Algebra Rules Review Sheet Page 11
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REVIEW OF ALGEBRA
11
Recall that the symbol
s1
means “the positive square root of.” Thus,
sr
s
|
2
means
s
r
and
s
0
. Therefore,
the equation
sa
2
a
is not always true. It is true
only when
a
0
.
If
a
0
, then
a
0
, so we have
2
a
. In view of (12), we then
sa
have the equation
13
sa
2
a
which is true for all values of .
a
Hints for the proofs of the following properties are given in the exercises.
Properties of Absolute Values
Suppose and are any real numbers and is an
a
b
n
integer. Then
a
a
n
n
1.
2.
3.
ab
a b
b
0
a
a
b
b
For solving equations or inequalities involving absolute values, it’s often very helpful
to use the following statements.
a
a
Suppose
a
0
. Then
_a
x
a
0
4.
x
a
if and only if
x
a
| x |
5.
x
a
if and only if
a
x
a
FIGURE 5
6.
x
a
if and only if
x
a
or
x
a
| a-b |
b
a
For instance, the inequality
x
a
says that the distance from
x
to the origin is less
than , and you can see from Figure 5 that this is true if and only if lies between
a
x
a
and .
a
| a-b |
If
a
and
b
are any real numbers, then the distance between
a
and
b
is the abso-
a
b
lute value of the difference, namely,
a
b
, which is also equal to
b
a
. (See Fig-
ure 6.)
FIGURE 6
Length of a line segment=| a-b |
EXAMPLE 22
Solve
2x
5
3
.
SOLUTION
By Property 4 of absolute values,
2x
5
3
is equivalent to
2x
5
3
or
2x
5
3
So
2x
8
or
2x
2
. Thus,
x
4
or
x
1
.
EXAMPLE 23
Solve
x
5
2
.
SOLUTION 1
By Property 5 of absolute values,
x
5
2
is equivalent to
2
x
5
2
Therefore, adding 5 to each side, we have
3
x
7
2
2
and the solution set is the open interval
3, 7
.
3
5
7
SOLUTION 2
Geometrically, the solution set consists of all numbers whose distance from
x
FIGURE 7
5 is less than 2. From Figure 7 we see that this is the interval
3, 7
.
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