Absolute Value in Algebra

Absolute Value means ...

... how far a number is from zero:

absolute value 6 either way

"6" is 6 abroad from aught,
and "−half dozen" is besides half-dozen away from cipher.

And then the absolute value of vi is vi,
and the absolute value of −vi is also 6

Absolute Value Symbol

To show we want the absolute value we put "|" marks either side (called "bars"), similar these examples:


vertical bar The "|" can be found just above the enter primal on well-nigh keyboards.

More Formal

More formally we have:

Absolute Value

Which says the absolute value of x equals:

  • x when x is greater than zero
  • 0 when x equals 0
  • −ten when ten is less than nix (this "flips" the number dorsum to positive)

So when a number is positive or zero we leave it alone, when it is negative nosotros modify it to positive using −x.

Example: what is |−17| ?

Well, it is less than zero, then we demand to calculate "−10":

− ( −17 ) = +17

(Because ii minuses make a plus)

Useful Backdrop

Here are some properties of absolute values that tin can exist useful:

  • |a| ≥ 0 ever!

    That makes sense ... |a| can never be less than nix.

  • |a| = √(a2)

    Squaring a makes it positive or zilch (for a equally a Existent Number). Then taking the square root will "undo" the squaring, but exit it positive or zero.

  • |a × b| = |a| × |b|

    Means these are the same:

    • the absolute value of (a times b), and
    • (the absolute value of a) times (the absolute value of b)

    Which tin can too be useful when solving

  • |u| = a is the same equally u = ±a and vice versa

    Which is often the key to solving nearly accented value questions.

Example: Solve |10+2| = 5

Using "|u| = a is the same as u = ±a":

this: |10+ii| = five

is the same as this: x+2 = ±5

Which has two solutions:

ten+2 = −v ten+ii = +5
10 = −7 x = 3

Graphically

Let us graph that instance:

|x+two| = 5

It is easier to graph when we take an "=0" equation, so subtract five from both sides:

|x+2| − 5 = 0

So at present we can plot y=|x+ii|−5 and find where it equals nil.

Here is the plot of y=|x+2|−5, but just for fun permit's brand the graph past shifting information technology around:

|x+2| - 5 = 0
First with y=|x| so shift it left to make
it y=|x+two| 		then shift it downward to make
it y=|x+two|−5

And the two solutions (circled) are −7 and +three.

Absolute Value Inequalities

Mixing Absolute Values and Inequalites needs a little care!

In that location are four inequalities:

< ≤ > ≥
less than less than
or equal to 		greater than greater than
or equal to

Less Than, Less Than or Equal To

With "<" and "≤" we get i interval centered on zip:

Case: Solve |x| < 3

This means the distance from 10 to zero must exist less than 3:

-3 to 3

Everything in between (but not including) -three and iii

Information technology tin exist rewritten every bit:

−3 < 10 < 3

As an interval it can be written as:

(−3, 3)

The aforementioned affair works for "Less Than or Equal To":

Example: Solve |x| ≤ iii

Everything in between and including -iii and three

It can exist rewritten equally:

−iii ≤ x ≤ three

Equally an interval it can be written every bit:

[−3, three]

How well-nigh a bigger example?

Example: Solve |3x-half dozen| ≤ 12

Rewrite it as:

−12 ≤ 3x−6 ≤ 12

Add 6:

−6 ≤ 3x ≤ xviii

Lastly, multiply by (1/3). Because we are multiplying by a positive number, the inequalities volition not change:

−2 ≤ ten ≤ 6

Done!

As an interval information technology can be written as:

[−ii, 6]

Greater Than, Greater Than or Equal To

This is unlike ... nosotros become two split intervals:

Instance: Solve |x| > 3

Information technology looks like this:

|x| > 3

Up to -iii or from 3 onwards

It can be rewritten as

x < −3 or ten > iii

As an interval it can be written as:

(−∞, −three) U (3, +∞)

Careful! Practise non write information technology as

−three > x > 3 no!

"x" cannot exist less than -3 and greater than three at the same fourth dimension

It is really:

10 < −3 or x > iii yes

"x" is less than −3 or greater than 3

The same matter works for "Greater Than or Equal To":

Case: Solve |x| ≥ 3

Can exist rewritten as

x ≤ −3 or 10 ≥ iii

As an interval it tin be written every bit:

(−∞, −three] U [iii, +∞)