Transcript 4.3 and 4.4: Solving Quadratic Equations

Vertex Form
Quadratic Function
in Vertex Form:
y = a(x – h)2 + k
The graph of
y = a(x – h)2 + k is
the graph of y = ax2
translated h units to
the right and k units
up.
Vertex Form
Quadratic Function
in Vertex Form:
y = a(x – h)2 + k
• Vertex: (h, k)
• Axis of Symmetry:
x=h
Exercise 1
What is the y-intercept of the graph of
y = a(x – h)2 + k?
Exercise 2
What are the x-intercepts of the graph of
y = a(x – h)2 + k?
Graphing Quadratic Functions, II
To graph y = a(x – h)2 + k use SRT
transformations.
• Scaling: a times the y-coordinate
• Reflecting: over the x-axis if a < 0
• Translating: Left/Right h units and
Up/Down k units
– Remember that the up/down movement is in
the direction of the sign of k, but for left/right,
it’s the opposite direction as the sign of h.
Exercise 3
Graph
y
1
2
 x  3  5
2
Exercise 4
Graph
y    x  2  3
2
Objective 1b
You will be able to graph a
quadratic function in intercept
form
𝑦 = −0.42 𝑥 − 16.9 𝑥 − 26.9
Intercept Form
Quadratic Function
in Intercept Form:
y = a(x – p)(x – q)
• x-intercepts: p and q
• Vertex: Average of p
and q
• Axis of Symmetry:
x
pq
2
Exercise 5
What is the y-intercept of the graph of
y = a(x – p)(x – q)?
Graphing Quadratic Functions, III
To graph y = a(x – p)(x – q):
1. Use the a-value to determine  or 
2. Plot the x-intercepts p and q
3. Find and plot the vertex (x = average of p
and q)
4. Graph the axis of symmetry
5. Find and plot a couple of other function
values and their reflections
Exercise 6
Graph
y   x  x  4
Exercise 7
Graph
y  2  x 1 x  4
𝑦 = −0.05𝑥 2 + 1.2𝑥 − 1.9
𝑦 = −0.05 𝑥 − 12.4
2
+ 5.7
𝑦 = −0.05 𝑥 − 1.6 𝑥 − 23.1
You will be able to rewrite your
quadratic function in standard,
vertex, or intercept form
Objective 2
Exercise 8a
Write each function in standard form.
1.
y  7  x  6 x 1
2.
y  2  x  3  9
2
Protip #1: Standard Form
To write your quadratic equation in standard
form is really easy. Just FOIL, distribute,
or expand as necessary, and then
combine like terms. Finally, make it look
like this:
y = ax2 + bx + c
Exercise 8b
Write each function in standard form.
1.
y  2  x  5 x  4
2.
y  3  x  5   1
2
Exercise 9a
Write each function in intercept form.
1.
y  x2  5x  14
2.
y  8x2  38x  10
Protip #2: Intercept Form
To write your equation in intercept form, you
have to do some factoring:
1. Factor out GCF.
That includes a −1
if the leading term
is negative
2. Factor the
trinomial that is left
over
y  6 x 2  x  15


y   6x 2  x 15
y  2 x  33x  5
Protip #2: Intercept Form
To write your equation in intercept form, you
have to do some factoring:
3. Each factor has to
look like (x ± p).
This means you
have to divide out
any coefficient of x
from each factor.
Whatever you divide
out becomes part of
the a-value.
y  6 x 2  x  15


y   6x 2  x 15
y  2 x  33x  5
Divide out a 2
Divide out a 3
Protip #2: Intercept Form
To write your equation in intercept form, you
have to do some factoring:
3. Each factor has to
look like (x ± p).
This means you
have to divide out
any coefficient of x
from each factor.
Whatever you divide
out becomes part of
the a-value.
y  6 x 2  x  15


y   6x 2  x 15
y  2 x  33x  5
3 
5

y  6 x   x  
2 
3

Exercise 9b
Write each function in intercept form.
1.
y  x 2  2 x  48
2.
y  12x 2  72x 105
Exercise 10a
Write each function in vertex form.
1.
y  x2  8x  17
2.
y  2x2  24x  25
Protip #3: Vertex Form
To put a quadratic equation in vertex form, you
have to complete the square on only one side of
the equation.
2
y  4x  24x  49
1. Separate the
constant term from
the variable terms
2. Factor out the
leading coefficient
from the variable
terms
y  4 x 2  24x

 49

y  4 x 2  6x  ____  ____ 49
Protip #3: Vertex Form
To put a quadratic equation in vertex form, you
have to complete the square on only one side of
the equation.
y  4x 2  24x  49
3. Complete the square
with the variable
terms
4. Whatever you add
on the inside of the
parenthesis, subtract
on the outside of the
parenthesis.
y  4 x 2  24x

 49

y  4 x 2  6x  ____
36  49
9  ____
Protip #3: Vertex Form
5. Remember that what
you add on the inside
of the parenthesis is
actually multiplied by
the leading coefficient
that you factored out in
Step 2. Make sure to
multiply by this
number before
subtracting on the
outside of the
parenthesis
y  4x 2  24x  49
y  4 x 2  24x

 49

y  4 x 2  6x  ____
36  49
9  ____
y  4  x  3  13
2
Exercise 10b
Write each function in vertex form.
1.
y  x2  6x  3
2.
y  3x 2  9 x  4
Solving Quadratic Equations
Objectives:
1. To solve a quadratic equation by factoring
Warm-Up
If the product of 𝐴 and 𝐵 equals zero, what
must be true about 𝐴 or 𝐵?
Zero Product Property
𝐴∙𝐵 =0
If the product of two
expressions is zero,
then at least of the
expressions equal
zero.
Maybe that’s zero
Or maybe this one’s zero
(Or maybe they’re both zero)
Vocabulary
𝑦-intercepts
𝑥-intercepts
Parabola
Roots (of a function)
Zeros (of a function)
Objective 1
You will be able to solve a
quadratic equation by
factoring
Exercise 1
Use your calculator
to graph the
equation
y = x2 – x – 6.
Where does the
parabola cross the
x-axis?
4
2
-5
5
-2
x-intercepts =
points where
parabola
crosses x-axis
-4
-6
Parabola
The graph of a
quadratic function
is a parabola.
x-intercepts:
where the
parabola
intersects the
x-axis
4
2
-5
5
-2
x-intercepts =
points where
parabola
crosses x-axis
-4
-6
Exercise 2
How many y-intercepts can the graph of the
quadratic function y = ax2 + bx + c have?
– Only one since it’s a function!
How many x-intercepts can the graph of the
quadratic function y = ax2 + bx + c have?
– Two, one, or none!
Exercise 3
If you wanted to find the
y-intercept of a
quadratic function,
what would you do?
Plug in zero for x
and solve for y
Exercise 4
Find the y-intercept of y = x2 – 6x – 7.
Exercise 5
If you wanted to find
the x-intercept of a
quadratic function,
what would you
do?
Plug in zero for y
and solve for x
Exercise 6
Find the x-intercept(s) of y = x2 – 6x – 7.
The problem here is, how do you solve
0 = x2 – 6x – 7 since you can’t just get x by
itself?
The answer is to use the Zero
Product Property
Solving Quadratic Equations
The standard form of a quadratic equation
in one variable is ax2 + bx + c = 0, where a
is not zero.
Solving a quadratic equation in standard
form is the same thing as finding the
x-intercepts of y = ax2 + bx + c.
Solving Quadratic Equations
The standard form of a quadratic equation
in one variable is ax2 + bx + c = 0, where a
is not zero.
We can use the zero product property to
solve certain quadratic equations in
standard form if we can write ax2 + bx + c
as a product of two expressions. To do
that, we have to factor!
Exercise 7
Find the x-intercept(s) of y = x2 – 6x – 7.
Let y = 0
0  x2  6 x  7
0   x  7  x  1
Set each
factor equal
to zero
x7  0
x7
Factor
x 1  0
x  1
x-intercepts: (7, 0) and (−1, 0)
Solving Quadratic Equations
To solve a quadratic
equation, try applying the
zero product property.
Factor
your1
Step
quadratic
Set each
factor
equal to
Step
2
zero and
solve
Exercise 8
Solve 0 = x2 – x – 42.
Same Thing as…
In solving a quadratic
function, you must
find the x-values
that make the
function equal to
zero.
Same as the
roots of the
quadratic
equation
Same as
finding the 𝑥values of the
𝑥-intercepts
Solving
Quadratics
Same as the
zeroes of the
quadratic
function
How Many Solutions?
There can be 2,
1, or 0
solutions to a
quadratic
equation,
depending
upon where it
is in the
coordinate
plane.
Exercise 9
Find the roots of each equation.
1. x2 + 2x = 0
2.
x2 – 12x +36 = 0
3.
x2 + 1 = 0
4.
x2 + 2x + 4 = 0
Exercise 10
Explain why you cannot use the zero
product property to solve every quadratic
equation.
Exercise 11
Find the zeros of each function.
1. y = 16x2 – 4
2.
y = 9x2 + 12x + 4
3.
y = 5x2 + 16x + 3
4.
y = 2x2 + x + 3
Exercise 12
Solve the equation.
1. 4x2 – 17x – 15 = 0
2. 3x2 + 22x + 60 = -14x – 48
Example 13
Find the value(s) of x.
Exercise 14
Find the value of x if the area of the triangle
is 42 square units.
Exercise 15
Use substitution to solve the system of
equations.
Solving Quadratic Equations
Objectives:
1. To solve a quadratic
equation by factoring
Warm-Up
Solve for x.
x2 – 10x + 25 = 35
Perfect Squares?
The previous exercise was an interesting
example, but not every trinomial is a
perfect square trinomial. However, we can
cleverly rearrange the terms to make any
trinomial into a perfect square.
This is called completing the square.
Objective 1
You will be
able to solve
quadratic functions
by completing the square
Algebra Tiles Activity!
In this activity, we will
be using algebra tiles
to learn how to
complete the square.
Don’t worry; unlike
most algebra tile
activities, this one is
actually quite useful.
But first, what is an
algebra tile?
Geometric representation of
x
x2:
x2
x:
x
1
x x
Unit Square:
1
1 1
Algebra Tiles Activity!
1. You’ll need some tiles, so cut them out.
2. Use your tiles to represent x2 + 6x.
x2
x x x x x x
3. What you want to accomplish is to make
your polynomial model into a square by
rearranging your tiles and adding the
correct number of unit squares.
Algebra Tiles Activity!
x2
xxx
xxx
x+3
x+3
1 1 1
1 1 1
1 1 1
To create this square,
put half of your x’s
on one side of the
square and the
2
= (x + 3)
other half of your
2
= x + 6x + 9
x’s on the other
side of the square.
Now fill in the
missing bits with
unit squares.
Algebra Tiles Activity!
4. Now, let’s try again, this time 5. Based on
your
with the following expressions.
investigation,
how could
you complete
the square
given
x2 + bx?
Algebra Tiles Activity!
xxx
1 1 1
1 1 1
1 1 1
= (x + 3)2
= x2 + 6x + 9
half
x2
xxx
x+3
x+3
3
In general then, to
complete the
square, you take
half of the middle
term and then
square it. The
number you get
becomes the
constant term.
Algebra Tiles Activity!
xxx
1 1 1
1 1 1
1 1 1
half
x2
xxx
x+3
x+3
Notice also that the
number you get
from taking half of
the middle term
= (x + 3)2
becomes the
= x2 + 6x + 9
constant of the
3
binomial that gets
squared.
Completing The Square
Exercise 1a
Find the value of c that makes x2 – 26x + c a
perfect square trinomial. The write the
expression as the square of a binomial.
Exercise 1b
Find the value of c that makes x2 + 7x + c a
perfect square trinomial. The write the
expression as the square of a binomial.
Exercise 1c
Find the value of c that the expression a
perfect square trinomial. The write the
expression as the square of a binomial.
1. x2 + 14x + c
2. x2 – 22x + c
3. x2 – 9x + c
Exercise 2
Solve by finding square roots.
1. x2 + 6x + 9 = 36
2. x2 – 10x + 25 = 1
Solving Quadratics
Completing the square
allows you to solve
almost any quadratic
equation, regardless of
whether it is a perfect
square. This is
basically an application
of the Addition Property
of Equality.
x2  8x  6  0
x2  8x ___  6
x2  8x  16  6  16
 x  4
2
 22
x  4   22
x  4  22
Solving Quadratics
Solving a Quadratic Equation by
Completing the Square:
1. Use addition/subtraction to write your
equation in the form x2 + bx = c.
2. Complete the square on the quadratic
side of the equation and add this number
to both sides of the equation.
3. Factor and take the square root.
Exercise 3a
Solve 3x2 – 36x +150 = 0 by completing the
square.
Exercise 3b
Solve the equation by completing the
square.
1. x2 + 6x + 4 = 0
2. x2 – 10x + 8 = 0
3. 2x2 – 4x – 14 = 0
4. 3x2 + 12x – 18 = 0
Exercise 4a
Solve x2 – 9x + 20 = 0 by completing the
square.
Exercise 4b
Solve the equation by completing the
square.
1. x2 + 3x − 5 = 0
2. x2 – 7x + 13 = 0
Exercise 5a
Solve 2x2 – 12x + 7 = 0 by completing the
square.
Exercise 5b
Solve the equation by completing the
square.
1. 3x2 + 24x + 41 = 0
2. x2 – x + 6 = 0
Complete The Square
xxx
1 1 1
1 1 1
1 1 1
= (x + 3)2
= x2 + 6x + 9
half
x2
xxx
x+3
x+3
Objectives:
1. To solve quadratic
equations by
completing the
square
3
Vocabulary
Real Number
Complex Number
Imaginary Unit
Imaginary Number
Pure Imaginary Number
Absolute Value
Complex Plane
Complex Conjugates
Objective 1
You will be able to simplify
square roots of negative
numbers
Exercise 1
Use a graphing
calculator to graph
y = x2 + 1.
What are the xintercepts?
Exercise 2
Solve the quadratic equation x2 + 1 = 0.
The problem here is that −1 is not a real
number since there is no real number that
you can square to get −1.
– This does not mean there is no solution; it’s
more complex than that.
Imaginary Unit
The imaginary unit i can be used to find the
square roots of negative numbers.
𝑖 = −1
𝑖 2 = −1
Exercise 3
A pattern exists as a
result of raising i,
an imaginary
number, to n, an
integer greater than
or equal to 1.
According to the
table, what is the
value of i raised to
the 16th power?
in
Solution
i1
−1
−1
i2
i3
i4
i5
i6
− −1
1
−1
−1
Taking Powers of i
As the previous example shows, there are
only 4 possible values for in.
i1  1  i
2
2
i  1  1
All other powers of 4 just
repeat this pattern.
i 3  i 2  i  1  i  i
i 4  i 2  i 2   1 1  1
So, how do you think you
would evaluate i101?


Taking Powers of i
As the previous example shows, there are
only 4 possible values for in.
i  1  i
2
i 2  1  1
1


i 3  i 2  i  1  i  i
i 4  i 2  i 2   1 1  1
Divide the exponent by 4,
then use the remainder to
find the result:
Remainder
Result
1
i
2
−1
3
−i
0
1
Exercise 4
Evaluate each of the following.
1. i54
2. i120
3. i89
4. i39
Protip: Divisible by 4
How can you tell if a number is divisible by
4?
Any multiple of 100 is divisible by
4.
Note that 100𝑛 is multiple of 4, where 𝑛 ∈ ℤ
100𝑛
100
=
𝑛 = 25𝑛
4
4
Since this is an integer
with no remainder, any
multiple of 100 is divisible
by 4.
Protip: Divisible by 4
How can you tell if a number is divisible by
4?
If the last two digits of a number are
divisible by 4, the whole number is
divisible by 4.
132
100 + 32
100 32
=
=
+
= 25 + 8 = 33
4
4
4
4
Since this is an integer with no remainder,
as long as the last two digits are divisible
by 4, the whole number is divisible by 4.
Protip: Divisible by 4
How can you tell if a number is divisible by
4?
If the last two digits of a number are
divisible by 4, the whole number is
divisible by 4.
228
200 + 28
200 28
=
=
+
= 50 + 7 = 57
4
4
4
4
Since this is an integer with no remainder,
as long as the last two digits are divisible
by 4, the whole number is divisible by 4.
Protip: Divisible by 4
How can you tell if a number is divisible by
4?
If the last two digits of a number are
divisible by 4, the whole number is
divisible by 4.
316
300 + 16
300 16
=
=
+
= 75 + 4 = 79
4
4
4
4
Since this is an integer with no remainder,
as long as the last two digits are divisible
by 4, the whole number is divisible by 4.
Exercise 5
Simplify each of the following.
1.
−36
2.
−13
3.
𝑖 5
2
Negative Square Roots
The Square Root of a Negative Number
Property
1.
2.
Example
If 𝑟 is a positive number,
then −𝑟 = 𝑖 𝑟
𝑖 𝑟
2
−5 = 𝑖 5
2
= −𝑟
𝑖 5 = 𝑖2 ∙ 5
= −5
Exercise 6a
Find the roots of each quadratic equation.
1.
x2 + 1 = 0
2. 2x2 + 18 = −72
Exercise 6b
Solve each equation.
1. x2 = −13
2. x2 + 11 = 3
3. 3x2 – 7 = −31
4. 5x2 + 33 = 3
Objective 2
You will be able to plot
complex numbers in the
complex plane and find their
absolute value
Complex Numbers
A complex number in standard form is
written
𝑎, 𝑏 ∈ ℝ
a  bi
Real Part
Imaginary Part
• All real numbers are complex numbers
– This happens when b = 0
• For imaginary numbers, b ≠ 0.
• For a pure imaginary number, a = 0.
Exercise 7
Draw a Venn
Diagram that
represents the set
of complex
numbers and
includes real,
imaginary, and pure
imaginary numbers.
ℂ: The set of all complex
numbers
Complex Plane
All complex numbers
are essentially
2-dimensional.
– When you graph a
real number, it
appears on a 1-D
number line
Complex Plane
All complex numbers
are essentially
2-dimensional.
– However, complex
numbers have
both a real and an
imaginary part.
Imaginary
Axis
Real Axis
Exercise 8
Plot the complex numbers on the same
complex plane.
1. 4 + 2i
2. −1 + 3i
3. −4i
4. 2 – 2i
Exercise 9
Plot the complex
number 4 + 3i in the
complex plane.
How could we find
its distance from the
origin? In other
words, how could
we find its absolute
value?
Absolute Value
Absolute Value of a Complex Number
The absolute value
of a complex
number z = a + bi,
denoted |z|, is a
nonnegative real
number defined as
z  a 2  b2
Exercise 10a
Find the absolute value of each complex
number.
1. 5 – 12i
2. 17i
Exercise 10a
Find the absolute value of each complex
number.
1. 4 – i
2. −3 – 4i
3. 2 + 5i
4. −4i
Objective 3
You will be able to
add, subtract, multiply,
and divide complex
numbers
Adding and Subtracting
To add or subtract two complex numbers,
simply add or subtract their real and
imaginary parts separately.
Sum
(a + bi) + (c + di) = (a + c) + (b + d)i
Difference (a + bi) – (c + di) = (a – c) + (b – d)i
Exercise 11a
Write the expression as a complex number
in standard form.
1. (12 – 11i) + (-8 + 3i)
2. (15 – 9i) – (24 – 9i)
3. 35 – (13 + 4i) + i
Exercise 11b
Write the expression as a complex number
in standard form.
1. (9 – i) + (−6 + 7i)
2. (3 + 7i) – (8 – 2i)
3. −4 – (1 + i) – (5 + 9i)
Multiplying
To multiply complex numbers, you have to
use a combination of the distributive
property and properties of the imaginary
unit.
2i 3  5i   6i  10i 2  6i 10  1  6i  10  10  6i
Exercise 12a
Write the expression as a complex number
in standard from.
1. −5i(8 – 9i)
2. (-8 + 2i)(4 – 7i)
Exercise 12b
Write the expression as a complex number
in standard from.
1. i(9 – i)
2. (3 + i)(5 – i)
Exercise 13
Multiply and classify the product.
1. (5i)(−5i)
2. (3 + 6i)(3 – 6i)
Dividing
To “divide” complex numbers, you have to
multiply by a complex conjugate
The complex numbers
a + bi and a – bi are
complex conjugates
The product of
complex conjugates is
always a real number
2  3i 1  i  2  3i 1  i  2  2i  3i  3i 2
1  5i




1 i 1 i
2
2
1  i 1  i 
Dividing
To “divide” complex numbers, you have to
multiply by a complex conjugate
The complex numbers
a + bi and a – bi are
complex conjugates
The product of
complex conjugates is
always a real number
2  3i 1  i  2  3i 1  i  2  2i  3i  3i 2
1 5


  i

1 i 1 i
2 2
2
1  i 1  i 
Exercise 14a
Write the quotient in standard form.
3  4i
5i
Exercise 14b
Write each quotient in standard form.
1.
5
1 i
2.
5  2i
3  2i