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12

ABSOLUTE VALUE

The geometrical meaning

THIS SYMBOL |x| denotes the absolute value of x, which is the number without its sign.  |+3| = 3.   |−3| = 3.

Here is the purely algebraic definition of |x|:

If x 0, then |x| = x; if x < 0, then |x| = −x.

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.   Evaluate the following.

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Do the problem yourself first!

   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

Problem 2.   Explain the following rules.

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.

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, 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 whatever appears within the vertical bars -- a in this example -- the argument of the absolute value.  Either the argument will be b, or it will be −b.

Example 1.   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 3.

a)  An absolute value equation has how many solutions?   Two.

b)  Write them for this equation:  |x| = 4.

x = 4,  or  x = −4.

Problem 4.   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 5.   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 6.   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 2.  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 3.   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.

We must 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 any value of x in that interval.

Problem 7.   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 8.   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 9.   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 4.  Absolute value greater than.

|a| > 3.

For which values of a will this be true?

Geometrically,

a > 3  or  a < −3.

This is the form of the solution, for any argument a:  

If

|a| > b  (and b > 0),

then

a > b  or  a < −b.

Problem 10.   Solve for x :

|x| > 5.

x > 5  or  x < −5.

Problem 11.   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 12.   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 13.   Solve for x :

|1 − 2x| > 9.

1 − 2x > 9,  or   1 − 2x < −9.

Solve those two equations.  On finally dividing by −2,
the senses will change.

−2x > 8   −2x < −10
 
x < −4     or   x > 5.

The geometrical meaning of |xa|

Geometrically,  |xa|  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 14.   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 15.   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 16.   |x − 5| < d.  State the geometrical meaning of that, and illustrate it on the number line.

x falls within d units of 5.

x therefore falls in the interval between 5 − d and 5 + d.

5 − d < x < 5 + d

Next Lesson:  Exponents


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