Download Solving a Quadratic Equation of the Form

Chapter 9
Section 1
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
9.1
1
2
3
Solving Quadratic Equations by the
Square Root Property
Solve equations of the form x2 = k, where
k > 0.
2
Solve equations of the form (ax + b) = k,
where k > 0.
Use formulas involving squared variables.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solving Quadratic Equations by the Square
Root Property
In Section 6.5, we solved quadratic equations by factoring.
Since not all quadratic equations can easily be solved by
factoring, we must develop other methods.
Recall that a quadratic equation is an equation that can be
written in the form
ax 2  bx  c  0
for real numbers a, b, and c, with a ≠ 0. We can solve the
quadratic equation x2 + 4x + 3 = 0 by factoring, using the
zero-factor property.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9.1 - 3
Solving Quadratic Equations Using the Zero-Factor
Property
Solve 2x2 − 3x + 1 = 0 by factoring.
Solution:
 2x 1 x 1  0
x 1  0
x 1
or
2 x 1  0
or
1
x
2
1 
 ,1
2 
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9.1 - 4
Objective 1
Solve equations of the form
2
x = k, where k > 0.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9.1 - 5
Solve equations of the form x2 = k, where k > 0.
We can solve equations such as x2 = 9 by factoring as follows.
x2  9
x2  9  0
 x  3 x  3  0
 x  3  0 or  x  3  0
x  3
x 3
We might also have solved x2 = 9 by noticing that x must be a
number whose square is 9. Thus,
x   9  3
or
x 9 3
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9.1 - 6
Solve equations of the form x2 = k, where
k > 0. (cont’d)
This can be generalized as the 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
 k . (± is read “positive or negative” or “plus or
minus.”)
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 © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9.1 - 7
EXAMPLE 1
Solving Quadratic Equations of
the form x2 = k
Solve each equation. Write radicals in simplified form.
z 2  49
2
z
 49
Solution:
7
z 7
x 2  169
x2  169

x 2  12
x 2  12
x2 3
3 x 2  8  88
2
3x
96

3
3
x2  32
x4 2
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
2 3
4 2
Slide 9.1 - 8
Objective 2
Solve equations of the form
2
(ax + b) = k, where k > 0.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9.1 - 9
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 © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9.1 - 10
EXAMPLE 2
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 
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9.1 - 11
EXAMPLE 3
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  1  7  1
or
5m  1  1   7  1
5m 1  7
or

5
5
1  7
or
m
5
 1  7 


 5 
5m 1  7

5
5
1  7
m
5
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9.1 - 12
EXAMPLE 4
Recognizing a Quadratic
Equation with No Real Solutions
Solve (7z + 1)2 = –1.
Solution:
7 z  1  1
or
7 z  1   1

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9.1 - 13
Objective 3
Use formulas involving squared
variables.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9.1 - 14
EXAMPLE 5
Finding the Length of a Bass
L2 g
Use the formula, w 
, to approximate the length
1200
of a bass weighing 2.80 lb and having girth 11 in.
Solution:
2
L 11
1200  2.80 
1200 
1200
3360 L211

11
11
305.5  L2
L  17.48
The length of the bass is approximately 17.48 in.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9.1 - 15