Absolute Value: Notation, Expressions, Equations

What Does Absolute Value Mean?

The term “Absolute Value” refers to the magnitude of a quantity without regard to sign. In other words, its distance from zero expressed as a positive number.

The notation used to indicate absolute value is a pair of vertical bars surrounding the quantity, sort of like a straight set of parentheses. These bars mean: evaluate what is inside and, if the final result (once the entire expression inside the absolute value signs has been evaluated) is negative, change its sign to make it positive and drop the bars; if the final result inside the bars is zero or positive, you may drop the bars without making any changes:

\lvert ~1-4~ \rvert\\*~\\*=~\lvert ~-3~ \rvert\\*~\\*=~3

Another example is:

\lvert ~4-1~ \rvert\\*~\\*=~\lvert ~3~ \rvert\\*~\\*=~3

Note that absolute value signs do not instruct you to make “all” quantities inside them positive. Only the final result, after evaluating the entire expression inside the absolute value signs, should be made positive.

\lvert ~1-4~ \rvert~\ne~\lvert ~1+4~\rvert~~\text{ Do not make this mistake!}

Absolute Value expressions that contain variables

Just as with parentheses, absolute value symbols serve as grouping symbols: the expression inside the bars must be evaluated and expressed as either zero or a positive quantity before the bars may be dropped. While this is fairly straightforward when working with constant values, as shown above, what happens when a pair of absolute value signs contains a variable?

\lvert x-4 \rvert

In such a case, we don’t know whether the absolute value will have any effect on the quantity until we know whether “x-4” evaluates to a negative value or not. And we won’t know that until we know the value of “x”. Once we do know a value for “x”, if the expression within the bars is not negative, we can drop the bars; if it is negative, we must change its sign (negating it, making it positive) before dropping the absolute value signs.

This same process can be also be used to solve absolute value equations.

Solving Equations Involving Absolute Values

Equations that involve one or more absolute value expressions can be solved by breaking them into piecewise functions. Each “piece” will start and end at a point where an expression inside absolute value bars changes sign. For example, if we are working with:

6~=~\lvert x-4\rvert

We can break this equation up into two equations that are part of a single piecewise function, neither of which involves an absolute value sign. The resulting piecewise function will have the exact same graph as the original function. How? By first identifying where the expression(s) inside the absolute value bars change(s) sign.

For this problem, the sign of the result inside the absolute value signs changes at x = 4 (set the expression inside the absolute value signs equal to 0, then solve for x). Let’s consider what effect the absolute value function has when x takes on values around 4.

when x < 4, “x-4” will be negative, and the absolute value function will change the sign of the result
when x = 4, “x-4” will equal zero, and the absolute value function will not change the result
when x > 4, “x-4” will be positive, and the absolute value function will not change the result

Therefore, we can group the x = 4 and x > 4 situations into a single “piece”, since neither of them will need to change the sign of the result. This analysis is reflected in a piecewise re-definition of the above function shown below.  Note that the absolute value signs have been replaced by parentheses below, with a negative sign before the left parentheses in the situation when the sign had to be changed:

\begin{cases}6~=~-(x-4),~ x<4\\*~\\*6~=~~~(x-4),~x\geq 4 \end{cases}

 This is now a set of equations each of which can be solved using familiar algebra techniques. When that process has been completed, double-check that the solution satisfies the domain restriction (restriction on possible values of x):

~~~~6~=~-(x-4),~ \text{when } x<4\\*~\\*~~~~6~=~-x+4\\*~\\*~~~~2~=~-x\\*~\\*-2~=~x~~~\text{which meets the condition }x<4

So, the first solution to the original problem is x = -2.  The second possible solution comes from solving the second “piece”:

~~6~=~(x-4),~ \text{when } x\geq 4\\*~\\*~~6~=~x-4\\*~\\*10~=~x~~~\text{which meets the condition }x \geq 4

So, the second solution is x = 10.  If you substitute both of these values, -2 and 10, for x in the original equation that contained the absolute value expression, you can determine for yourself that both values lead to true statements, therefore both are solutions to this equation.

Solving equations involving multiple absolute values

If an equation contains multiple absolute value expressions, each absolute value expression must be considered independently to determine the points at which the piecewise function will change to a different function definition.  For example:

6~=~\lvert x-4\rvert +\lvert 2-x\rvert

The left-hand absolute value above starts/stops having an effect at x = 4, and the right-hand one at x = 2. This problem therefore has three regions we must consider: x less than 2, x between 2 and 4, and x greater than 4. Think carefully about the domain restrictions to ensure that they do not overlap, and also think carefully about whether each expression that was inside absolute value signs needs to be negated for all x values within this domain (or not):

\begin{cases}6~=~-(x-4)+(2-x),~ x<2\\*~\\*6~=~-(x-4)-(2-x),~2\leq x\leq 4 \\*~\\*6~=~~~(x-4)-(2-x),~x>4\end{cases}

For this problem:

  • when x is less than 2 the first absolute value expression will always be negative, and thus must be negated for all x values less than 2.  The second absolute value expression will always be positive when x is less than 2, so it never needs to be negated within this domain.
  • when x is equal to 2 or 4, one or the other of the absolute value expressions will equal zero. In this case, it will not matter if they are negated or not, as negative zero is still zero. Therefore, we can consider the cases when x equals 2 or 4 at the same time as the cases when x lies between 2 and 4 (below).
  • when x is between 2 and 4, the first absolute value expression will still always be negative, and must therefore be negated. However, the second absolute value expression will also always be negative, and thus must also be negated for this “piece” of our piecewise function.
  • when x is greater than 4, the first absolute value expression will always be positive, so the absolute value signs can be dropped without negating the expression. The second absolute value expression will always be negative when x is greater than 4, so it must be negated when converting the absolute value signs into parentheses.

We now have three linear equations that can be solved to arrive at the solution to the original equation. In solving these equations, then verifying that the answers satisfy the accompanying domain restrictions, you should find that the answers are 0 and 6. Note that no value of x between 2 and 4 can satisfy the middle equation, which when simplified produces a statement that is always false (such as 6 = 2).

Invalid Solutions

Suppose you are asked to solve an equation such as

-2~=~\lvert x-4 \rvert

If you follow the procedure outlined above, you will end up with:

\begin{cases}-2~=~-(x-4),~ x<4\\*~\\*-2~=~~(x-4),~x\geq 4 \end{cases}

The solution to the first equation is:

-2~=~-x+4,~ x<4\\*~\\*-6~=~-x\\*~\\*~~~~6~=~x

And the solution to the second is:

-2~=~~x-4,~x\geq 4\\*~\\*~~~2~=~~x

Note that in this case, both solutions violate their associated domain restrictions, therefore there is no solution to this problem. This makes intuitive sense if you revisit the original problem… if the result of an absolute value expression must always be zero or a positive number, how could that result ever equal a negative two?

Non-linear Absolute Value Equations

Non-linear absolute value equations may be solved using the same approach as outlined above. Some non-linear situations will result in multiple solutions for each equation, each of which should be checked against the domain restriction (or tested by substituting it into the original absolute value equation).

Advertisements

Published by

Whit Ford

Math Tutor in Yarmouth, Maine

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s