Equations with absolute value

To solve an equation that contains the absolute value (or modulus) of the unknown variable x, written in the form $$ | P(x) | = Q(x) $$, the absolute value must be eliminated by rewriting the equation in an equivalent form.

There are two standard techniques, each suited to a specific type of problem.

Technique 1

If the equation is simple and the constant term n∈R is an arbitrary real number,

$$ |P(x)| = n $$

the solution depends entirely on the sign of the constant term n.

• If n>0, we can apply the fundamental property of absolute values |x|=|y|⇔x=±y.
  
  Since |P(x)|=n, it follows that P(x)=n or P(x)=-n, that is $$ |P(x)|=\pm n $$ Solving these equations yields all solutions of the original problem.

  Example. Consider the equation $$ |x+2| = 5 $$ The constant term is n=5, which is positive.
  Therefore, there are two cases to consider: x+2=±5, namely $$ x+2=5 $$ and $$ x+2=-5 $$ Solving each equation gives $$ x = 5-2 = 3 $$ and $$ x = -5-2 = -7 $$ Hence, the solutions of the original equation |x+2|=5 are x=3 and x=-7.

• If n<0, the equation has no solutions, because the absolute value of any real number is always nonnegative, that is |P(x)|≥0. In this case, the condition for existence is not satisfied, and the equation admits no solution.

  Example. Consider the equation $$ |x+2| = -5 $$ The constant term is n=-5, which is negative. For any real value of x, the absolute value is either zero or positive. The smallest possible value of the absolute value is zero, which occurs when x=-2. Therefore, the equation has no solutions.

Technique 2

If the equation involves one or more algebraic expressions,

$$ |P(x)| = Q(x) $$

the solution proceeds according to the following steps.

1. Determine the sign of the expression P(x) inside the absolute value.
2. Once the intervals in which P(x) is nonnegative have been identified, split the problem into two systems of equations. One system corresponds to the case P(x)≥0: $$ \begin{cases} P(x) = Q(x) \\ \\ P(x) \ge 0 \end{cases} $$ The second system corresponds to the case P(x)<0. In this situation, by the definition of absolute value, the opposite expression -P(x) must be used: $$ \begin{cases} -P(x) = Q(x) \\ \\ P(x) \lt 0 \end{cases} $$
3. The union of the solution sets of the two systems provides the complete solution set of the original equation |P(x)|=Q(x).

A worked example

In the following equation, the unknown x appears inside the absolute value,

$$ |x-4| = 2x-1 $$

First, analyze the sign of the expression inside the absolute value to determine when it is positive, zero, or negative.

$$ x-4 \ge 0 $$

The expression P(x)=x-4 is nonnegative precisely when x≥4.

$$ x \ge 4 $$

We now construct two systems, one for x≥4 and one for x<4.

The first system

If x≥4, then P(x)=Q(x) applies directly, without the absolute value, because P(x) is zero or positive.

When the expression inside the absolute value is nonnegative, the absolute value can be removed without affecting the associated inequality.

$$ \begin{cases} x \ge 4 \\ \\ x-4=2x-1 \end{cases} $$

$$ \begin{cases} x \ge 4 \\ \\ x-2x=-1+4 \end{cases} $$

$$ \begin{cases} x \ge 4 \\ \\ -x=3 \end{cases} $$

$$ \begin{cases} x \ge 4 \\ \\ x=-3 \end{cases} $$

This system has no solutions, because x=-3 does not satisfy the constraint x≥4.

The second system

If x<4, the equation must be rewritten as -P(x)=Q(x), again without the absolute value.

In this case, the opposite expression -P(x) is used because P(x), without the absolute value, is negative.

$$ \begin{cases} x \lt 4 \\ \\ -(x-4)=2x-1 \end{cases} $$

$$ \begin{cases} x \lt 4 \\ \\ -x+4=2x-1 \end{cases} $$

$$ \begin{cases} x \lt 4 \\ \\ -x-2x=-1-4 \end{cases} $$

$$ \begin{cases} x \lt 4 \\ \\ -3x=-5 \end{cases} $$

$$ \begin{cases} x \lt 4 \\ \\ -3x \cdot (-1) = -5 \cdot (-1) \end{cases} $$

$$ \begin{cases} x \lt 4 \\ \\ 3x=5 \end{cases} $$

$$ \begin{cases} x \lt 4 \\ \\ x=\frac{5}{3} \end{cases} $$

This system has a solution, since x=5/3 satisfies the condition x<4.

In summary, the first system has no solutions, whereas the second system has exactly one solution.

Therefore, the original equation |x-4| = 2x-1 has a single solution, namely x=5/3.

$$ x = \frac{5}{3} $$

Check. Substitute x=5/3 into the original equation: $$ |x-4| = 2x-1 $$ $$ |\frac{5}{3}-4| = 2(\frac{5}{3})-1 $$ $$ |\frac{5-12}{3}| = \frac{10-3}{3} $$ $$ |\frac{-7}{3}| = \frac{7}{3} $$ $$ \frac{7}{3} = \frac{7}{3} $$ The equality holds, confirming that x=5/3 is indeed a solution.

And so on.