12
ABSOLUTE VALUE
The geometrical meaning of |x − a|
THIS SYMBOL |x| denotes the absolute value of x, which is the number without its sign. |+3| = 3. |−3| = 3.
The following is the purely algebraic definition of |x|:
If x ≥ 0, then |x| = x; if x < 0, then |x| = −x.
(If you are not viewing this page with Internet Explorer 6, your browser may not be able to display the symbol ≥, "is greater than or equal to;" or ≤, "is less than or equal to.")
Geometrically, |x| is the distance of x from 0.
Both 3 and −3 are a distance of 3 units from 0. |3| = |−3| = 3. Distance, in mathematics, is never negative.
Problem 1. Explain the following rules.
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a) |−x| = |x|
Both −x and x are the same distance from 0. Neither one is ever negative.
b) |2 − x| = |x − 2|
2 − x is the negative of x − 2. (Lesson 7). Therefore, according to part a), they are equal.
c) |x|² = x²
We may remove the absolute value bars because the left-hand side is never negative, and neither is the right-hand side.
Problem 3. Evaluate the following.
| a) | |6| = 6 | b) | |−6| = 6 | c) | |0| = 0 | d) | |2 − 7| = 5 | |||
| e) | |8| + |−4| = 8 + 4 = 12 | f) | |−3| − |−2| = 3 − 2 = 1 | |||||||
| g) | 1 − |−1| = 1 − 1 = 0 | h) | −8 + |−7| = −8 + 7 = −1 | |||||||
| i) | −4 |−4| |
= | −4 4 |
= −1 | j) (−4)|−4|= (−4)· 4 = −16 |
Absolute value equations
|a| = 5
What values could a have?
a could be either 5 or −5. For, if we replace a with either of those values, the statement -- the equation -- will be true.
And so, any equation that looks like this --
|a| = b
-- has the two solutions
a = b, or a = −b.
We call a the argument of the absolute value. Either the argument will be b, or it will be −b.
The argument is whatever appears within the vertical bars.
Example. Solve for x:
|x − 2| = 8
Solution. x − 2 is the argument. Either that argument will be 8, or it will be −8.
x − 2 = 8, or x − 2 = −8.
We must solve these two equations. The first implies
x = 8 + 2 = 10.
The second implies
x = −8 + 2 = −6.
These are the two solutions: x = 10 or −6.
Problem 4.
a) An absolute value equation has how many solutions? Two.
b) Write them for this equation: |x| = 4.
x = 4, or x = −4.
Problem 5. Solve for x.
|x + 5| = 4
Solve these two equations:
| x + 5 | = | 4 | x + 5 | = | −4 | ||
| x | = | 4 − 5 | x | = | −4 − 5 | ||
| x | = | −1 | or | x | = | −9 | |
Problem 6. Solve for x.
|1 − x| = 7
| 1 − x | = | 7 | 1 − x | = | −7 | ||
| −x | = | 7 − 1 | −x | = | −7 − 1 | ||
| −x | = | 6 | −x | = | −8 | ||
| x | = | −6 | or | x | = | 8 | |
Problem 7. Solve for x.
|2x + 5| = 9
| 2x + 5 | = | 9 | 2x + 5 | = | −9 | ||
| 2x | = | 9 − 5 | 2x | = | −9 − 5 | ||
| 2x | = | 4 | 2x | = | −14 | ||
| x | = | 2 | or | x | = | −7 | |
Absolute value inequalities
There are two forms of absolute value inequalities. One with less than, |a|< b, and the other with greater than, |a|> b. They are solved differently. Here is the first case.
Example 1. Absolute value less than.
|a| < 3.
For that inequaltiy to be true, what values could a have?
Geometrically, a is less than 3 units from 0.
Therefore,
−3 < a < 3
This is the solution. The inequality will be true if a has any value between −3 and 3.
In general, if an inequality looks like this --
|a| < b
-- then the solution will look like this,
−b < a < b
for any argument a.
Example 2. For which values of x will this inequality be true?
|2x − 1| < 5
Solution. The argument, 2x − 1, will fall between −5 and 5:
−5 < 2x − 1 < 5
To isolate x, first add 1 to each term of the inequality:
−5 + 1 < 2x < 5 + 1
−4 < 2x < 6
Now divide each term by 2:
−2 < x < 3
The inequality will be true for values of x in that interval.
Problem 8. Solve this inequality for x :
|x + 2| < 7
−7 < x + 2 < 7
Subtract 2 from each term:
−7 − 2 < x < 7 − 2
−9 < x < 5
Problem 9. Solve this inequality for x :
|3x − 5| < 10
−10 < 3x − 5 < 10
Add 5 to each term:
−5 < 3x < 15
Divide each term by 3:
| − | 5 3 |
< x < 5 |
Problem 10. Solve this inequality for x :
|1 − 2x| < 9
−9 < 1 − 2x < 9
Subtract 1 from each term:
−10 < −2x < 8
Divide each term by −2. The sense will change.
5 > x > −4
That is,
−4 < x < 5
Example 3. Absolute value greater than.
|a| > 3.
For which values of a will this be true?
Geometrically,
a > 3 or a < −3.
This is form of the solution, for any argument a:
If
|a| > b
then
a > b or a < −b.
Problem 11. Solve for x :
|x| > 5
x > 5 or x < −5.
Problem 12. For which values of x will this be true?
|x + 2| > 7
x + 2 > 7, or x + 2 < −7.
The first equation implies x > 5. The second, x < −9.
Problem 13. Solve for x :
|2x + 5| > 9
2x + 5 > 9, or 2x + 5 < −9.
Solve these two equations:
| 2x | > | 4 | 2x | < | −14 | ||
| x | > | 2 | or | x | < | −7 | |
Problem 14. Solve for x :
|1 − 2x| > 9
1 − 2x > 9, or 1 − 2x < −9.
Solve these two equations. On dividing by −2,
the senses will change.
| −2x | > | 8 | −2x | < | −10 | ||
| x | < | −4 | or | x | > | 5 | |
The geometrical meaning of |x − a|
Geometrically, |x − a| is the distance of x from a.
|x − 2| means the distance of x from 2. And so if we write
|x − 2| = 4
we mean that x is 4 units aways from 2.
x therefore is equal either to −2 or 6.
On the other hand, if we write
|x − 2| < 4
we mean x is less than 4 units away from 2.
This means that x could have any value in the open interval between −2 and 6.
Problem 15. What is the geometric meaning of |x + a|?
The distance of x from −a. For, |x + a| = |x −(−a)|.
|x + 1|, then, means the distance of x from −1. For example, if
|x + 1| = 2,
then x is 2 units away from −1.
x = −3, or x = 1.
Problem 16. What is the geometrical meaning of each of the following? And therefore what values has x?
a) |x| = 2
x is 2 units away from 0. For, |x| = |x − 0|. x therefore is equal to 2 or −2.
b) |x − 3| = 1
x is 1 unit away from 3. x therefore is equal to 2 or 4.
c) |x + 3| = 1
x is 1 unit away from −3. x therefore is equal to −4 or −2.
d) |x − 5| ≤ 2
x is less than or equal to 2 units away from 5. x therefore may take any value in the closed interval between 3 and 7.
e) |x + 5| ≤ 2
x is less than or equal to 2 units away from −5. x therefore may take any value in the closed interval between −7 and −3.
Problem 17. |x − 5| < d. State the geometrical meaning of that, and illustrate it on the number line.
x falls within d units of 5.
Thus x falls in the interval between 5 − d and 5 + d.
5 − d < x < 5 + d
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