Transcript Document

Quadratic Functions
Copyright © Cengage Learning. All rights reserved.
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4.4
Solving Quadratic Equations by the Square Root
Property and Completing the Square
Copyright © Cengage Learning. All rights reserved.
Objectives


Solve a quadratic equation using the square
root property.
Solve a quadratic equation by completing the
square.
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Solving from Vertex Form
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Solving from Vertex Form
One characteristic of numbers to be cautious about is that
when we square a real number, squaring will effectively
remove any negative sign.
We know that 5 squared is 25.
Negative 5 squared is also 25.
x2 = 25
(5)2 = 25 (–5)2 = 25
x=5
x = –5
Both must be given as answers
to this equation.
x2 = 25
When using a square root, we
must use the plus/minus symbol
to represent both answers.
x=
x=5
x = –5
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Solving from Vertex Form
When we write a square root, we are referring to the
positive solution, which is called the principal square root.
Therefore,
= 5.
Because the principal square root only accounts for
positive solutions we lose possible negative results, so we
must use a plus/minus symbol () to show that there are
two possible answers.
Using the square root and plus/minus is an example of the
square root property.
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Solving from Vertex Form
With this property in mind, we are going to look at two basic
ways to solve quadratic equations: the square root property
and completing the square. First, we will use the square
root property to solve quadratics when they are given in
vertex form.
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Solving from Vertex Form
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Example 1 – Solving quadratic equations using the square root property
Solve 10 = 2(x – 4)2 – 8.
Solution:
10 = 2(x – 4)2 – 8
+8
+8
Isolate the squared variable expression.
18 = 2(x – 4)2
9 = (x – 4)2
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Example 1 – Solution
3 = x – 4
3 = x – 4 or –3 = x – 4
7=x
cont’d
Use the square root property.
Don’t forget the plus/minus
symbol.
Rewrite as two equations and
solve.
or 1 = x
10 ≟ 2(7 – 4)2 – 8
10 = 10
Check both answers.
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Example 1 – Solution
cont’d
10 ≟ 2(1 – 4)2 – 8
10 = 10
Both answers work.
x = 7 and x = 1 are both valid answers to this equation.
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Completing the Square
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Completing the Square
If a quadratic equation has both a second-degree term and
a first-degree term (ax2 + bx + c), the square root property
cannot be easily used to solve.
If we try to get the squared variable expression by itself, we
will have a variable term in the way of the square root. To
handle this problem, we use a technique called completing
the square.
Completing the square will transform the equation so that it
has a perfect square that can be solved using the square
root property.
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Completing the Square
In general, taking half of the coefficient of the first-degree
term and squaring it will give us a constant that will make
these expressions a perfect square trinomial.
x2 + bx
The constant that will complete
the square:
The factored form will include half of b as one of the terms
in the binomial.
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Completing the Square
We will start the process of completing the square by
simplifying one side of an equation.
This process is easiest when the coefficient of the squared
term is 1. If this coefficient is not 1, we will divide both sides
of the equation by the coefficient to make it 1.
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Example 4 – Solving quadratic equations by completing the square
Solve by completing the square of 5x2 + 30x – 35 = 0.
Solution:
Step 1: Isolate the variable terms on one side of the
equation.
To isolate the variable terms, we move the
constant term to the other side of the equation.
5x2 + 30x – 35 = 0
35
35
5x2 + 30x = 35
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Example 4 – Solution
cont’d
Step 2: If the coefficient of x2 is not 1, divide both sides of
the equation by the coefficient of x2.
Divide both sides by the coefficient, 5.
x2 + 6x = 7
Step 3: Take half of the coefficient of x, then square it. Add
this number to both sides of the equation.
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Example 4 – Solution
cont’d
To complete the square on the left side of the
equation, we add a number that will make the
resulting trinomial a perfect square trinomial.
To find this number, we take half of the coefficient
of x and square it. This number is then added to
both sides of the equation. In this equation, the
coefficient of x is 6, so we get
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Example 4 – Solution
cont’d
Add this constant to both sides of the equation.
x2 + 6x + 9 = 7 + 9
x2 + 6x + 9 = 16
Step 4: Factor the quadratic into the square of a binomial.
The left side will factor as a perfect square.
x2 + 6x + 9 = 16
(x + 3)2 = 16
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Example 4 – Solution
cont’d
Step 5: Solve using the square root property.
(x + 3)2 = 16
Use the square root property.
x+3=4
x+3=4
x + 3 = –4
x=1
x = –7
Rewrite as two equations
and solve.
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Example 4 – Solution
cont’d
Step 6: Check your answers in the original equation.
5(1)2 + 30(1) – 35 ≟ 0
(5 + 30 – 35) ≟ 0
0=0
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Example 4 – Solution
cont’d
(5(–7)2 + 30(–7) – 35) ≟ 0
(5(49) – 210 – 35) ≟ 0
(245 – 210 – 35) ≟ 0
Both answers work.
0=0
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Completing the Square
Completing the square will allow us to solve any quadratic
equation using the square root property.
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Converting to Vertex Form
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Converting to Vertex Form
To put a quadratic function into vertex form, use the
technique of completing the square.
The process is very similar, but instead of adding the
constant that will complete the square to both sides, you
will add and subtract the constant on one side.
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Example 6 – Converting a quadratic function into vertex form
Convert to vertex form.
a. f(x) = 2x2 + 20x – 6
b. g(x) = 5x2 + 3x + 20
Solution:
a. Converting a function to vertex form will be similar to
completing the square, as we did earlier. Because we
are not solving, we will keep everything on one side of
the equation, leaving the function notation alone.
f(x) = 2x2 + 20x – 6
Group the variable terms.
f(x) = (2x2 + 20x) – 6
Factor out the coefficient
of the squared term.
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Example 6 – Solution
cont’d
f(x) = 2(x2 + 10x) – 6
Now we want to complete the square of the variable term
in parentheses. Find the constant that will complete the
square.
52 = 25
Add and subtract 25 inside the parentheses. Then take
out the subtracted 25, keeping the multiplication by 2.
Finally, you can simplify and factor.
f(x) = 2(x2 + 10x + 25 – 25) – 6
Add and subtract 25 inside
the parentheses.
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Example 6 – Solution
f(x) = 2(x2 + 10x + 25) – 2(25) – 6
f(x) = 2(x2 + 10x + 25) – 50 – 6
cont’d
Bring out the subtracted
25, keeping the multiply by
2 from the parentheses.
Simplify and factor.
f(x) = 2(x + 5)2 – 56
b. g(x) = 5x2 + 3x + 20
g(x) = (5x2 + 3x) + 20
Group the variable terms.
Factor out the coefficient
of the squared term.
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Example 6 – Solution
cont’d
Find the constant that will complete the square.
Add and subtract
inside the parentheses.
Bring out the subtracted
keeping the multiply by 5
from the parentheses.
,
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Example 6 – Solution
cont’d
Simplify.
Factor.
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Graphing from Vertex Form
with x-Intercepts
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Graphing from Vertex Form with x-Intercepts
Now that we know how to solve a quadratic in vertex form,
we can find the horizontal intercepts of a quadratic in vertex
form, and these can be part of our graphs.
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Example 7 – Graphing a quadratic function in vertex form including horizontal intercepts
Sketch the graph of f(x) = –1.5(x – 2.5)2 + 45.375. Give the
domain and range of the function.
Solution:
Step 1: Determine whether the graph opens up or down.
The value of a is –1.5. Because a is negative, the
graph will open downward.
Step 2: Find the vertex and the equation for the axis of
symmetry.
Because the quadratic is given in vertex form, the
vertex is (2.5, 45.375).The axis of symmetry is the
vertical line through the vertex x = 2.5.
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Example 7 – Solution
cont’d
Step 3: Find the vertical intercept.
To find the vertical intercept, we make the input
variable zero.
f(x) = –1.5(x – 2.5)2 + 45.375
f(0) = –1.5(0 – 2.5)2 + 45.375
f(0) = –1.5(–2.5)2 + 45.375
f(0) = –1.5(6.25) + 45.375
f(0) = 36
Therefore, the vertical intercept is (0, 36).
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Example 7 – Solution
cont’d
Step 4: Find the horizontal intercepts (if any).
Horizontal intercepts occur when the output
variable is equal to zero, so substitute zero for the
output variable and solve.
Isolate the squared
variable expression.
30.25 = (x – 2.5)2
Use the square root
property.
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Example 7 – Solution
cont’d
 5.5 = x – 2.5
5.5 = x – 2.5
8=x
–5.5 = x – 2.5
Write two equations
and solve.
–3 = x
Therefore, we have the horizontal intercepts (–3, 0)
and (8, 0).
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Example 7 – Solution
cont’d
Step 5: Plot the points you found in steps 2 through 4. Plot
their symmetric points and sketch the graph.
(Find an additional pair of symmetric points if
needed.)
The vertical intercept (0, 36)
is 2.5 units to the left of the
axis of symmetry x = 2.5, so
its symmetric point will be
2.5 units to the right of the
axis of symmetry at (5, 36).
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Example 7 – Solution
cont’d
Connect the points with a smooth curve. The input
variable x can be any real number without causing
the function to be undefined, so the domain of the
function is all real numbers or (
, ).
The graph decreases to
negative infinity, and the
highest output, y = 45.375,
is at the vertex so the range
is (
, 45.375] or
y  45.375.
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