**Solving Absolute Value Equations**

**Introduction**

There are several ways to solve absolute value
equations. One way we can take advantage
that |x|^2 = x^2, but only if we are dealing with real numbers. To see why, please see the link below:

However, the surest way to solve equations involving
absolute value equations. Steps:

1. Isolate the
expression with absolute value one side of the equation. Hence to the equation
to read something like this: |f(x)| =
g(x).

2. Solve two
equations: f(x) = +g(x) and f(x) =
-g(x).

Some examples of how the method works:

**Example 1:**

|x| + x = 5

|x| = 5 – x

The next step to solve two equations: x = +(5 – x) and x = -(5 – x)

x = +(5 – x)

x = 5 – x

2x = 5

x = 5/2

x = -(5 – x)

x= -5 + x

0 = -5, but 0 ≠ -5, so no solution in this case.

In our final analysis, x = 5/2

**Example 2:**

3*|x-2| = 6 + 4x

|x-2| = 2 + 4/3 * x

Now we need to solve both x – 2 = +(2 + 4/3*x) and x – 2 =
-(2 + 4/3*x)

x – 2 = 2 + 4/3*x

-4 = 1/3 * x

-12 = x

x – 2 = -(2 + 4/3*x)

x – 2 = -2 – 4/3*x

0 = -7/3*x

0 = x

Both valid, so our solutions are x = -12 and x = 0.

**Example 3**:

|x^2 + 5*x + 6| = x

We know the drill, solve x^2 + 5*x + 6 = x and x^2 + 5*x + 6
= -x.

x^2 + 5*x + 6 = x

x^2 + 4*x + 6 = 0

x = (-4 ± √(16 – 24))/2

x = (-4 ± √(-8))/2

x = -2 ± i*√2

x^2 + 5*x + 6 = -x

x^2 + 6*x + 6 = 0

x = ( -6 ± √(36 – 24))/2

x = (-6 ± √12)/2

x = -3 ± √3

Hope you find this helpful, in the near future I want to
tackle other common problems found in algebra.

Eddie

This blog is property of Edward Shore, 2016.