Solving Absolute Value Equations

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:

http://edspi31415.blogspot.com/2016/05/absolute-value-does-x2-x2.html

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:

http://www.sosmath.com/algebra/solve/solve3/s31/s31/s31.html

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.