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Unit 8
Quadratic and Exponential Functions
Introduction
In Unit 7 we learned how to solve polynomial equations. Most of these equations
could be called quadratic equations. In a quadratic equation the highest exponent is
a square term. An example of a quadratic equation would be y = x2 + 6x + 9.
The name “quadratic” and the term “square” come from geometry. A square is a special
kind of rectangle. To find the area of a rectangle we multiply the length times the width.
For a square, the length and width are equal so we actually multiply the side times itself.
The examples below illustrate this.
4 cm
The area for this square is 4 . 4 = 42
4cm
x
The area for this square is x . x = x2
x
This is why the exponent of ‘2’ is called a “square”.
We call the equations “quadratic” because they can represent the area of a four sided
figure. The prefix “quad” means four!
The area of the four sided figure below can be found using binomial multiplication.
2x + 1
(2x + 1)(x + 3) = 2x2 + 6x + x + 3
x+3
2x2 + 7x + 3
This expression is the area of a four sided figure.
That is why we call it a quadratic.
Unit 8
Vocabulary and Concepts
Exponential Equation An equation where the variable ‘x’ is used as a power.
Line of Symmetry A line that divides a parabola into two mirror images.
Parabola
The shape of a quadratic equation’s graph (a U-shape).
Quadratic Equation An equation of order 2.
Square Root
The square root of a number is a factor that times itself gives us
the original number.
Vertex
The highest or lowest point on a parabola’s graph.
Key Concepts for Exponents and Polynomials
Reversing a Square
The reverse of a square operation is a square root.
The Square Root Principle A method for solving quadratic equations based on
isolating the square and taking a square root on both
sides of the equation. This method will give us a
positive and negative answer.
Completing the Square
The Quadratic Formula
A method for solving quadratic equations based on
finding a binomial square so we can use the square
root principle.
A formula to solve a quadratic equation based on the
constants and coefficients ax2 + bx + c = 0.
x=
Exponential Growth
 b  b 2  4ac
2a
A situation or pattern that is formed by multiplying by
the same value to go from the previous term to the next term.
Unit 8 Section 1
Objective

The student will simplify and estimate square roots.
Every operation in mathematics has a reverse. Addition is reversed by subtraction,
multiplication is reversed by division, evaluating is reversed by solving. The operation
of squaring a number must have a reverse. The problems below will help us understand
how the reverse operation works.
If 42 = 16 then the reverse operation with 16 must give us 4.
If 52 = 25 then the reverse operation with 25 must give us 5.
If 102 = 100 then the reverse operation with 100 must give us 10.
This reverse operation is called a square root. The symbol
, is used for the reverse
operation. This reverse operation is called a “square root”. We can interpret this symbol
as asking us to find the reverse of a square. We can put this into words as the question
shown below.
Find the number that times itself gives us the value in the symbol.
The examples below show us how the question works.
9
36
144
asks us “What number times itself gives us 9?” The answer is 3.
asks us “What number times itself gives us 36?” The answer is 6.
asks us “What number times itself gives us 144?” The answer is 12.
This operation can also be done with variables and expressions.
asks us “What number times itself gives us x2?” The answer is x.
Because x2 is equal to x . x , so ‘x’ is the number that times itself equals x2.
x2
( x  2) 2 asks us “What expression times itself gives us (x+2)2?”
The answer is x+2. Because (x+2)2 is equal to (x+2)(x+2), so (x+2) is the
expression that times itself equals (x+2)2.
While there are some values that will work out evenly, there are many values that
do not work out evenly. The examples below show us the difference in the two types
of problems.
Example A
9
Example B
43
Example A has an answer of 3 while example B must be between 36 and 49
which means the answer is between 6 and 7. We can find estimates for the square
roots that don’t work out evenly. The examples below show us the method we can
use to estimate square roots with reasonable accuracy.
Example A
Estimate the
18
2
7
16
25
4
18 to the nearest tenth.
The 18 does not work out to be a whole number. The
square roots that do have whole number answers, and
are on either side of 18 on the real number line, are 16
and 25. The difference between 18 and 16 is 2. The
difference between 18 and 25 is 7.
5
Since the square root of 16 is 4 and the square root of 25 is 5 the answer for the square
root of 18 must be between 4 and 5. The value ‘18’ is much closer to 16 than to 25 so
we can estimate where it is on the real number line below the square roots. The dotted
arrow shows where we would estimate the 18 belongs on the real number line.
Given that the arrow points much closer to 4 than 5 our estimate should be about
4.2. We can check our estimate by multiplying our estimate times itself. So
(4.2)(4.2) = 17.64 which is reasonably close to 18.
Example B
Estimate the
98
17
81
9
2
100
98 to the nearest tenth.
The 98 does not work out to be a whole number. The
square roots that do have whole number answers, and
are on either side of 98 on the real number line, are 81
and 100. The difference between 81 and 98 is 17. The
difference between 98 and 100 is 2.
10
Since the square root of 81 is 9 and the square root of 100 is 10 the answer for the
square root of 98 must be between 9 and 10. The value 98 is much closer to 100 than
to 81 so we can estimate where it is on the real number line below the square roots.
The dotted arrow shows where we would estimate the 98 belongs on the real number
line. Given that the arrow points much closer to 10 than 9 our estimate should be about
9.9. We can check our estimate by multiplying our estimate times itself. So
(9.9)(9.9) = 98.01 which is very close to 98.
Square roots are frequently irrational numbers. Irrational numbers are non-terminating,
non-repeating decimal values. For instance the
2 is 1.414213562 when we round
it to nine decimal places. This value, however, could have infinitely more digits behind
the decimal point. This 2 will never stop and never have a pattern. Fractions, which
are rational numbers will have a pattern. The fraction 4/3 as a decimal is 1.3333…
or 1.3 repeating. Square roots of any positive real number can be written as a decimal
number or approximation. However, if we try to take the square root of a negative value
we cannot find a real answer. The problem below shows our difficulty.
 4 does not have a real answer. If we try to use -2, the answer then
when we check the answer by multiplying (-2)(-2), we get positive 4. Trying
to use 2 an the answer also does not work since (2)(2) is positive 4. This
means there is no real answer, because there is no way to get a negative.
We should also be able to evaluate expressions that contain square roots. All the
operations inside the square root symbol must be done before any of the operations
on the outside. In a way, we can treat the square root symbol ( also called a radical )
as a grouping symbol. The problem below demonstrates how to substitute and simplify
with these expressions.
x  2 – 1 with x = 23
Evaluate 3
x2 –1
3
23  2 – 1
3
3
25 – 1
3(5) – 1
15 – 1
14
Exercises Unit 8 Section 1
Find the square roots. If there is no real answer then write “no real number”.
1.
121
2.
4
3.
 36
4.
169
5.
 64
6.
1
7.
196
8.
49
Estimate the square roots to the nearest tenth. Check your answers by multiplying
your estimate times itself. Show your work.
9.
13.
79
106
10.
20
11.
52
12.
12
14.
172
15.
94
16.
35
Use your calculator and find the square roots to the nearest hundredth.
17.
79
18.
20
19.
52
20.
12
21.
106
22.
172
23.
94
24.
35
Simplify the square roots.
25.
26.
a2
( x  4) 2
27.
x4
Evaluate the following. Show your work.
28. Evaluate
4
x 1 + 2
29. Evaluate
-2
3x  6 – 4
with x = 10
x2  9 – 6
30. Evaluate
31. Given f(x) =
with x = 5
2x  9
with x = 4
Find f(5) and f(12.5), show your work.
Solve the following
32. x2 = 7x + 8
33. (a + 11)(2a – 13) = 0
34. (a2 – 4)(a2 + 9a – 10) = 0
35. (x – 5)3 = 0
Solve the listed systems using substitution.
36.
x=y+2
3x + 2y = 21
37. -x + y = -8
x = 2y + 7
38.
2y = -2x + 4
-2x + 3y = -9
39. The best estimate ( this means you can’t use the calculator ) for the
a. 11
b. 10.9
c. 10.7
119
d. 11.1
40. The area of a square garden is 85 ft2 the best estimate for the length of the
side of the square is
a. 9.2
b. 10.2
c. 9.1
d. 9.5
41. A triangle has a height of .5x and a base of 4x. If the area of the triangle
is 25 cm2 then
a. Find the value of x and explain how you found it.
b. Find the height and base.
is
Unit 8 Section 2
Objective

The student will solve quadratics using the square root principle.
There are several key ideas we use when solving equations. These are: the goal of
solving an equation is to isolate the variable; the method is to reverse operations in the
equation; the process is to do the same operation on both sides of the equation. Since
we now have a reverse operations for squares we can apply these ideas to solving
quadratic equations.
When we apply the Order of Operations used for evaluation we often do the exponents
first. Since solving is the reverse of evaluation, the square root (the reversing of the
the exponent) is done last in many instances. This means that we must isolate the
square term on one side of the equation before we can take the square root.
The examples that follow show us how to isolate the square and solve the equation.
We are reversing the square and using a
square root on both sides of the equation.
x2 = 81
We used the “plus and minus” symbol
x 2 =  81
because a positive value squared or a
x = -9, 9
negative value squared will both produce a
positive 81.
as an answer.
We can confirm that the correct answer is both the positive and negative nine
by solving the equation using factoring as a method. The two methods must
both give us the same answer.
Example A
Solve
Solve
x2 = 81
x2 = 81
x2 = 81
- 81 -81
2
x – 81 = 0
(x + 9)(x – 9) = 0
The Subtraction Property of Equality is used.
The equation is in standard form.
We used the difference of perfect squares factoring.
x = -9, 9
Example B
Solve
(x + 1)2 = 25
(x +
1)2
= 25
=  25
x + 1 = 5
( x  1)
x + 1 = -5
2
x+1=5
The constant in the square changes
the process slightly. Since the square
is isolated we can take the square
root on both sides. The positive and
negative result means we must break
the equation into two new equations
to get our final answers.
Example B
(continued)
Because of the ‘  ’ symbol we need to
solve two equations; one for the plus,
and one for the minus.
x + 1 = 5
x + 1 = -5
-1 -1
x+1=5
-1 -1
x = -6
x=4
Our answer then is x = -6, 4
Example C
Solve a2 + 5 = 38 Round the answers to the nearest hundredth.
a2 + 5 = 38
-5 - 5
We use the Subtraction Property of Equality.
a2 = 33
a2
Example D
=  33
a = -5.74, 5.74
Solve 2(y - 1)2 - 4 = 51 Round the answers to the nearest hundredth.
2(y - 1)2 - 4 = 51
+4 +4
2(y - 1)2 = 55
2
2
(y - 1)2 = 27.5
( y  1) 2
y–1
Example E
We take square roots on both sides.
We use the Addition Property of Equality.
We use the Division Property of Equality.
=  27.5
=  5.24
We take square roots on both sides.
We get two equations because of the  .
y – 1 = -5.24
+1 +1
y – 1 = 5.24
+1 +1
y = -4.24
y = 6.24
Solve (y + 6)2 = -9 Round the answers to the nearest hundredth.
(y + 6)2 = -9
=  9
The square root of -9 is not a real number. We have no real solution.
( y  6) 2
If we are asked to list our answer in an exact form then we can not use a decimal
approximation. This means our answer must contain a radical. The examples below
show us how to find exact answers.
Example F
Solve the equation (x +3)2 = 23 List your answer in an exact form.
(x +3)2 = 23
( x  3) 2
= 
First we take square roots on both sides
23
Next we get two equation due to the 
x+3=
-3
x + 3 =  23
-3
-3
23
-3
x = -3 +
x = -3 -
23
We have our two answers x = -3 +
23
23 or -3 -
We can also list our answer as x = -3 
23
23
Example G Solve the equation 2(y – 5)2 + 1 = 15 List your answer in an exact form.
2(y – 5)2 + 1 = 15
-1 = -1
First we reverse operations outside
the square.
2(y – 5)2 = 14
2
2
(y – 5)2 = 7
( y  5)
2
= 
Now we reverse the square with
7
y–5 =  7
+5
+5
y =5  7
a square root.
Now we reverse what was inside
the square root.
So our answer can be written as y = 5 
7
When we are solving quadratics using the Square Root Principle, we must reverse all
the operations on the outside of the radical or root first. Then we take square roots
on both sides. If necessary, we then set up two equations; one for the plus, and one
for the minus. Finally, we reverse the operations inside the radical or root to get
our two solutions.
VIDEO LINK: Khan Academy Solving Quadratics by the Square Root Principle
Exercises Unit 8 Section 2
Find the square roots. If there is no real answer then write “no real number”.
1.
144
2.
9
3.
25
4.
 81
5.
4
6.
1
7.
225
8.
100
Estimate the square roots to the nearest tenth. Check your answers by multiplying
your estimate times itself. Show your work.
9.
90
10.
62
11.
12.
27
45
Simplify the square roots.
13.
z2
14.
(a  5) 2
15.
(2 x  3) 2
16.
b6
17.
(7  y ) 2
18.
( a  b) 2
Solve the following equations using the Square Root Principle.. Round your answers to the
nearest hundredth when needed. Write “no real answer” if there is no solution.
Show your work.
19.
20. a2 = 36
21. y2 + 5 = 30
22. 3x2 = 30
23. 77 = 5w2 – 3
24. 21 = 2x2 + 2
25. b2 = 100
26. y2 + 5 = 5
27. 0 = z2 + 6
28. (x + 4)2 = 25
29. (y – 5)2 = 4
30. 55 = (a + 2)2
31. 121 = (2x + 1)2
32. (4 – z)2 = 36
33. 90.25 = (z + 3)2
34. 72 = (x – 7)2 – 9
35. 15 = 2(a + 6)2
36. 416.25 = 3(w + 1)2 – 52
38. 8 = (2a + 11)2 – 17
39. 5(x + 2)2 + 7 = 7
37.
x2 = 121
( y  9) 2
=8
2
Solve the following equations and leave your answers in radical form. ( List your
answers in an exact form. ) Write “no real answer” if there is no solution. Show your work.
2 2
)
3
40. (x + 1)2 = 21
41. (y – 11)2 = 2
42. 14 = (z +
43. 3(x + 4)2 = 39
44. -2(y – 9) = -10
45. 12 = 5(z – 7)2 – 3
46. 7(x + 2)2 + 4 = 25
47.
48.
8(z +1)2 + 6 = 16
-12 = -3(2x + 5)2 + 12
49. Explain why
50.
 16 cannot be a real number.
If a square has a side of x + 2 and an area of 100 in2 then
a. Set up an equation that will help us find x. Explain how you created
the equation.
b. Solve the equation you set up in part a. Explain each step you use
to solve the equation.
c. What is the length of the side of the square. List the proper units and
Explain how you found the answer.
51.
What are the solutions to 4(x – 8)2 + 5 = 45?
a. 8  10
b. 10  8
c. -8  11
d. 8  11
Unit 8 Section 3
Objective

The student will solve quadratics by completing the square.
To complete the square for a quadratic, we need to examine the relationship of the
constants in a binomial to the coefficients in the product. The problems below will
help us see this relationship.
(x
(x
(x
(x
+
+
+
+
3)2
4)2
5)2
6)2
=
=
=
=
(x
(x
(x
(x
+
+
+
+
3)(x
4)(x
5)(x
6)(x
+
+
+
+
3)
4)
5)
6)
=
=
=
=
x2
x2
x2
x2
+
+
+
+
3x
4x
5x
6x
+
+
+
+
3x
4x
5x
6x
+
+
+
+
6 = x2 + 6x + 9
16 = x2 + 8x + 16
25 = x2 + 10x + 25
36 = x2 + 12x + 36
We will use these problems to create a table.
Constant in
the Binomial
3
4
5
6
-7
-8
Coefficient of
the Middle Term
6
8
10
12
-14
-16
Last
Term
9
16
25
36
49
64
The question we must ask ourselves is, “What arithmetic do we have to do with
the value in the first column to get the other two values or columns?” The conclusion
we can reach is that we multiply the constant value by 2 to get the coefficient, and we
square the constant value to get the last term. In symbols the arithmetic would be:
Constant in
the Binomial
n
Coefficient of
the Middle Term
2n
Last
Term
n2
We can also ask ourselves, “How can we start with the middle term’s coefficient and
find the other two values?” To find the constant in the binomial we would divide by 2.
And then to find the last term, we would square the result of the division (the quotient).
Symbolically, we would show the arithmetic as:
Constant in
the Binomial
m
2
Coefficient of
the Middle Term
m
Last
Term
( m )2
2
This means if we know the coefficient of the middle term, we can find the constant in the
binomial and the value that would be the last term. The examples below show us how
to find these values.
Example A
Given the coefficient of the middle term of a binomial square is 20,
find the constant that belongs in the binomial and the last term.
20x
20 ÷ 2 = 10
102 = 100
So
The first
term is
always the
square.
Example B
This would be the middle term.
This is the constant in the binomial.
This is the last term.
The first term is the square.
x2 + 20x + 100 = (x + 10)2
This is
the given
middle
term.
This is the
square or
last term.
We square
the 10.
102 = 100
The constant in the binomial
is always calculated first.
This was the 20 ÷ 2 = 10.
Given the coefficient of the middle term of a binomial square is -24,
find the constant that belongs in the binomial and the last term.
-24x
-24 ÷ 2 = -12
(-12)2 = 144
So
Example C
This would be the middle term.
This is the constant in the binomial.
This is the last term.
The first term is the square.
x2 – 24x + 144 = (x - 12)2
Given the coefficient of the middle term of a binomial square is 7,
find the constant that belongs in the binomial and the last term.
7x
7 ÷ 2 = 3.5
(3.5)2 = 12.25
So
x2 + 7x + 12.25 = (x + 3.5)2
This would be the middle term.
This is the constant in the binomial.
This is the last term.
The first term is the square.
In order to solve more complex quadratics using the Square Root Principle we will need
to complete the square in expressions and equations. When we complete the square
we must be sure not to change the overall value of the expression. If we start with an
expression and we add and subtract the same number from the expression we can
keep the value of an expression unchanged. This principle is demonstrated below.
Example A
If we start with 3x as our expression and we add and subtract 5, the
expression will look different but it could be simplified to the original
problem.
3x
(3x + 5) – 5
These two expressions are equal but they look different due to adding
and subtracting five.
Example B
If we start with x2 as our expression and we add and subtract 10, the
expression will look different but it could be simplified to the original
problem.
x2
(x2 + 10) – 10
These two expressions are equal but they look different due to adding
and subtracting ten.
Example C
If we start with a2 + 8a as our expression and we add and subtract 16,
the expression will look different but it could be simplified to the original
problem.
a2 + 8a
(a2 + 8a + 16) – 16
These two expressions are equal but they look different due to adding
and subtracting 16. However, if we consider just the trinomial in the
parentheses we can factor the trinomial.
(a2
a2 + 8a
+ 8a + 16) – 16
(a + 4)(a + 4) – 16
(a + 4)2 – 16
Example C shows us how to complete the square for any expression with an order of 2.
Example D
If we start with y2 – 12y as our expression and we add and subtract 36,
the expression will look different but it could be simplified to the original
problem. The reason we pick the number 36 is that half of -12 is -6
and (-6)2 = 36.
y2 – 12y
(y2 – 12y + 36) – 36
These two expressions are equal but they look different due to adding
and subtracting 36. However, if we consider just the trinomial in the
parentheses we can factor the trinomial.
y2 – 12y
(y2 – 12y + 36) – 36
(y – 6)(y – 6) – 36
(y – 6)2 – 36
Example E
If we start with z2 + 22z as our expression and we add and subtract 121,
the expression will look different but it could be simplified to the original
problem. The reason we pick the number 121 is that half of 22 is 11
and (11)2 = 121.
z2 + 22z
(z2 + 22z + 121) – 121
These two expressions are equal but they look different due to adding
and subtracting 121. However, if we consider just the trinomial in the
parentheses we can factor the trinomial.
z2 + 22z
(z2 + 22z + 121) – 121
(z + 11)( z + 11) – 121
(z + 11)2 – 121
There is a pattern to these problems that can be quite helpful to us. The table below shows
us how the coefficients and constants relate.
expression
constant
square
completed square
Example C
x2 + 8x
8÷2=4
42 = 16
(x + 4)2 – 16
Example D
y2 – 12y
-12 ÷ 2 = -6
(-6)2 = 36
(y – 6)2 – 36
Example E
z2 + 22z
22 ÷ 2 = 11
(11)2 = 121
(z + 11)2 – 121
To complete the square we divide the linear term’s coefficient by two and square the
quotient. Then the completed square is ( x  coefficient ÷ 2) – square.
Example F
Given x2 + 26x, complete the square.
( x + (26 ÷ 2) )2 – (26 ÷ 2)2
(x + 13)2 – 169
Example G
Given x2 – 17x complete the square.
( x + (-17 ÷ 2) )2 – (-17 ÷ 2)2
(x – 8.5)2 – 72.25
We can combine our method for completing the square with our adding and subtracting
the same value that maintains the value of any expression.
The steps below outline how to use this process to solve equations.
Step 1. Use the Division Property of Equality to make the
coefficient of the square term equal to one.
Step 2. Divide the linear term’s coefficient by 2. This will be
the constant in the binomial square.
Step 3. Square the constant found in Step 2.
Step 4. Add and subtract the square on the side of the equation
containing the square term.
Step 5. Simplify and solve the resulting equation using the
Square Root Principle.
The examples below use these steps.
Example A
Solve x2 + 20x + 25 = 0
This will not factor so we must use completing the square.
x2 + 20x + 25 = 0
20 ÷ 2 = 10
(10)2 = 100
Step 1. The coefficient of the square term is 1
to start with.
Step 2. Dividing linear term’s coefficient by 2
Step 3. Squaring the quotient
x2 + 20x + 100 – 100 + 25 = 0
Step 4. Add and subtract the square
(x + 10)2 – 75 = 0
Step 5. Simplifying the equation and solving
+75 = +75
(x + 10)2 = 75
x + 10 = 75
x + 10 = 8.66
-10 = -10
x = -1.34
The 75 is approximately 8.66
x + 10 = - 75
found by using a calculator and
x + 10 = -8.66
rounded off to the nearest
-10 = -10
hundredth.
x = -18.66
x = -1.34, -18.66
Example B
Solve x2 – 9x – 15.75 = 0
This will not factor so we must use completing the square.
x2 – 9x – 15.75 = 0
-9 ÷ 2 = -4.5
(-4.5)2 = 20.25
Step 1. The coefficient of the square term is 1 to
start with.
Step 2. Dividing linear term’s coefficient by 2
Step 3. Squaring the quotient
x2 – 9x + 20.25 – 20.25 – 15.75 = 0
Step 4. Add and subtract the square
(x – 4.5)2 – 36 = 0
Step 5. Simplifying the equation and solving
+36 = +36
(x – 4.5)2 = 36
x – 4.5 = 36
x – 4.5 = 6
+4.5 = +4.5
x = 10.5
x – 4.5 = - 36
x – 4.5 = -6
+4.5 = +4.5
x = -1.5
x = 10.5, -1.5
Example C
Solve x2 + 4x = 50.76
This will not factor so we must use completing the square.
x2 + 4x = 50.76
Step 1. The coefficient of the square term is 1 to
start with.
Step 2. Dividing linear term’s coefficient by 2
Step 3. Squaring the quotient
4÷ 2=2
(2)2 = 4
x2 + 4x + 4 – 4 = 50.76
Step 4. Add and subtract the square
(x + 2)2 – 4 = 50.76
+4 = +4
(x +2)2 = 54.76
x + 2 = 54.76
x + 2 = 7.4
-2 = -2
x = 5.4
Step 5. Simplifying the equation and solving
x + 2 = - 54.76
x + 2 = -7.4
-2 = -2
x = -9.4
x = 5.4, -9.4
VIDEO LINK: Youtube: Solving quadratics by completing the square
Exercises Unit 8 Section 3
Find the square roots. If there is no real answer then write “no real number”.
1.
2.
225
49
4
3.
4.
100
Estimate the square roots to the nearest tenth. Check your answers by multiplying
your estimate times itself. Show your work.
5.
6.
104
46
Simplify the square roots.
7.
a2
8.
( x  11) 2
9.
(3 x  7) 2
Solve the following equations using the Square Root Principle. Round your answers to
the nearest hundredth when needed. Show your work.
10.
x2 = 144
11. a2 = 25
12. y2 + 7 = 43
13. -3x2 = 96
14. 57 = 4w2 – 3
15. 6 = -2x2 + 14
16. (x + 2)2 = 49
17. (y – 1)2 = 8
18. 100 = (2a + 3)2
19. 16 = (x – 1)2 – 9
20. 248.43 = 3(a + 7)2
21.
22. x2 – 16x
23. a2 + 6a
24. y2 + 10y
25. x2 – 15x
26. b2 + 5b
27. x2 – 30x
28 a2 + 8.8a
29. 14w + w2
30. z2 – z
Complete the square.
( y  5) 2
+2=2
3
Solve the following by completing the square. Round your answers to the nearest
hundredth where needed. Show your work.
31. x2 + 4x – 4 = 0
32. a2 – 6a – 4 = 0
33. 0 = y2 + 18y – 9
34. x2 – 14x = 12
35. a2 = a + .75
36. 8y + 4.25 = y2
37. 42 – x2 = 12x
38. 5y + 115.8525 = y2
39. z2 + 24z = 20
40. y2 – 2y = 11
41. 11w – w2 + 3.25 = 0 42. a2 = -12a + 1
43. Given the rectangle to the right as labeled,
if the rectangle has an area of 40 cm2 then
a. Find the value of x by solving a quadratic
equation using the method of completing
square. Show your work algebraically
and explain how you found your answer.
x+8
x-1
b. Find the length and width of the rectangle and check your answer.
Unit 8 Section 4
Objective

The student will solve quadratics with leading coefficients by
completing the square.
We have seen the algorithm for solving quadratics by completing the square. That
algorithm or set of steps is given below.
The steps below outline how to use this process to solve equations.
Step 1. Use the Division Property of Equality to make the
coefficient of the square term equal to one.
Step 2. Divide the linear term’s coefficient by 2. This will be
the constant in the binomial square.
Step 3. Square the constant found in Step 2.
Step 4. Add and subtract the square on the side of the equation
containing the square term.
Step 5. Simplify and solve the resulting equation using the
Square Root Principle.
In this section we will use the Division Property of Equality to help complete the square.
The examples that follow show us how this is done.
Example A
Solve 4x2 – 12x = 44 Round your answer to the nearest hundredth.
4x2 – 12x = 44
4
4
4
2
x – 3x = 11
Step 1. Use the
Division Property of
Equality to get a
leading coefficient of
one.
-3 ÷ 2 = -1.5
(-1.5)2 = 2.25
Step 2. Divide the Step 3. Square
the quotient.
linear coefficient
by 2.
x2 – 3x + 2.25 – 2.25 = 11
(x2 – 3x + 2.25) – 2.25 = 11
(x – 1.5)2 – 2.25 = 11
+2.25 +2.25
(x – 1.5)2 = 13.25
x – 1.5
x – 1.5
+1.5
x
= 13.25
= 3.64
+1.5
= 5.14
x2 – 3x + 2.25 – 2.25 = 11
Step 4. Add and Subtract
the square.
Step 5. Simplify the
equation and complete
the square.
Then solve the equation.
x – 1.5 = - 13.25
x – 1.5 = -3.64
+1.5 +1.5
x = -2.14
x = 5.14, -2.14
Example B
Solve 2x2 + 16x = 47.38 Round your answer to the nearest hundredth.
2x2 + 16x = 47.38
2
2
2
2
x + 8x = 23.69
Step 1. Use the
Division Property of
Equality to get a
leading coefficient of
one.
8÷2=4
Step 2. Divide the
linear coefficient
by 2.
(4)2 = 16
x2 + 8x +16 – 16 = 23.69
Step 3. Square
the quotient.
x2 + 8x +16 – 16 = 23.69
(x2 + 8x +16) – 16 = 23.69
(x + 4)2 – 16 = 23.69
+16 +16
(x+ 4)2 = 39.69
x + 4 = 39.69
x + 4 = 6.3
-4 -4
x = 2.3
Step 4. Add and Subtract
the square.
Step 5. Simplify the
equation and complete
the square.
Then solve the equation.
x + 4 = - 39.69
x + 4 = -6.3
-4 -4
x = -10.3
x = 2.3, -10.3
Example C
Solve -3x2 = 15x + 45 Round your answer to the nearest hundredth.
-3x2 = 15x – 45
+3x2 +3x2
0 = 3x2 + 15x – 45
0 = 3x2 + 15x – 45
3
3
3
3
0 = x2 + 5x - 15
Step 1. Use the
Division Property of
Equality to get a
leading coefficient of
one.
5 ÷ 2 = 2.5
Step 2. Divide the
linear coefficient
by 2.
We need to get the square term
positive and on the same side as
the linear term.
(2.5)2 = 6.25 0=x2 + 5x +6.25 – 6.25 - 45
Step 3. Square
the quotient.
Step 4. Add and Subtract
the square.
Step 5. Simplify the
x2 + 5x + 6.25 – 6.25 - 45 = 0
equation and complete the
(x2 + 5x + 6.25) – 51.25 = 0
square.
(x + 2.5)2 – 51.25 = 0
Then solve the equation.
+51.25 +51.25
(x + 2.5)2 = 51.25
x + 2.5 = 51.25
x + 2.5 = 7.16
-2.5
-2.5
x = 4.66
x + 2.5 = - 51.25
x + 2.5 = -7.16
-2.5 -2.5
x = -9.66
x = 4.66, -9.66
VIDEO LINK: Khan Academy Example 3 Completing the Square
Exercises Unit 8 Section 4
Solve the following equations using the Square Root Principle. Round your answers to the
nearest hundredth when needed. Show your work.
2. a2 = 400
3. y2 – 11 = 53
4. 2x2 = 200
5. 0 = 4w2 + 20
6. 16 = 3x2 + 10
7. (x – 3)2 = 196
8. 11 = (4a + 1)2 – 14
1.
x2 = 81
Solve the following equations using the Square Root Principle. All answers should be
exact. Show your work. Write “no real answer” if there is no solution.
9. (x + 12)2 = 17
10. (y – 3)2 + 13 = 15
11. -38 = -4(a = 15)2 + 6
12. (z – 5)2 + 2 = -9
13. 5(x + 8)2 = 3
14. 2(3x + 6)2 + 4 = 4
15. x2 – 40x
16. a2 + 12a
17. y2 + 22y
18. x2 – 9x
19. b2 + 21b
20. x2 – 3.2x
Complete the square.
Solve the following by completing the square. Round your answers to the nearest
hundredth where needed. Show your work.
21. x2 + 10x – 6.36 = 0
22. a2 – 16a – 95 = 0
23. 202.76 = y2 + 20y
24. x2 + x = -0.21
25. -a2 = 2a – 152.76
26. 8y + 4.25 = y2
27. 2y - y2 = -88
28. 4w + w2 = 186.44
29. a2 = 22a + 35.25
Solve the following by completing the square. Round your answers to the nearest
hundredth where needed. Show your work.
30. 2x2 – 18x = 450.5
31.
3x2 + 12x – 151.47 = 0
32. 5z2 + 10z – 35 = 0
33. 4y2 = 10y – 12
34. -2a2 = 28a – 14
35.
36. 0 = 4z2 + 20z – 100
37.
12 – 6x2 = 21x
3y2 + 9y = 317.73
38. Given the equation 3x2 + 30x = 15 solve the equation by completing the square.
Show your work algebraically and explain how you found your solutions as if you
were tutoring a classmate who was absent from class instruction.
Unit 8 Section 5
Objective

The student will use the quadratic formula to solve quadratics.
There is one more way to solve quadratic equations. This method requires that we
derive a formula using the Square Root Principle. To derive a formula we need to
represent any quadratic equation with variables. The equation below can represent
any quadratic.
ax2 + bx + c = 0
In this equation the ‘a’ represents any coefficient or number we can put in front of the
‘x2’ term. The ‘b’ represents any coefficient or number we can put in front of the ‘x’
term. The ‘c’ represents and constant we could add or subtract to get zero.
We can use the same process from section 4 to solve this equation that we used to
solve equations with real numbers for ‘a’, ‘b’, and ‘c’.
Deriving of the quadratic formula.
ax2 + bx + c = 0
a
a
a a
x2 
First we divide the equation by ‘a’, using the
Division Property of Equality.
b
c
x 
 0
a
a
Next we divide the coefficient of the
linear term by 2
b
b 2 b 1
b
2     
a
a 1 a 2 2a
( 2ba )
2
is the square of the answer to the last step.
( 2ba ) – ( 2ba )
x2 
b
x +
a
(x 
b
2a
) – ( 2ba )
(x 
b
2a
)
2
2
=
2
2
+
( 2ba )
2
2
+
c
=0
a
–
c
a
c
= 0
a
We add and subtract
the square.
We complete the square
and start to solve.
We need to isolate ‘x’ so
we move terms to the
other side of the equation.
Deriving of the quadratic formula. (continued)
(x 
b
2a
)
2
(x 
b
2a
)
2
b
x 
2a
)
)
(
(x 
b
2a
–
c
a
–
c
a
We multiply out the square
so we can combine terms.
b2
4a 2
2
=
4ac
b2
–
2
4a 2
4a
2
=
b 2  4ac
4a 2
=

b
x 
2a
b
2a
x
x
2
=
b 

x 

2a 

x 
( 2ba )
=
2
=

=

=
b
2a
=
c 4a

to
a 4a
get common denominators.
We multiplied
We subtracted the fractions.
b 2  4ac
4a 2
b 2  4ac
4a 2
We take square roots on both sides.
We simplify the square roots.
b 2  4ac
2a

b 2  4ac
2a
 b  b 2  4ac
2a
We isolate ‘x’.
We combine the fractions.
The derivation of the formula is somewhat long. The formula itself is, however,
very easy to use. The examples that follow show us how to substitute and
simplify with the formula.
Solve the equation x2 + 6x – 40 = 0 using the Quadratic Formula.
Example A
a = 1, b = 6, c = -40
x
=
These are the values for a, b, and c.
This is the Quadratic Formula. We should
write it before we start in order to do the
substitution correctly.
 b  b 2  4ac
2a
 6  6 2  4(1)( 40)
x
=
x
=
x
=
 6  196
2
x
=
 6  14
2
2(1)
After substituting, we need to simplify.
 6  36 160
2
We can now get our answers. There will be two answers because
of the plus-minus sign.
=
x
 6  14
2
x
=
 6  14
2
x
=
 6  14
2
x
=
8
2
x
=
 20
2
x = 4 or -10
Solve the equation 3x2 – 7x = 30
Example B
3x2 – 7x = 30
-30 -30
using the Quadratic Formula.
First we must get the equation into
into standard form.
3x2 – 7x – 30 = 0
a = 3, b = -7, c = -30
=
x
x
=
x
=
x
These are the values for a, b, and c.
This is the Quadratic Formula. We should
write it before we start in order to do the
substitution correctly.
 b  b 2  4ac
2a
 (7)  (7) 2  4(3)( 30)
2(3)
After substituting we need to simplify.
7  49  360
2
=
7  409
2
We can now get our answers. There will be two answers because
of the plus-minus sign.
x
x
=
7  409
2
= 3 .5 
x = 3.5 
409
2
x
= 3.5 
409
or 3.5 
2
409
2
409
2
When the square root does not work out to an rational number, we can
leave it as a square root.
Example C
Solve the equation 4x2 + x + 12 = 0 using the Quadratic Formula.
a = 4, b = 1, c = 12
 b  b 2  4ac
x
=
x
=
 1  (1) 2  4(4)(12)
2(4)
x
=
 1  1  192
8
x
=
 1   191
2
2a
These are the values for a, b, and c.
This is the Quadratic Formula. We should
write it before we start in order to do the
substitution correctly.
After substituting we need to simplify.
The  191 is not a real number. This means there is NO solution
to the equation. Since there is no solution we can write this
symbolically as O. This symbol stands for the empty set.
VIDEO LINK: Khan Academy Using the Quadratic Formula
Exercises Unit 8 Section 5
Solve the following equations using the square root principle.. Round your answers to
the nearest hundredth when needed. Show your work.
1.
x2 = 441
4. 0 = 5w2 – 20
2. y2 + 11 = 63
3. 2x2 = 310
5. 12 = (3a + 1)2 – 24
Complete the square.
6. x2 – 30x
7. a2 + 19a
8. y2 – y
Solve the following by completing the square. Round your answers to the nearest
hundredth where needed. Show your work.
9. x2 + 18x – 5 = 0
10. a2 – 6a = 31
11. 243.39 = y2 + 5y
Solve the following by completing the square. Round your answers to the nearest
hundredth where needed. Show your work.
12. 3x2 – 18x = 33
13.
4x2 + 24x – 12 = 0
14. -2z2 + 20z – 36 = 0
15. 7y2 = 28y + 12.32
Use the Quadratic Formula to solve the following. Show your work. Do NOT find
decimal approximations leave square roots in your answers where needed.
16. x2 + 21x + 68 = 0
17. a2 – 23a + 112 = 0
18. 6x2 + 11x – 10 = 0
19. 12y2 – 3x = 42
20. 2x + 15 = 24x2
21. 2x2 + 7x + 4 = 0
22. y2 + 4y + 1 = 0
23. b2 – 12b = -5
24. 8a2 – 3 = 2x
25.
26. x2 + 3x + 9 = 0
27. w2 – w + 5 = 0
28. 4x2 + 26x + 36 = 0
29. 2x2 = 9x – 44
30. 2x2 + 5x - 9 = 0
31. –y2 + 6y = 3
15z2 +31z + 10 = 0
32. Given the equation x2 + 2x = 48
a. Solve the equation by factoring. Show your work algebraically.
b. Solve the equation by completing the square. Show your work algebraically.
c. Solve the equation using the quadratic formula. Show your work algebraically.
Unit 8 Section 6
Objectives

The student will graph quadratics using substitution.

The student will identify and find key features of quadratics.
Graphing by substitution is a process we have used before in Unit 3. When we use this
process we pick values for ‘x’, substitute these values into an equation, and generate the
values for ‘y’. Once we have a set of ordered pairs we can plot the points and connect
the points to form a shape. The shapes we saw in previous units could have been lines,
V-shapes, or parabolas. In graphing quadratics we will see the parabola shapes again.
The examples below show us how to graph several parabolas.
Example A
Graph y = x2 + 4x – 3
using substitution.
In order to make sure we see all of the shape we must pick a wide selection of x-values.
As a start we will use all the integers from -5 to 5.
x
y
2
-5
2
2
-4
-3
y = x2 + 4x – 3
y = (-5) + 4(-5) – 3 = 25 – 20 – 3 = 2
y = (-4) + 4(-4) – 3 = 16 – 16 – 3 = -3
2
y = (-3) + 4(-3) – 3 =
9 – 12 – 3 = -6
-3
-6
2
4 – 8 – 3 = -7
-2
-7
2
1 – 4 – 3 = -6
-1
-6
y = (-2) + 4(-2) – 3 =
y = (-1) + 4(-1) – 3 =
2
=
0–0–3
= -3
0
-3
2
=
1+4–3 = 2
1
2
2
=
4+8–3 = 9
2
9
2
= 9 + 12 – 3 = 18
3
18
2
= 16 + 16 – 3 = 29
4
29
2
= 25 + 20 – 3 = 42
5
42
y = (0) + 4(0) – 3
y = (1) + 4(1) – 3
y = (2) + 4(2) – 3
y = (3) + 4(3) – 3
y = (4) + 4(4) – 3
y = (5) + 4(5) – 3
We picked a wide set of x-values but some of the y-values were too large to fit on the
graph so we simply left them off. The 7 points we were able to plot defined the shape of
the parabola quite readily.
Graph y = (x – 1)2 - 2 using substitution.
Example B
In order to make sure we see all of the shape we must pick a wide selection of x-values.
As a start we will use all the integers from -5 to 5.
y = (x – 1)2 - 2
x
y
2
-5
34
2
-4
23
2
-3
14
2
-2
7
y = (-5 - 1) – 2 = 36 – 2 = 34
y = (-4 - 1) – 2 = 25 – 2 = 23
y = (-3 - 1) – 2 = 16 – 2 = 14
y = (-2 - 1) – 2 = 9 – 2 = 7
2
y = (-1 - 1) – 2 = 4 – 2
=2
-1
2
y = (0 - 1)
2
–2 = 1–2
= -1
0
-1
y = (1 - 1)
2
–2
= 0–2
= -2
1
-2
y = (2 - 1)
2
–2
= 1–2
= -1
2
-1
y = (3 - 1)
2
–2 = 4–2
= 2
3
2
y = (4 - 1)
2
–2 = 9–2
= 7
4
7
y = (5 - 1)
2
–2
= 14
5
14
= 16 – 2
We picked a wide set of x-values but some of the y-values were too large to fit on the
graph so we simply left them off. The 7 points we were able to plot defined the shape of
the parabola quite readily.
Example C
Graph y = -.5(x + 1)2
using substitution.
In order to make sure we see all of the shape we must pick a wide selection of x-values.
As a start we will use all the integers from -5 to 5.
x
y
y = -.5(-5 + 1)2 = -.5(16) = -8
-5
-8
y = -.5(-4 + 1)2 = -.5(9)
= -4.5
-4
-4.5
y = -.5(-3 + 1)2 = -.5(4)
= -2
-3
-2
y = -.5(-2 + 1)2 = -.5(1)
= -.5
-2
-.5
y = -.5(-1 + 1)2 = -.5(0)
= 0
-1
0
y = -.5(0 + 1)2
= -.5(1)
= -.5
0
-.5
y = -.5(1 + 1)
2
= -.5(4)
= -2
1
-2
y = -.5(2 + 1)
2
= -.5(9)
= -4.5
2
-4.5
y = -.5(3 + 1)
2
= -.5(16) = -8
3
-8
y = -.5(4 + 1)
2
= -.5(25) = -12.5
4
-12.5
y = -.5(5 + 1)
2
= -.5(36)
5
-16
y = -.5(x + 1)2
= -16
The three preceding examples of all the equations were quadratic. Quadratics can be
written in two forms and we need to remember what those two forms are and how to
convert between them.
When we start with a square term we can complete the square and write the equation
in a form that is best for graphing.
y = x2 + 12x – 4
y = x2 + 12x + 36 – 36 – 4
y = (x + 6)2 – 40
y = x2 – 10x + 3
y = x2 – 10x + 25 – 25 + 3
y
= x2 – 8x + 16 – 16 + 3
2
y = (x – 5)2 – 22
y = 2x2 – 16x + 6
y = 2(x – 4)2 – 26
In all three of the graphs in our examples we can see the U-shape of a parabola. A
parabola has key features that can help us when we need to graph a quadratic equation.
The equation for this parabola is y = .5(x + 1)2 - 2
The parabola to the left has three features
that are important to all parabolas. The
first is the vertex point, and the second
feature is a line of symmetry. The third
is the roots where the equation = 0.
Roots
Line of Symmetry
Vertex Point
Definition:
The vertex of a parabola is the high or low point of the graph.
Definition:
A line of symmetry divides a graph into two mirror images.
Definition:
A root is where the y-coordinate is 0 and the graphh intersects the x-axis.
The vertex for this parabola is at (-1, -2) and the equation was y = .5(x + 1)2 – 2.
The vertex point can be found from the equation. The x-coordinate of the vertex is the
opposite of the constant in the parentheses and the y-coordinate is the last constant.
The roots ar at (1,0) and (-3, 0) and the line of symetry is x = -1.
If we know where the vertex is located we can pick x-values that are close to the vertex
and make our work much shorter. The examples below show us how to find the vertex
point in a quadratic.
Example A
Given y = (x + 3)2 – 1 the vertex point is (-3, -1). This is because the
opposite of the ‘+ 3’ in the parentheses is -3, and the last constant is -1.
Example B
Given y = (x – 4)2 + 7 the vertex point is (4, 7). This is because the
opposite of the ‘– 4’ in the parentheses is 4 and the last constant is 7.
Example C
Given y =-3(x + 1)2 + 5 the vertex point is (-1, 5). This is because the
opposite of the ‘+ 1’ in the parentheses is -1 and the last constant is 5.
Example D
Given y = x2 + 14x – 5 we must first complete the square. So
y = x2 + 14x – 5 converts to y = x2 + 14x + 49 – 49 – 5 which is
y = (x + 7)2 – 54. The vertex point is (-7, -54). This is because the
opposite of the ‘+ 7’ in the parentheses is -7 and the last constant is -54.
Picking points for substitution can be much easier. The example below shows us how to
use this information.
Example A
Graph y = (x + 3)2 – 6
The vertex is at (-3, -6) so we can set up our table as shown below.
We can put the
vertex point into
the table.
x
y
-3 -6
We put two x-values
above and below the
the vertex.
x
y
We calculate the y-values
y = (x + 3)2 - 6
-5
y = (-5 + 3)2 – 6 = 4 – 6 = -2
-4
y = (-4 + 3)2 – 6 = 1 – 6 = -5
-3 -6
y = (-3 + 3)2 – 6 = 0 – 6 = -6
-2
y = (-2 + 3)2 – 6 = 1 – 6 = -5
-1
y = (-1 + 3)2 – 6 = 4 – 6 = -2
Fill in the table
x
y
-5 -2
-4 -5
-3 -6
-2 -5
-1 -2
When we plot the five points we can see
the U-shape of the parabola and connect
the points.
Identifying the vertex means we pick five
points and calculate 4 y-values. If we can
draw a line of symmetry we know we
have the correct shape.
The line of symmetry is x = -3. The vertex is (-3, -6).
The roots are approximately (-.5, 0) and (-6.5,0).
Example B
Graph y = x2 – 4x + 3
We start by completing the square to find the vertex.
y = x2 – 4x + 3
y = x2 – 4x + 4 – 4 + 3
y = (x – 2)2 – 1
The vertex is at (2, -1) so we can set up our table as shown below.
We can put the
vertex point into
the table.
x
y
We put two x-values
above and below the
the vertex.
x
y
0
-1
2
3
4
Fill in the table
y = (x - 2)2 - 1
x
y
2
0
3
2
1
0
2
2 -1
2
3
0
2
4
3
y = (0 – 2) – 1 = 4 – 1 = 3
y = (1 – 2) – 1 = 1 – 1 = 0
1
2
We calculate the y-values
-1
y = (2 – 2) – 1 = 0 – 1 = -1
y = (3 – 2) – 1 = 1 – 1 = 0
y = (4 – 2) – 1 = 4 – 1 = 3
When we plot the five points we can see
the U-shape of the parabola and connect
the points.
Identifying the vertex means we pick five
points and calculate 4 y-values. If we can
draw a line of symmetry we know we
have the correct shape.
The line of symmetry is x = 2. The vertex
is located at (2, -1) and the roots are (1, 0)
and (3, 0)
Graph y = -2x2 - 12x + 4
Example C
We start by completing the square to find the vertex.
y = -2x2 + 12x – 16
y
= x2 – 6x + 8
2
y
= x2 – 6x + 9 – 9 + 8
2
y
= (x – 3)2 – 1
2
y = -2(x – 3)2 + 2
The vertex is at (3, 2) So we can set up our table as shown below.
We can put the
vertex point into
the table.
x
We put two x-values
above and below the
the vertex.
y
x
y
1
2
3
4
5
y = -2(x - 3)2 + 2
y = -2(1 – 3)2 + 2 = -8 + 2 = -6
y = -2(2 – 3)2 + 2 = -2 + 2 = 0
2
3
We calculate the y-values
2
y = -2(3 – 3)2 + 2 = 0 + 2 = 2
y = -2(4 – 3)2 + 2 = -2 + 2 = 0
y = -2(5 – 3)2 + 2 = -8 + 2 = -6
Fill in the table
x
y
1
-6
2
0
3
2
4
0
5
-6
When we plot the five points we can see
the U-shape of the parabola and connect
the points.
Identifying the vertex means we pick five
points and calculate 4 y-values. If we can
draw a line of symmetry we know we
have the correct shape.
The line of symmetry is x = 3. the vertex is
at (3,2) and the roots are (2,0) and (4,0)
Our process for graphing can be described as:
Step
Step
Step
Step
Step
Step
1.
2.
3.
4.
5.
6.
Complete the square if necessary.
Identify the vertex point.
Build the table.
Calculate the y-values.
Plot the points and graph the parabola.
Check the shape.
Looking at our last three examples we can again see how to find the vertex and line
of symmetry.
Equation
Vertex
Line of Symmetry
Example A
y = (x + 3)2 – 6
(-3, -6)
x = -3
Example B
y = (x – 2)2 – 1
(2, -1)
x= 2
Example C
y = -2(x – 3)2 + 2
(3, 2)
x=3
We find the vertex of a parabola using the opposite of the constant in the
parentheses as the x-coordinate and the last constant as the y-coordinate.
The line of symmetry should be the vertical line passing through the vertex.
The equation will be x = the opposite of the constant in the parentheses.
VIDEO LINK: Khan Academy Graphing a Parabola With a Table of Values
Exercises Unit 8 Section 6
Complete the square.
1. x2 + 40x
2. a2 + 18a
3. y2 – 3y
Complete the square in the following equations
4. y = x2 + 6x + 7
5. y = x2 – 8x + 5
6. y = x2 – x – 2
7. y = 3x2 – 24x + 27
8. y = -2x2 + 4x + 12
9. y = .5x2 + 3x
List the vertex and line of symmetry for each of the quadratic equations below.
10. y = (x – 4)2 + 9
11. y = 2(x + 5)2 + 1
12. y = -2(x – 6)2
13. y = x2 + 4
14. y = 3(x – 8)2 – 2
3
1
17. y = (x + )2 + 2
4
3
15. y = -(x – 7)2 – 3
16. y = -11(x – 2.5)2 – 3.6
18. y = x2
Complete the square then list the vertex and line of symmetry for each quadratic.
19. y = x2 – 20x + 90
20. y = x2 + 14x + 50
21. y = x2 – 21x
Graph the following quadratic equations using substitution. Show your tables and graphs.
If you completed a square show the work converting the equation’s form. List the
"roots" of the equations if they exist.
22. y = (x + 1)2 – 2
23.
24. y = (x – 3 )2 – 5
25. y = -2(x + 3)2 + 4
26. y = (x + 4)2 – 3
1
28. y = (x – 1)2 + 3
2
30. y = x2 – 4x
27. y = -(x – 5)2 – 1
32.
y = (x – 2)2 + 1
29. y = 3(x + 2)2 – 7
31. y = x2 + 6x + 3
y = x2
33. y = -x2 + 8x - 14
Solve the equations below using the Quadratic Formula. Show your work. Do NOT find
decimal approximations leave square roots in your answers where needed.
34. x2 + 7x + 10 = 0
35. 0 = 12x2 – 22x – 4
36. 2x2 + 8x – 5 = 0
37. x2 + 7x = 5
38. Which of the graphs below represents the quadratic function f(x) = x2 + 2x - 3 ?
a.
b.
c.
-4
4
-8
-4
4
-8
d.
8
8
8
-4
8
4
-8
-4
4
-8
39. Given the quadratic function y = x2 + 4x - 5 then
a. Solve the equation by factoring and explain in detail how you found the answer.
b. Make a table of x and y values for the ingeter values of x from -6 to 2.
c. Using the Table from part 'b' graph the function, then list the vertex and
line of symetry.
d. Using the graph find the solutions to the equation 0 = x2 + 4x - 5
Unit 8 Section 7
Objectives

The student will apply quadratics to contextual problems.
Quadratic equations can model many kinds of situations. These include: area problems,
ballistics problems, finding focal points for antennas, motion problems involving
acceleration, fuel consumption and many more. We will examine a few of the
applications.
Example A
The sum of two positive numbers is 20. The sum of the squares of the
numbers is 208. Find the two numbers.
Let
x
be the first number
then 20 – x is the second number
We find the quadratic as the squares of each number are added together.
x2 + (20 – x)2 = 208
x2 + 400 – 40x + x2 = 208
We multiply out the binomial square.
2x2 – 40x + 400 = 208
- 208 -208
Next we combine terms and get the
quadratic into standard form.
2x2 – 40x + 192 = 0
2x2 – 40x + 192 = 0
2(x2 – 20x + 96) = 0
2(x – 8)(x – 12) = 0
x = 8, 12
We could use factoring, the Quadratic
Formula, or the Square Root Principle to
solve the equation. We will try factoring
first.
Now we check our answers. According to the original problem the two
numbers must add up to 20. So, 8 + 12 = 20, which this is correct.
The squares of the two numbers must add up to 208. We can check
this too.
(8)2 + (12)2 = 64 + 144 = 208
This is also correct. We have the correct answers.
Example B
Twice the square of a positive number is 14 more than 29 times the
number.
Let x be the first number
We create the quadratic by translating the sentence.
2x2 = 27x + 14
-27x -27x -14
-14
First we get the equation into
standard form.
We could use factoring, the Quadratic
Formula, or the Square Root Principle
to solve the equation. We will try the
Quadratic Formula here.
2x2 – 27x – 14 = 0
 b  b 2  4ac
x
=
x
=
x
=
x
=
27  841
4
x
=
27  29
4
x
=
27  29
4
or
x = 14 or -
1
2
2a
 ( 27)  ( 27) 2  4(2)( 14)
After substituting we need to simplify.
2(2)
27  729  112
4
x
=
27  29
4
Since the problem asked for a positive number we don’t need to
test the negative one-half to see if it is a correct answer. But we
still must check the 14.
2x2 =
2(14)2 =
2(196) =
392 =
392 =
27x + 14
27(14) + 14
27(14) + 14
378 + 14
392
This checks so 14 is our answer.
Example C
The rectangle below has a length of 3x – 4 and a width of 2x + 1.
if the area of the rectangle is 255 sq ft., find the length and width.
3x - 4
2X + 1
Since the area of a rectangle has a formula of A = l . w, we will
substitute all our values and expressions into the formula.
A=l. w
255 = (3x – 4)(2x + 1)
255 = 6x2 – 5x – 4
-255
-255
First we multiply out the binomials.
We then get the equation into standard
form.
0 = 6x2 – 5x – 259
x
=
 b  b 2  4ac
We can use the Quadratic Formula.
2a
 ( 5)  ( 5) 2  4(6)( 259)
After substituting, we simplify.
2(6)
x
=
x
=
x
=
5  6241
12
x
=
5  79
12
x
=
5  79
12
5  25  6216
12
or
x
=
5  79
12
x = 7 or -6.16
Since the length and width can’t be negative, only the 7 can be the answer
for x. With x = 7, the length is 3(7) – 4 = 17 and the width is 2(7) + 1 = 15.
We can check our answers by multiplying to find the area, (17)(15) = 255.
Our answers are length = 17’ and width = 15’.
Exercises Unit 8 Section 7
Solve the following. Show your equation and the method of solving the equation.
1. A positive number squared minus 570 is 13 times the number. Find the number.
2. Twice the square of a positive number is 100 more than 35 times the number.
Find the number.
3. Seventy-two less than four times the square of a number is twenty-eight times the
number. Find the two solutions.
4. The product of two consecutive positive integers is 576. Find the numbers.
5. The sum of two positive numbers is 25. The product of the numbers is 154.
Find the numbers.
6. The sum of two positive numbers is 31 and the sum of the squares of the two
numbers is 593. Find the numbers.
7. If a number is subtracted from its square the result is 110.
8. A positive number is 240 less than its square.
9. A rectangle has sides of x + 11 and x – 2. If the area of the rectangle is 30 cm2
Find the value of x.
10. A square has a side of 3x – 1. If the area of the square is 196 m2 find the value of x.
11. A rectangle has sides of 5x + 2 and 3x – 6. If the area of the rectangle is 756 in2,
Find the length and width of the rectangle.
12. A triangle has a height of x – 7 and a base of 2x + 4. If the area of the triangle
is 140 cm2 then find the dimensions of the triangle.
13. If a rectangle’s length is 9 cm greater than its width and its area is 532 ft2,
find the dimensions of the rectangle.
14. If a rectangle has a perimeter of 54 ft and an area of 176 ft2, find the dimensions
of the rectangle.
15. The height of a rocket launched straight up into the air is modeled by the equation
h = 136t – 16t2. In this equation ‘h’ stands for height and ‘t’ stands for time.
At what time or times is the rocket at 800 ft.
Unit 8 Section 8
Objectives

The student will graph and apply exponential equations.
There are many different types of equations and all of them can model certain
contextual situations. The problem below is an illustration for a new type of problem.
Example A
The metaphagic computer virus has just been released. The virus
will infect 4 new computers in the first 24 hours, and then destroy
the hard drive of the infected computer. On day one 4 computers
are infected. On day two each of the 4 infected computers from
day 1 will infect 4 more computers and then be destroyed. The
pattern below shows how the virus spreads.
Day 1
4
Day 2
16
Day 3
64
Day 4
256
Day 5
1024
This pattern can be shown in two other ways. The pattern can be
shown as a product.
Day 1
4
Day 2
(4)(4)
Day 3
(4)(4)(4)
Day 4
(4)(4)(4)(4)
Day 5
(4)(4)(4)(4)(4)
We can see how the products work by multiplying out all the terms.
(4)(4) = 16, (4)(4)(4) = 64, (4)(4)(4)(4) = 256 and so on.
We can turn the product for of the pattern into a pattern with exponents.
Day 1
41
Day 2
42
Day 3
43
Day 4
44
Day 5
45
The equation that models this pattern is y = 4x. If we look closely at
the last representation of the pattern we see a correspondence between
the number of the day and the exponent of the term. In the equation,
the exponent stands for the number of the day, so ‘x’ is the number of
the day. The ‘y’ in the equation represents the number of computers
that would be infected on that day. Once we have the equation, we
can make predictions with the equation. If we want to know the number
of computers infected on day 10, we can substitute as shown below.
y = 4x
y = 410
y = 1048576
This tells us why it is important to have anti-virus software. Over a
million computers could be infected in just 10 days.
The type of equation that modeled our computer virus is called an exponential
equation. Whenever the ‘x’ variable is a power, then the equation is called
exponential. The situations that are modeled by exponential equations are often
known as “exponential growth” problems. These problems are characterized by
phrases such as doubling, tripling, or compounding.
The example below is a case of exponential growth.
Example B
Tom recently posted a video on the internet. The first day 2 people
saw the video. The second day 4 people viewed the video. The
third day there were 8 people, the fourth day had 16 people.
a. Create a pattern that shows how the number of people viewing
the video is growing.
Day1
2
Day 2
4
Day 3
8
Day 4
16
We listed the days and number of views in order.
b. How many people will see the video on day 5 if the pattern
continues?
2 . 16 = 32
Since the pattern was based on doubling, we multiplied by 2.
c. Prove the equation that models this growth is y = 2x.
y
y
y
y
=
=
=
=
21
22
23
24
=
=
=
=
2
4
8
16
If we can generate the sequence or pattern by substituting the
natural numbers as the exponent, then the equation models the growth.
d. How many people will see the video on the 15th day?
y = 215 = 32768 people.
e. Graph the equation or pattern.
Views
20
15
10
5
0
0
1
2
3
4 Days
Exponential patterns can be recognized by the fact that we use multiplication by the
same number to generate the terms. In the first example, we multiplied by 4 to get
from one term to another. In the second example, we multiplied by 2 to get from
term to term.
As with any type of equation, an exponential equation can be graphed by substitution.
The examples below show exponential equations being graphed by substitution.
Example A
x
Graph y = 2x – 1
y
y
y
y
y
y
-2
-1
0
1
2
x
x
y
-2
-1
0
1
2
-.75
-.5
0
1
3
Next we calculate
the y-values for
our table.
First we set
up our table.
Generally we
pick values
from -2 to 2.
Example B
y = 2x – 1
= 2-2 – 1 = -.75
= 2-1 – 1 = -.5
= 20 – 1 = 0
= 21 – 1 = 1
= 22 – 1 = 3
Finally we plot our
points and connect
them to form the
graph.
Graph y = -2(3x) + 6
y
-2
-1
0
1
2
First we set
up our table.
Generally we
pick values
from -2 to 2.
y
y
y
y
y
y = -2(3x) + 6
= -2(3-2) + 6 = 5.7
= -2(3-1) + 6 = 5.3
= -2(30) + 6 = 4
= -2(31) + 6 = 0
= -2(32) + 6 = -12
x
y
-2
-1
0
1
2
5.7
5.3
4
0
-12
Next we calculate
the y-values for
our table.
Finally we plot our points and
connect them to form the graph.
We had to leave out one of the
points because the y-value would
not fit on the grid.
VIDEO LINK: Khan Academy Graphing Exponential Function
Exercises Unit 8 Section 8 Set A
1. A bacteria is being cultured in a lab. In the first hour the culture reached a count of
300. In the second hour the bacteria reached a level of 600, the third hour the count
was 1200 and the fourth hour of the study the count reached 2400.
a. List the pattern we see as the count grows hour by hour.
1st hour
2nd hour
3rd hour
4th hour
b. Describe in words how the count is growing with each hour.
c. What level will the bacteria count reach in the 5th hour?
d. Prove that the bacteria growth is modeled by the equation y = 150(2x).
e. Use the above equation to predict the bacteria count at the 10th hour.
f. Create a bar graph of the pattern.
2. Jason is running for mayor of a large city. He wants to start an email campaign.
He sends an email to 10 of his friends extolling his virtues and making campaign
promises. The email asks each of them to forward the message to 5 more people.
So, the first time the email is sent to 10 people. If each one of these people passes
it on to 5 people the second cycle will have 50 messages sent. If each one of the
50 people forwards the email to 5 more people the third cycle will have 250 emails
sent.
a. List the pattern we see as the number of emails sent grows with each cycle.
1st cycle
2nd cycle
3rd cycle
4th cycle
b. Describe in words how the number of emails is growing with each cycle.
c. How many emails will be sent in the fifth cycle?
d. Prove that the growth of the emails is modeled by the equation y = 2(5x).
e. Use the above equation to predict the number of emails in the 8th cycle.
f. Create a LINE graph of the pattern.
3. Given the visual pattern below, answer the questions that follow.
a. Create a numeric pattern that shows us how the number of squares in each
element is growing.
1st element
2nd element 3rd element
4th element
b. Describe in words how to generate the numeric pattern going from term to
term.
c. How many squares will the 5th element in the pattern have?
d. Prove that the equation y = 3x models the way the numeric pattern grows.
e. Use the above equation to predict how many squares would be in the
11th element of the pattern.
f. Create a LINE graph that shows how the number of squares is growing
in the pattern.
Find the square roots. If there is no real answer then write “no real number”.
4.
100
5.
6.
9
 25
7.
225
Estimate the square roots to the nearest tenth. Check your answers by multiplying
your estimate times itself. Show your work.
8.
9.
41
83
Use your calculator and find the square roots to the nearest hundredth.
10.
123
11.
203
12.
13.
160
Simplify the square roots.
14.
y2
15.
(a  3) 2
16.
(3x  1) 2
2
Evaluate the following. Show your work.
17. Evaluate
2
x 1 + 3
with x = 26
Solve the following equations using the Square Root Principle. Round your answers to the
nearest hundredth when needed. Show your work.
18.
x2 = 144
19. y2 – 12 = 52
20. -2x2 = -24
21. 142 = 6w2 – 8
22. y2 – 2 = -2
23. 1 = z2 + 5
24. (x – 3)2 = 49
25. 2(y – 5)2 + 1 = 19
26. 30 = (a – 4)2
Complete the square then list the vertex and line of symmetry for each quadratic.
27. y = x2 – 10x + 35
28. y = x2 + 12x + 40
29. y = x2 – 7x
Graph the following quadratic equations using substitution. Show your tables and graphs.
30. y = (x – 1)2 + 2
31.
y = (x + 2)2 + 1
Exercises Unit 8 Section 8 Set B
Evaluate the following.
1. Given f(x) = 2x + 5 find f(3)
2. Given f(x) = -(2x)
find f(4)
3. Given g(x) = 2x-1
find g(6)
4. Given g(x) = 3(4x)
find g(2)
5. Given f(x) = 2x
find f(-1)
6. Given g(x) = 2x + 1 find g(0)
7. Given f(x) = 2x – 1
find f(-2)
Graph the following exponential equations using substitution. Show your tables.
8. y = 2x + 1
9. y = -(2x ) + 3
10. y = 3x – 2
11. y = 3(2x ) – 5
12. y = -(3x) + 4
13. y = 2(x+1)
14. Mary tells her two best friends a secret on the first day. On the second day,
each of her two best friends tell three other people the secret. On day three,
each of the these people tell three more people.
a. List the pattern we see as the number of people told grows day by day.
1st day
2nd day
3rd day
4th day
b. Describe in words how the number of people told grows day by day.
c. How many people will be told the secret on the seventh day?
d. Prove that this situation is modeled by the equation y = 2(3x).
f. Create a line graph of the pattern.
Evaluate the following. Show your work.
15. Evaluate
-2
x 3
16. Evaluate
.5
x 6 + 1
with x = 13
with x = 42
Solve the following equations using the Square Root Principle. Round your answers to
the nearest hundredth when needed. Show your work.
17.
x2 = 196
18. y2 – 1 = 39
19. -3x2 = 24
20. 57 = 4w2 – 7
21. y2 + 11 = 11
22. 6 = z2 + 5
23. (x + 5)2 = 25
24. 3(y + 2)2 + 4 = 151
25. 20 = (a + 1)2
Complete the square, then list the vertex and line of symmetry for each quadratic.
26. y = x2 – 16x + 60
27. y = x2 + 8x + 21
28. y = x2 – 9x
Graph the following quadratic equations using substitution. Show your tables and
graphs.
29. y = -2(x – 3)2 + 5
30.
y = (x + 1)2 - 5