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.
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:
|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:
Solve these two equations. On dividing by −2,
the senses will change.
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
Next Lesson: Exponents
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