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.
