of x
b
g
x=0
or
x=0
or
x=0
or
3x − 7 = 0
3x = 7
7
x=
3
Check and verify these roots.
2
b) 2x – 16x = 0
Solution
2x(x – 8) = 0 Factor out common factor of 2x
2x = 0 or
x = 0 or
x −8= 0
x=8
Check and verify roots.
Example 3
Solve:
2
a) 3x + 4x = 4
Solution
3x 2 + 4 x − 4 = 0
b3x − 2gbx + 2g = 0
3x − 2 = 0 or
2 or
x=
3
Factor [Recall Trinomial Factoring
from Principles of Math 10 or
x +2=0
Module 1, Section 2.]
Check these roots.
Module 2
Set up in general form
x = −2
Principles of Mathematics 11
Principles of Mathematics 11
Section 2, Lesson 2
119
2
b) x + 9 = 6x
Solution
x 2 − 6x + 9 = 0
bx − 3gbx − 3g = 0
x −3 = 0
or
x −3 = 0
x =3
or
x =3
The answer is 3. When the same number repeats as a root or
a zero, it is called a double root or double zero.
2
If you graph x – 6x + 9 = 0, notice that the vertex is on the
x-axis and the root is the same number as the x-coordinate of
the vertex.
2
c) 2x + 4x – 16 = 0
Solution
d
i Always factor the common factor out first
2b x + 4gb x − 2g = 0 Factor
Either 2b x + 4g = 0 or x − 2 = 0
2 x 2 + 2x − 8 = 0
x=2
x = −4
Check roots.
3
2
d) x + 5x – 6x = 0
Solution
d
i
x b x + 6gb x − 1g = 0
x x 2 + 5x − 6 = 0
x=0
or
x=0
or
Factor out the common factor of x
x +6=0
x = −6
or
or
x −1 = 0
x =1
This question is not quadratic but its method of solution is
similar. Its properties will be examined later.
Module 2
120
Section 2, Lesson 2
3
Principles of Mathematics 11
2
e) 4x – 6x – 4x = 0
Solution
2x (2x2 − 3x− 2)= 0
2x ( 2x + 1)( x− 2) = 0
2x + 1= 0
2x = 0
x=0
x= −
x− 2= 0
1
2
x= 2
Example 4
The sum of the squares of two consecutive, odd integers is 74.
Find the integers.
Solution:
Let x be the value of the lesser integer. The greater would, then,
be x + 2.
x2 + (x + 2)2 = 74
x2 + x2 + 4x + 4 = 74 (expanding (x + 2)2)
2x2 + 4x – 70 = 0
(simplifying and putting in general form)
x2 + 2x – 35 = 0
(dividing both sides by 2)
(x + 7 )(x – 5 ) = 0
(trinomial factoring)
Therefore x = –7 (in which case x + 2 = –7 + 2 = –5).
or
x = 5 (in which case x + 2 = 5 + 2 = 7).
So the two integers are –7 and –5 or 5 and 7!
In the activity which follows you will get lots of practice on
quadratic word problems. Be sure to define your variables carefully
and draw a diagram if the unknowns are lengths. With that in
mind please move on to the self-marking activity and be sure to
check your answers with the answer key which follows Module 2
when it is complete.
Module 2
Principles of Mathematics 11
Section 2, Lesson 2
121
Self-Marking Activity
1. Solve these equations by factoring. Check your solutions.
2
a) x – x – 12 = 0
2
b) x – x – 20 = 0
2
c) –x – 2x + 3 = 0
2
d) x + 9x + 18 = 0
2
e) 2x + 3x – 2 = 0
2. Rearrange each of the following equations and solve by
factoring, if possible. Check your roots.
a) 10 x 2 = 7x + 12
b) 5x 2 + 21x = 54
b
g b
g
c) 3x x − 2 − x x + 1 + 5 = 0
9
1
x−2 =0
2
2
2
e) x + 9 = 0
d) x 2 +
f) 16 x 2 − 64 = 0
g) 3x 2 + 16x + 10 = 0
h) x 2 + 12x − 28 = 0
3. One leg of a right triangle is 7 m longer than the other leg.
The hypotenuse is 17 m long. Find the length of each leg.
2
4. Two square checkerboards together have an area of 169 cm .
One has sides that are 7 cm longer than the other. Find the
length of the sides of each.
5. Find 3 consecutive odd integers such that the product of the
second and the third is 63.
6. Two positive numbers differ by 4 and the sum of their
squares is 136. Find the numbers.
7. A concrete pathway x metres wide is being built around a
2
40-m x 30-m garden. The area of the pathway is 984 m . Find
the width of the pathway.
Module 2
122
Section 2, Lesson 2
Principles of Mathematics 11
8. A rectangular piece of card is 5 cm longer than it is wide. A
3-cm by 3-cm square is cut out of each corner, and the four sides
3
are folded up to form an open box with a volume of 450 cm .
What were the length and width of the original piece of
cardboard?
Check your answers in the Module 2 Answer Key.
Module 2
Principles of Mathematics 11
Section 2, Lesson 3
123
Lesson 3
Solving Quadratic Equations by Completing
the Square and by the Quadratic Formula
Outcomes
When you complete this lesson, you will be able to
• solve a quadratic equation by completing the square
• solve a quadratic equation by using the quadratic formula
• recognize and solve equations in quadratic form
Overview
Some quadratic equations are easier to solve by factoring than
by graphing. However, not all quadratic equations are
factorable. One technique used to solve quadratic equations is
called completing the square. It follows a similar pattern to
what you did in Module 2, Section 1 to complete the square.
When completing the square to solve a quadratic equation you
must preserve the equality. When you add a constant to one
side of an equation, be sure to add the same constant to the
other side of the equation.
2
There are two cases regarding the solution to ax + bx + c = 0 by
completing the square.
1. When the leading coefficient is 1.
Example 1
2
Solve by completing the square: x – 4x + 2 = 0
Solution
x 2 − 4x + 2 = 0
x 2 − 4 x = −2
Subtract 2 from each side to isolate
the terms with variables
2
2
x 2 − 4 x + (−2 ) = −2 + (−2 ) Add the square of
to each side
(x − 2) = −2 + 4
2
(x − 2) = 2
2
x −2 = ± 2
x = 2± 2
1
2
the coefficient of x
Simplify
Take the square root of each side
Add 2 to each side
Module 2
124
Section 2, Lesson 3
Principles of Mathematics 11
Remember:
2 + 2 and 2 − 2 are the roots of the quadratic equation
2 + 2 and 2 − 2 are the zeros of the function
2 + 2 and 2 − 2 are the x-intercepts of the graph
e2 +
e
j
j
2 , 0 and 2 − 2 , 0 are the points where the graph
crosses the x-axis
2. When the leading coefficient is not 1.
If the leading coefficient of the quadratic is not 1, you divide
both sides of the equation by the coefficient before you
complete the square.
Example 2
2
Solve by completing the square: 2x – 6x – 7 = 0
Solution
2x2 − 6 x − 7 = 0
7
x2 − 3x =
2
FG −3IJ
H 2K
FG x − 3IJ
H 2K
FG x − 3IJ
H 2K
2
x2 − 3x +
x−
Add seven to each side and divide by 2
FG IJ
H K
=
7
3
+ −
2
2
=
14 9
+
4 4
=
23
4
2
2
2
Add the square of
1
2
(–3) to each side
Simplify
Take the square root
3
23
=±
2
2
x=
3
23 Simplify
±
2
2
These are the exact roots of the equation. Using substitution
to check solutions that have fractions and square roots is
cumbersome. If you have a graphing calculator, you can use
it to do a very efficient check. For instance:
3
23
+
≈ 3.9
2
2
Module 2
3
23
−
≈ −0.9
2
2
Principles of Mathematics 11
Section 2, Lesson 3
125
2
On the graph of 2x – 6x – 7 = 0, 3.9 and –0.9 are the
approximate x-intercepts.
To derive a general formula for solving quadratic equations, you
can complete the square for the general quadratic equation
2
ax + bx + c = 0, a ≠ 0.
Standard form
ax 2 + bx + c = 0
b
c
x2 + x + = 0
Divide by a
a
a
b
− c Isolate terms with x on one side
x2 + x =
a
a
FG IJ
H K
FG x + b IJ
H 2a K
FG x + b IJ
H 2a K
b
b
x + x+
a
2a
2
x+
2
2
FG IJ Complete the square by adding
H K b ⋅ g to each side
−c
b
=
+
a
2a
2
1
2
b 2
a
b2
c
=
−
2
4a
a
Simplify
b2 − 4ac
4a 2
Simplify
2
=
b
b2 − 4ac Take square root of each side
=±
2a
4a 2
x=−
x=
b
b2 − 4ac Subtract
±
2a
2a
−b ± b2 − 4ac
2a
b
2a
from each side
Simplify
When solving an equation using the quadratic formula, it is
important to remember that − b ± b2 − 4 ac is all divided by 2a.
Think of the results of the quadratic formula as two fractions.
x=−
b
b2 − 4ac
x=−
+
2a
2a
x=−
b
2a
b
b2 − 4ac
−
2a
2a
z
b2 − 4ac
2a
z
line of symmetry
Module 2
126
Section 2, Lesson 3
Principles of Mathematics 11
b
, the first term of the fractional expressions, is
2a
the x-coordinate of the vertex (Module 2, Section 1). So the Axis of
b
Symmetry is x = –
. The two fractional solutions are values
2a
on the x-axis, on opposite sides of the Axis of Symmetry,
Recall that –
b 2 – 4ac units away.
2a
You can use the quadratic formula to solve any quadratic
equation. When using the quadratic formula, you must set up
2
the equation in the general form of ax + bx + c = 0, a ≠ 0, so
that a, b, and c can be correctly identified.
Example 1
2
Using the quadratic formula, solve 2x + 5x – 3 = 0.
a = 2, b = 5, c = –3
Solution
x=
x=
−b ± b 2 − 4ac
2a
−5 ± 52 − 4 (2 )(−3 )
2 (2 )
−5 ± 25 + 24
4
−5 ± 49
=
4
−5 ± 7
=
4
=
−5 + 7
−5 − 7
or x =
4
4
1
x=
or x = −3
2
x=
Note: This equation could have been solved by factoring.
Module 2
Principles of Mathematics 11
Section 2, Lesson 3
127
Example 2
2
Solve x + 3x – 9 = 0.
Solution
a = 1, b = 3, c = –9
x=
=
− b ± b2 − 4 ac
2a
a fa f
− 3 ± 3 2 − 4 1 −9
af
21
=
−3 ± 9 + 36
2
=
−3 ± 45
2
=
−3 ± 3 5
2
x=
−3 + 3 5
2
−3 − 3 5
2
or x =
Note: This equation does not factor. Therefore, either the
quadratic formula or the method of completing the square would
have to be used.
There are equations that at first glance do not seem to be
quadratic. But a closer look reveals that they actually have a
quadratic form or pattern.
Example 3
4
2
Solve: 2x – 5x + 2 = 0
Solution
Notice that the equation can be written in quadratic form:
2 2
2 1
2(x ) – 5(x ) + 2 = 0
2
Temporarily replace x by another letter p to obtain a quadratic
equation:
2
2p – 5p + 2 = 0
Module 2
128
Section 2, Lesson 3
Factoring, you get
a2 p − 1fa p − 2f = 0
p=
1
or p = 2
2
2
Replacing p by x you get
x2 =
1
or x 2 = 2
2
x=±
1
2
x=±
2
(rationalized)
2
or x = ± 2
2
2
,−
, 2 , and − 2
2
2
The answers are:
Answers should be checked in the original equation.
Check:
x= 2
e j − 5e 2 j
2 2
4
8 − 10 + 2 = 0
2
+2=0
Yes
x = − 2 also works because powers involved are even
x=
2
2
F 2I
2G
H 2 JK
4
F 2I
− 5G
H 2 JK
2
+2=0
2 5
− +2=0
4 2
1 5 4
− + = 0 Yes, it works
2 2 2
− 2
x=
also works because powers involved are even
2
Module 2
Principles of Mathematics 11
Principles of Mathematics 11
Section 2, Lesson 3
129
Example 4
FG 2 IJ
H x − 3K
FG 2 IJ + 3 = 0
H x − 3K
2
Solve: 5
+8
Solution
2
, you can rewrite
This displays a quadratic pattern. If p =
x−3
the equation as
5 p2 + 8 p + 3 = 0
b5 p + 3gb p + 1g = 0
p=
−3
or p = −1
5
Replace p by
a
2
x−3
2
−3
2
=
or
= −1
x−3
5
x−3
f
a
f
−3 x − 3 = 10
−1 x − 3 = 2
−3 x + 9 = 10
−x + 3 = 2
−3 x = 1
x=−
− x = −1
1
3
x=1
Check to see that roots work in the original equation.
As always we check our solutions in the original equation. This will be
left as an exercise for you.
Module 2
130
Section 2, Lesson 3
Principles of Mathematics 11
Self-Marking Activity
1. For each quadratic equation below, state the values of a, b,
and c.
a) x 2 − 2 x − 5 = 0
b) 3 x 2 − 2 x + 5 = 0
c) 5 x 2 − 3 x = 8
d) 2 (x 2 − 2 x )− 1 = 0
e) 5 x 2 = 9 x
f) 4 − 2 x 2 = 9 x
2. Solve these equations using the quadratic formula. Be sure
to state the formula before substituting values into it.
a) x 2 + 2 x − 15 = 0
b) 2 w2 − 3w + 1 = 0
c) 7 w2 − 3w = 0
d) 1 = 5 x 2
e) x 2 − 0.1x − 0.06 = 0
f) − x 2 − 7 x − 1 = 0
3. Use the quadratic formula to find the roots of each equation
below.
a) 3 x 2 − 6 x − 5 = 0
b) 2 x 2 − 4 x − 1 = 0
c) 9 x 2 − 8 x − 7 = 0
d) 2 x 2 − x − 3 = 0
4. Find the zeros of the function f defined by
a) f : x → 5 x 2 − x − 3
af
b) f x = 2 x 2 + 6 x − 1
2
5. Find the roots of the quadratic equation 3x – 5x – 1 = 0 to
one decimal place.
2
6. Find the roots of the quadratic equation 6x + 5x – 6 = 0
a) using the quadratic formula
b) by factoring
Which method do you prefer and why?
2
7. Find the zeros of the quadratic function f(x) = x + 8x + 15 by
a) factoring
b) using the quadratic formula
c) graphing
Module 2
Principles of Mathematics 11
Section 2, Lesson 3
131
8. If possible, for each of the following, write an equivalent
2
equation in the pattern a • ‡ + b‡ + c = 0 for some real
numbers a, b, and c, and for some expression ‡ involving x.
a) 5 (x + 3 ) + 3 (x + 3 ) + 8 = 0
2
b) 2x – 3 + 3 2 x − 3 = 3
c)
(x
2
)
2
− 1 − 7 = 6x2 − 6
1
d) 6 x 2 + x = 4
e) x 6 + 3 x 2 + 2 = 0
9. Find the real number solution set for
a)
(x
2
+ 2x
) − 2 (x
2
2
)
+ 2x − 3 = 0
b) 2 x −2 + 4 x − 1 + 3 = 0
c) x = 6 x − 2
2
 x −1
 x −1
d) 
− 3

+2 =0
 x 
 x 
Check your answers in the Module 2 Answer Key.
Module 2
132
Notes
Module 2
Section 2, Lesson 3
Principles of Mathematics 11
Principles of Mathematics 11
Section 2, Lesson 4
133
Lesson 4
The Nature of Roots
Outcome
When you complete this lesson, you will be able to
• use the discriminant of a quadratic equation to describe the
nature of its roots
Overview
You have found the roots of a quadratic equation by graphing,
by factoring, by completing the square and by the quadratic
formula. You have obtained a variety of answers some of which
were real numbers and others being imaginary numbers.
You have found that any quadratic equation of the form
2
ax + bx + c = 0 can be solved using the quadratic formula
x=
−b ± b2 − 4ac
2a
2
The expression b – 4ac is called the discriminant and it enables
you to observe the nature of the roots without actually finding
the roots.
In this lesson, you will consider the value of the discriminant of
a given quadratic equation, the roots of the quadratic equation,
a description of the roots, and the graph of the corresponding
quadratic function.
Example 1
2
When b – 4ac > 0 is a number that is a perfect square.
Solution
2
Equation: x – 6x + 5 = 0
Value of Discriminant:
2
b – 4ac
a = 1, b = –6, c = 5
2
(–6) – 4(1)(5) = 36 – 20 = 16
2
Note that the value of b – 4ac is a
perfect square.
Module 2
134
Section 2, Lesson 4
Principles of Mathematics 11
Roots of Quadratic:
r1 =
b g
− −6 + 16
2
b g
− −6 − 16
=
+6 + 4
=5
2
6−4
=1
2
2
Description of Roots:
r1 and r2 are referred to as two real
rational roots
r2 =
=
2
Graph of the Quadratic Function: y = x – 6x + 5
y
z
ze r o
z
z
1
2
3
4
5
x
ze r o
z
Description of Zeros: The zeros are referred to as two real
rational zeros at 1 and 5.
Example 2
2
When b – 4ac > 0 and its value is a non-perfect square.
Solution
2
Equation: x – 6x + 7 = 0
Value of Discriminant:
2
b – 4ac
a = 1, b = –6, c = 7
2
(–6) – 4(1)(7) = 36 – 28 = 8
2
Note that the value of b – 4ac is a
non-perfect square.
Module 2
Principles of Mathematics 11
Section 2, Lesson 4
135
Roots of Quadratic:
r1 =
− (−6 )+ 36 − 28 6 + 8 6 + 2 2
=
=
= 3+ 2
2 (1)
2
2
r2 =
− (−6 )− 36 − 28 6 − 8 6 − 2 2
=
=
= 3− 2
2 (1)
2
2
r1 and r2 are referred to as two real
irrational roots.
Description of Roots:
2
Graph of the Quadratic Function: y = x – 6x + 7
y
e3 +
e3 −
j
2, 0
z
j
2, 0
x
b3, − 2g
Description of Zeros: Two real irrational zeros at
3 + 2 and 3 − 2
Example 3
2
When b – 4ac = 0.
Solution
2
Equation: x – 6x + 9 = 0
Value of Discriminant:
2
b – 4ac
a = 1, b = –6, c = 9
2
(–6) – 4(1)(9) = 36 – 36 = 0
2
Note that the value of b – 4ac is a
perfect square.
Module 2
136
Section 2, Lesson 4
Roots of Quadratic:
r1 =
r2 =
b g
− −6 + 36 − 36
b g
bg
21
=
6
=3
2
− −6 − 36 − 36 6
= =3
21
2
bg
Description of Roots: r1 and r2 are the same real root.
r1 and r2 are referred to as double roots
2
Graph of the Quadratic Function: y = x – 6x + 9
y
z
z
(3 , 0 )
x
Description of Zeros: It has one zero (the graph is tangent to
the x-axis).
Example 4
2
When b – 4ac < 0.
Solution
2
Equation: x – 6x + 14 = 0
Value of Discriminant:
Module 2
2
b – 4ac
a = 1, b = –6, c = 14
2
(–6) – 4(1)(14) = 36 – 56 = –20
Principles of Mathematics 11
Principles of Mathematics 11
Section 2, Lesson 4
137
Roots of Quadratic:
r1 =
r2 =
− (−6 ) + 36 − 56
2
− (−6 ) − 36 − 56
2
=
+6 + −20
2
=
6 − −20
2
Description of Roots: r1 and r2 are not real roots.
2
Graph of the Quadratic Function: y = x – 6x + 14
y
z
(3 , 5 )
x
Description of Zeros: There are no real zeros since the graph
does not cross or touch the x-axis.
Module 2
138
Section 2, Lesson 4
Principles of Mathematics 11
Example 5
2
For what values of m will x + 8x – m = 0 have real and unequal
roots.
Solution
2
For real and unequal roots, b – 4ac > 0.
a = 1, b = 8, c = –m
b gb g
8 2 − 4 1 −m > 0
64 + 4m > 0
4m > −64
m > −16
The value m must be greater than –16 for real and unequal roots.
You were not asked for an equation but an example that satisfies
2
the restriction, which would be x + 8x – 14 = 0.
Example 6
Write a quadratic equation that has two different real solutions.
Solution
A quadratic equation in standard form has two real number
2
solutions when b – 4ac > 0.
Choose any values of a and c, then find a value of b that satisfies
this inequality.
If a = 3, c = 11
b2 − 4ac > 0
b2 − 4 (3 )(11 ) > 0
b2 − 132 > 0
b2 > 132
2
One possible value of b would be 12 because 12 > 132.
Either b > √132 or b < –√132. One possible equation is
2
3x + 12x + 11 = 0.
Module 2
Principles of Mathematics 11
Section 2, Lesson 4
139
Self-Marking Activity
1. State whether each statement is true or false. If the statement
is false, rewrite it so it is true.
2
a) The discriminant of 2x + 5x + 6 is 23.
2
2
2
2
b) For each quadratic equation ax + bx + c = 0 if b – 4ac > 0,
then the equation has two real roots.
c) For each quadratic equation ax + bx + c = 0 if b – 4ac < 0,
then the equation has no real number solutions.
d) −6 + 32 = −3 + 8 2
10
5
2
e) The equation x + 2x + 1 = 0 has two distinct real roots.
2. If the discriminant of a quadratic equation has the given
value, state the characteristics of the roots.
a) –15
b) 25
c) –9
d) 0
e) 50
2
3. How many times would the graph of y = ax + bx + c (with a, b,
and c as real numbers) intersect the x-axis if the value of the
discriminant is
a) negative?
b) zero?
c) positive?
4. Determine the nature of the roots by calculating the
discriminant for each equation.
2
b) a + 2a + 7 = 0
2
d) 2x + x = 5
a) x – 8x + 16 = 0
c) p – 16 = 0
2
2
5. Determine the characteristics of the roots of the following
equations:
x2
+ 4x + 4 = 0
2
c) 2 x 2 − 3 = 4 x
a)
d
i
x −1
− x2 − 3 = 0
2
d) 6x 2 − x + 2 = 0
b)
e) 4 x 2 − 12 x + 9 = 0
2
6. Given 3x – mx + 3 = 0, for what values of m would the roots
not be real?
Module 2
140
Section 2, Lesson 4
Principles of Mathematics 11
7. Find value(s) of k so that each equation has one real root.
2
2
a) kx – 6x + 2 = 0
b) x + (k – 8)x + 9 = 0
8. For what values of k will the equation
2
2
2x + 4x + (2 – k – k ) = 0 have exactly one root?
2
9. State the nature of the roots of ax + bx + c = 0 given the
2
graph of y = ax + bx + c:
a)
b)
y
c)
y
5
y
5
5
x
5
5
5
5
5
x
10. Sketch the graph of a quadratic function with no real zeros.
11. Sketch the graph of a quadratic function whose
discriminant is positive and whose vertex is in the second
quadrant.
12. Find the value of k for each of the following situations:
2
a)
3x – 2x + k = 0 has two real roots
b)
x + kx + (k + 2) = 0 has a double root
c)
kx + 8x = 4 has no real roots
2
2
Check your answers in the Module 2 Answer Key.
Module 2
x
Principles of Mathematics 11
Section 2, Lesson 5
141
Lesson 5
Radical Equations
Outcome
When you complete this lesson, you will be able to
• solve radical equations
Overview
A radical equation is one that contains radicals or rational
exponents. To solve radical equations you try to eliminate the
radicals and obtain a linear or quadratic equation, which you
have already had experience in solving.
The following property plays a key role in the simplification
process.
n
n
If a = b, then a = b .
It means that both sides of an equation can be raised to the
same power as shown in the following examples.
Remember:
Square roots and all other even roots are only defined if the
radicand (the expression under the radical sign) is nonnegative.
For example:
4
x is defined only for x ≥ 0
x − 5 is defined only for x − 5 ≥ 0 or x ≥ 5
Example 1
Solve:
x =4
Solution
x =4
e xj
2
= 42
x = 16
Square each side to remove the square root
sign.
Check: 16 = 4
Remember that
1
1
x = x 2 and if you square x 2 , you get x.
Module 2
142
Section 2, Lesson 5
Principles of Mathematics 11
Example 2
Solve:
3
x −4 = 0
Solution:
Before raising both sides of an equation to the nth power, you must
isolate the radical expression on one side of the equation.
Rearrange
x =4
3
e xj
3
3
Cube each side remembering that
= 43
x = 64
3
1
e j
1
x = x 3 and x 3
3
1 ⋅3
= x 3 = x1
Check:
3
64 − 4 = 0
4−4 =0
The following is an equation containing an exponent.
Example 3
3
Solve: x 2 = 8
Solution
Raise both sides to the power
3
x2 = 8
e j
3
x2
2
3
x=
e 8j
3
x = 22
x=4
Check:
3
42 = 8
e 4j
3
Simplify
2
= 83
=8
23 = 8
Module 2
2
2
3
Principles of Mathematics 11
Section 2, Lesson 5
143
Raising both sides of an equation to the nth power may
introduce extraneous or false solutions. So when you use the
procedure, it is critical that you check each solution in the
original equation. For instance, squaring both sides of the
equation.
x = −1 produces x = 1
The result is extraneous. The original equation has no real
number solution.
Example 4
Solve: x + 2 x − 3 = 6
Solution
Isolate the radical
2 x −3 = 6− x
e2
j = b6 − x g
4b x − 3g = 36 − 12x + x
x −3
2
Square both sides
2
2
4 x − 12 = 36 − 12x + x 2
x 2 − 16x + 48 = 0
bx − 4gbx − 12g = 0
Rearrange
x = 4 or x = 12
Check:
4+2 4−3 = 6
12 + 2 12 − 3 = 6
4 + 2 ⋅1 = 6
12 + 2 ⋅ 3 = 6
6=6
18 ≠ 6
Answer: x = 4
e
j = b6 − x g
x − 3 = −b6 − x g.
When you squared both sides of the equation 2 x − 3
2
2
you introduced a second equation, that being 2
Thus extraneous roots may be introduced and a check of the
roots is always necessary.
Module 2
144
Section 2, Lesson 5
Principles of Mathematics 11
If a radical equation has more than one term with a variable in the
radicand, you may have to isolate and raise it to a power more than
once as shown in the following example.
Example 5
Solve:
2x + 3 − x + 1 = 1
Solution
Rearrange
2x + 3 = 1 + x + 1
2
( 2x + 3 ) = (1 +
= (1 +
2
)
x + 1 )• (1 +
x +1
Square each side
x +1
)
= 1 + x + 1 + x + 1 + (x + 1 )
2x + 3 = 1 + 2 x + 1 + x + 1
x +1 = 2 x +1
2
(x + 1)
(
= 2 x +1
)
2
x 2 + 2x + 1 = 4 (x + 1 )
x 2 + 2x + 1 = 4 x + 4
x 2 − 2x − 3 = 0
(x − 3 )(x + 1) = 0
x = 3 or − 1
Check:
bg
2 3 + 3 − 3 +1 = 1
b g
3−2 =1
2 − 1 + 3 − −1 + 1 = 1
1− 0 =1
The solution set is {3, –1}.
Module 2
Simplify
Rearrange
Square again
Principles of Mathematics 11
Section 2, Lesson 5
145
Self-Marking Activity
1. State whether each statement is true or false. If the
statement is false, rewrite it so that it is true.
a) For each real number x, x is a real number if and only if
x ≥ 0.
b) For each real x, x − 3 is a real number if and only if
x ≥ 0.
c) For each real x,
x 2 = x.
x 2 = −3 is
d) The real number solution set of
2
(
)
2
{ 3, − 3 }.
e) The equation (x – 1) =
2x − 1 is equivalent to the
sentence (x – 1) = 2x − 1 or (x − 1 ) = − 2 x − 1.
2
f) For each real number x, if x = 16, then x =
g) Squaring both sides of
16.
2
x + 2 = 3x yields x + 4 = 9x .
2. Simplify each of the following expressions.
a)
b)
c)
( 2x − 1 )
(5 + x )
(2 + x − 5 )
2
2
2
3. Find the real number solution set for each of the following
equations. Check your solutions.
a) x + 2 = 2x + 7
b) x = 2 − 2x − 5
c)
d)
x2 − 3 +1 = 0
e) x = 3x − 2 + 2
f)
x 2 + 6x = 2
g)
h)
2x + 3 − x + 1 = 1
3x + 2 = 3 x − 2
1− x + x = x +1
Check your answers in the Module 2 Answer Key.
Module 2
146
Notes
Module 2
Section 2, Lesson 5
Principles of Mathematics 11
Principles of Mathematics 11
Section 2, Lesson 6
147
Lesson 6
Rational and Absolute Value Equations
Outcomes
When you complete this lesson, you will be able to
• solve rational equations
• solve absolute value equations
Overview
You have solved linear equations, quadratic equations, radical
equations, and non-rational equations. You can solve rational
equations by using many of the techniques and properties you
have used to solve other equations.
An equation involving rational expressions is a rational
equation.
The algebraic method of solution is based on the least common
multiple. The least common multiple (LCM) of two or more
polynomials is the simplest polynomial that is a multiple of
each of the original polynomials.
Example 1
Find the least common multiple of the following polynomials.
You factor each polynomial and use each factor the greatest
number of times it occurs in each polynomial.
Polynomials
Least Common Multiple
a) x and x – 4
x(x – 4) (each factor occurs
once)
2
2xx = 2x (x occurs twice in
first term)
2
2
(x + 3)(x – 3)
b) x and 2x
c) x – 9 and x – 3
(x – 3)(x + 3) and (x – 3)
2
2
d) (x – 4x + 4) and x – 4
(x – 2)(x – 2) and (x – 2)(x + 2)
(x – 2)(x – 2)(x + 2)
The least common denominator (LCD) of two or more
fractions is the least common multiple of the denominator of the
fractions.
Module 2
148
Section 2, Lesson 6
To solve a rational equation:
• if the denominator can be factored, factor it first.
• look in the denominator for restrictions: values of the
variable which are forbidden because they make the
denominator equal zero. (Remember, you can’t divide by
zero.)
• multiply each term on both sides of the equation by the least
common denominator of the terms.
• simplify and solve the resulting equation. Then check!
Example 2
2 1 3
+ =
x 2 x
Solution
Denominators are fully factored, so choose lowest common
denominator to be 2x. The restriction is x ≠ 0.
2x ⋅
2
1
3
+ 2x ⋅ = 2x ⋅
x
2
x
4+x =6
Multiply both sides by 2x
Simplify
x=2
Check in original equation.
2 1 3
+ =
2 2 2
3 3
=
2 2
Check is okay.
You can also use your graphing calculator if you have one.
Press Y=
Enter
y1 = 2 x + 1 2
y2 = 3 x
Press GRAPH
Press 2nd
Module 2
CALC 5: Intersect
Principles of Mathematics 11
Principles of Mathematics 11
Section 2, Lesson 6
149
1. Adjust your WINDOW so that Xmin = –2, Xmax = 3, Ymin = –2, Ymax = 3
2. Press ENTER to confirm the 1st Curve.
3. Press ENTER to confirm the 2nd Curve.
4. “Arrow” left or right to place the cursor as close as possible to the
point of intersection of the two curves.
5. Press ENTER to confirm your Guess.
6. Record the x value at the bottom of the screen. This is your
solution, 2.
Point of intersection: (2, 1.5)
Example 2
Solve:
3x
6
= 2+
x −2
x−2
Solution:
Denominators do not need factoring.
Restricted value in denominator: x ≠ 2.
Lowest common denominator: x – 2
bx − 2gFGH x3−x2 IJK = 2bx − 2g + bx − 2gFGH x 6− 2 IJK
Multiply by x – 2
3 x = 2x − 4 + 6
x=2
When you check your answer, 2 is not a permissible value as the
denominator becomes 0.
∴ There is no solution to the equation.
Example 3
Solve:
3x
12
= 2
+2
x +1 x −1
Solution
One denominator needs to be factored to (x + 1)(x – 1) first.
Restricted value: x ≠ –1, 1
Lowest common denominator: (x – 1)(x + 1)
Module 2
150
Section 2, Lesson 6
Principles of Mathematics 11
bx − 1gbx + 1gFGH x3+x1IJK = bx − 1gbx + 1gFGH bx − 112gbx + 1gIJK + bx − 1gbx + 1gb2g
b g
b gb g
Simplify
− 3x = 12 + 2d x − 1i
3x x − 1 = 12 + 2 x − 1 x + 1 Multiply by LCD
3x 2
2
3x 2 − 3x = 12 + 2x 2 − 2
x 2 − 3x − 10 = 0
bx + 2gbx − 5g = 0
Factor
x = −2 or x = 5
Check:
x = −2
b g
3 −2
12
=
+2
2
−2 + 1
−2 − 1
b g
12
−6
=
+2
−1 4 − 1
6= 4+2
6=6
x =5
bg=
35
12
+2
5 +1 5 −1
15 12
=
+2
6 24
5 1
= +2
2 2
5 5
=
2 2
2
∴ x = –2 or x = 5
The last type of equation to be examined in this lesson is the
absolute value equation.
The absolute value of a number can be thought of as its distance
from 0 on a number line. Any positive number and any negative
number are each a positive distance from 0. The absolute value is
concerned with distance rather than direction.
Definition
The absolute value of any real number a, written as |a|, is defined
as
a = a if a ≥ 0
a = −a if a < 0
Example:
− 2 = − (−2 ) = +2
Module 2
Principles of Mathematics 11
Section 2, Lesson 6
151
There are two values that have a distance of 2 from 0. They are
2 and –2.
So, if |x| = 2, then x = 2 or x = –2.
distance is 2
distance is 2
2
0
2
x
Example 4
Solve and graph: |x – 3| = 2
Solution
bx − 3g = 2
or
x −3 = 2
bx − 3g = −2
x − 3 = −2
x =1
x =5
Check:
x =1
x =5
or
5−3 = 2
1−3 = 2
2=2
−2 =2
2=2
The solutions of |x – 3| = 2 are translations of the solutions of
|x| = 2, 3 units to the right.
z
z
2
0
1
2
5
x
or points which are a distance of 2 from 3
distance of 2
z
0
1
z
3
5
x
Module 2
152
Section 2, Lesson 6
Example 5
2
Solve: |x – 3x| = –4x + 6
Solution
Since the variable expression in the absolute value signs can be
positive or negative, you must solve two equations.
x 2 − 3 x = −4 x + 6
2
x +x–6 =0
d
i
− x 2 − 3 x = −4 x + 6
− x 2 + 3 x = −4 x + 6
(x + 3 )(x – 2 ) = 0
x 2 − 7x + 6 = 0
x = –3 or + 2
bx − 6gbx − 1g = 0
x = 6 or 1
Check:
x = −3
b−3g − 3b−3g = b−4gb−3g + 6
2
9 + 9 = 12 + 6
18 = 18
x=2
2
bg
bg
2 − 3 2 = −4 2 + 6
− 2 = −8 + 6
− 2 ≠ −2
x=6
2
bg
bg
6 − 3 6 = −4 6 + 6
18 ≠ −18
x =1
2
bg
bg
1 − 3 1 = −4 1 + 6
−2 = 2
2=2
x = –3 or x = 1.
Module 2
Principles of Mathematics 11
Principles of Mathematics 11
Section 2, Lesson 6
153
Example 6
A certain number is 11 greater than its reciprocal. Find the
number.
Solution
1
Let x be the number and its reciprocal.
x
1
x – = 11
x
1
x (x ) – x   = x (11 )
x
2
x – 1 = 11x
x 2 – 11x – 1 = 0
x=
11 ±
(–11 ) – 4 (1 )(–1 ) 11 ±
=
2 (1 )
2
So there are two solutions,
121 + 4 11 ± 125 11 ± 5 5
=
=
2
2
2
11 + 5 5
11 – 5 5
(11.09 approx.) and
2
2
(–.09 approx.).
Checking these two solutions will be left as an exercise for you.
Now, for some practice, please complete the following Assignment.
Be sure to check your answers when you are finished.
Module 2
154
Section 2, Lesson 6
Principles of Mathematics 11
Self-marking Activity
1. State whether each statement is true or false. If the statement
is false, rewrite it so it is true.
gFGH x4−x2 IJK = 4 x.
F 6 + x IJ = 6 + x.
If x ≠ 3, then 2b x − 3gG
H x − 3 2K
b
a) If x ≠ 2, then x − 2
b)
c) If x ≠ 0, then the equations x −
1
= 0 and x 2 − 1 = 0
x
are equivalent.
x −1 x − 2
=
is equivalent to
x +3 x +4
x −1 x + 4 = x + 3 x − 2 .
d) If x ≠ −3 and x ≠ −4 , then
b gb
g b
gb
g
2. Solve each of the following rational equations. Check your
solutions.
−2
x −3
−1
c) x − 4 =
x
2x
1
3x + 9
e)
+
+ 2
=0
x − 3 2x + 3 2x − 3x − 9
a) x =
2x − 9 x
5
+ =
x −7 2 x −7
2x + 1
3x 2
d)
−2=
3x + 1
3x + 1
2
x + 12
7x
f)
=
x −3
x −3
b)
3. Solve the following absolute value equations.
a)
3x = 12
b) 2x − 1 = 17
c)
5 x + 2 = −3
d)
x 2 + 4 x − 12 = 0
e)
x 3
1
−
=
2 4
12
f)
x − 5 = 3x + 7
4. Two positive integers differ by 4 and their reciprocals differ
by
1
. Find the integers.
15
5. The sum of a number and its reciprocal is 4 and the number is
greater than 2. Find the number.
Module 2
Principles of Mathematics 11
Section 2, Lesson 6
155
Check your answers in the Module 2 Answer Key.
Review Module 2 before attempting the review questions
beginning on the next page. These questions should help you
consolidate your knowledge as you prepare for the Module 2
Section Assignment 2.2.
Module 2
156
Notes
Module 2
Section 2, Lesson 6
Principles of Mathematics 11
Principles of Mathematics 11
Section 2, Review
157
Review
1. Solve the following quadratic equations by factoring:
2
a) 3x = 6x
2
b) 2x – 5x = 18
3
c) 32x – 8x = 0
2
d) –6x – 5x + 4 = 0
4
e) x – 1 = 0
f) 3x(x – 2) – x(x + 1) + 5 = 0
2. Find the roots of the equation by graphing:
2
a) x – 2x – 8 = 0
2
b) x – 8x = –15
2
c) x – 4 = 0
3. Solve by completing the square:
2
a) x + 6x – 2 = 0
2
b) 2x – 3x – 8 = 0
4. Solve by using the quadratic formula:
2
a) –1x – 7x – 1 = 0
2
b) 2x + 4x + 1 = 0
5. Without solving the equation, determine the nature of the
roots of:
a) 4 x 2 − 12x + 9 = 0
d
i
b) 2 x 2 − 3 = 4 x
x −1
− x2 − 3 = 0
2
d) 6x 2 − x + 2 = 0
c)
6. Find the values of k so that the equation,
2
2
3x + 6x + (3 – k – k ) = 0, has real and equal roots.
2
7. For what values of k will the equation 3x + kx + 12 = 0 have
two different real roots?
Module 2
158
Section 2, Review
Principles of Mathematics 11
2
8. a) For what values of k will the equation 2x + 5x = k have
no real roots?
b) What does this tell you about the graph of the function
for those values of k?
2
9. What is the discriminant for 7x + 5 = 10x?
10. Solve the following equations:
a)
1
1
2
+
= 2
x + 3x + 2 x − 1 x − 1
2
b)
2x + 5 − 2 2 x = 1
c)
3x − 2 − 2x − 3 = 1
d)
5
2
6
+ =
x + 3 x x +1
e)
dx
2
+ 2x
i + dx
2
2
i
+ 2x = 12
f)
x 2 − 2x − 8 = 0
g)
x + 5 = 2x − 3
11. The product of two consecutive positive integers is 72. Find
the integers.
12. If (6 – x),(13 – x), and (14 – x) are the length of the sides of a
right triangle, find the value of x, if (14 – x) is the
hypotenuse.
13. Two positive numbers differ by 1. The sum of their
reciprocals is 2 2/3. Find the numbers.
Check your answers in the Module 2 Answer Key.
Now do the Section Assignment which follows this section.
Module 2
Principles of Mathematics 11
Section Assignment 2.2
161
PRINCIPLES OF MATHEMATICS 11
Section Assignment 2.2
Module 2
162
Module 2
Section Assignment 2.2
Principles of Mathematics 11
Principles of Mathematics 11
Section Assignment 2.2
163
Section Assignment 2.2
Algebra
Total Value: 50 marks
(Mark values in brackets)
Answer the following questions in the space provided. To receive
full marks, solutions should be legible, well-presented, and
mathematically correct.
2
(1)
1. Solve by factoring: 3x – x = 10
(2)
2. Find the roots of the following questions by using the
quadratic formula.
2
a) 2x + x – 12 = 0
Version 04
Module 2
164
Section Assignment 2.2
Principles of Mathematics 11
(3)
2
b) 4x – 2x = 7
(3)
2
3. Solve by completing the square: 2x + 6x + 1 = 0
Module 2
Version 04
Principles of Mathematics 11
(3)
Section Assignment 2.2
165
4. Solve by any method.
a) 6x2 – 5x – 4 = 1
(3)
b) 3x2 + 3 = 7x
(2)
c) x3 – 3x + 2x2 = 0
Version 04
Module 2
166
Section Assignment 2.2
Principles of Mathematics 11
5. Solve for x:
(4)
a) x − x + 1 − 5 = 0
b)
Module 2
dx
2
+ 2x
i + dx
2
2
i
(5)
+ 2x = 12
Version 04
Principles of Mathematics 11
(5)
(4)
Section Assignment 2.2
b
4 x+3
x−2
−3=
x +3
x −2
c)
167
g
2
d) |x + 6x| = 7
Version 04
Module 2
168
Section Assignment 2.2
Principles of Mathematics 11
6. a) Give the discriminant for the quadratic equation,
(2)
2
4x + 3x + 1 = 0
b) State the nature of the roots.
2
7. If the discriminant of the quadratic equation, 3x – x + c = 0,
is 25, find
(2)
a) the value of c
b) without solving for the roots, state the nature of the roots.
Module 2
Version 04
(1)
Principles of Mathematics 11
Section Assignment 2.2
169
(3)
8. Find the values of k so that the equation 2x2 + 6x + k = 2 has two
different real roots.
(3)
9. The sum of a number and 15 times its reciprocal is equal to
8. Find the numbers.
Version 04
Module 2
170
Section Assignment 2.2
Principles of Mathematics 11
10. How wide a uniform border should be left on a plot of land
(4)
2
7 m by 11 m to have a rectangular area of 45 m available
for flowers.
(Total: 50)
Module 2
Version 04