Download Solving Quadratic Equations by the Square Root Property

Chapter 11
Section 1
11.1 Solving Quadratic Equations by the Square Root Property
Objectives
1
Review the zero-factor property.
2
Solve equations of the form x2 = k, where k > 0.
3
Solve equations of the form (ax + b)2 = k, where k > 0.
4
Solve quadratic equations with solutions that are not real numbers.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Solving Quadratic Equations by the Square Root Property
Quadratic Equation
An equation that can be written in the form
ax 2  bx  c  0,
where a, b, and c are real numbers, with a ≠ 0, is a quadratic
equation. The given form is called standard form.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 11.1- 3
Objective 1
Review the zero-factor property.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 11.1- 4
Review the zero-factor property.
Zero-Factor Property
If two numbers have a product of 0, then at least one of the numbers
must be 0. That is, if ab = 0, then a = 0 or b = 0.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 11.1- 5
CLASSROOM
EXAMPLE 1
Solving Quadratic Equations by the Zero-Factor Property
Solve each equation by the zero-factor property.
Solution:
2x2 − 3x + 1 = 0
 2x 1 x 1  0
x 1  0 or 2x 1  0
1
x  1 or x 
2
1 
 ,1
2 
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x2 = 25
x 2  25  0
 x  5 x  5  0
x  5  0 or x  5  0
x  5
or
x5
5,5
Slide 11.1-6
Objective 2
Solve equations of the form
x2 = k, where k > 0.
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Slide 11.1-7
Solve equations of the form x2 = k, where k > 0.
We might also have solved x2 = 9 by noticing that x must be a number
whose square is 9. Thus,
x 9 3
x   9  3.
or
This can be generalized as the square root property.
Square Root Property
If k is a positive number and if x2 = k, then
x  k or x  


k
The solution set is  k , k , which can be written
(± is read “positive or negative” or “plus or minus.”)
 k .
When we solve an equation, we must find all values of the variable that
satisfy the equation. Therefore, we want both the positive and negative
square roots of k.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 11.1-8
CLASSROOM
EXAMPLE 2
Solving Quadratic Equations of the Form x2 = k
Solve each equation. Write radicals in simplified form.
Solution:
z 2  49
x 2  15
z 2  49
z 7
x 2  15
7
2 x 2  90  0
2 x 2  90
x 2  45
x  45
x3 5
3 5
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x  15
 15
3 x 2  8  88
3x 2 96

3
3
x2  32
x4 2
4 2
Slide 11.1-9
CLASSROOM
EXAMPLE 3
Using the Square Root Property in an Application
An expert marksman can hold a silver dollar at forehead level, drop it,
draw his gun, and shoot the coin as it passes waist level. If the coin
falls about 4 ft, use the formula d = 16t2 to find the time that elapses
between the dropping of the coin and the shot.
Solution:
d = 16t2
4 = 16t2
1 2
t
4
By the square root property,
1
t
2
1
or t  
2
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Since time cannot be negative, we
discard the negative solution.
Therefore, 0.5 sec elapses between
the dropping of the coin and the
shot.
Slide 9.1- 10
Objective 3
Solve equations of the form
(ax + b)2 = k, where k > 0.
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Slide 11.1-11
Solve equations of the form (ax + b)2 = k, where k > 0.
In each equation in Example 2, the exponent 2 appeared with a
single variable as its base. We can extend the square root property
to solve equations in which the base is a binomial.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 11.1-12
CLASSROOM
EXAMPLE 4
Solving Quadratic Equations of the Form (x + b)2 = k
Solve (p – 4)2 = 3.
Solution:
p4 3
or
p4   3
p44  3 4
or
p44   34
or
p  4 3
p  4 3
4  3 
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Slide 11.1-13
CLASSROOM
EXAMPLE 5
Solving a Quadratic Equation of the Form (ax + b)2 = k
Solve (5m + 1)2 = 7.
Solution:
5m  1  7
or
5m  1   7
5m  1  7
or
5m  1  7
or
1  7
m
5
1  7
m
5
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 1  7 


 5 
Slide 11.1-14
Objective 4
Solve quadratic equations with
solutions that are not real
numbers.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 11.1- 15
CLASSROOM
EXAMPLE 6
Solve for Nonreal Complex Solutions
Solve the equation.
Solution:
x 2  17
x  17
or
x   17
x  i 17
or
x  i 17
The solution set is
i

17 .
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 11.1- 16
CLASSROOM
EXAMPLE 6
Solve for Nonreal Complex Solutions (cont’d)
Solve the equation.
Solution:
 x  5
2
 100
x  5  100
or
x  5   100
x  5  10i
or
x  5  10i
x  5 10i
or
x  5 10i
The solution set is
5  10i.
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Slide 11.1- 17