Absolute Value Equations #
Getting a Desired Form #
First we have to make sure our equation is in the form \( |ax+b| = c \). If we have something such as
\[ 2|5x+2|+4 = 18 \]we must first simplify:
\[ \begin{aligned} 2|5x+2|+4 &= 18 \\ 2|5x+2| &= 14 \\ |5x+2| &= 7 \end{aligned} \]Solving for Both Cases #
We break down the previous equation into two cases: the positive and negative. This is because the absolute value function always makes our input positive, erasing any trace of its old sign. The left side could be equal to 7 or -7, as the absolute value of either is 7.
Positive case:
\[ \begin{aligned} |5x+2| &= 7 \\ 5x+2 &= 7 \\ x &= 1 \end{aligned} \]Solution: \( x = 1 \)
Negative case:
\[ \begin{aligned} |5x+2| &= 7 \\ 5x+2 &= -7 \\ x &= \frac{-9}{5} \end{aligned} \]Solution: \( x = \frac{-9}{5} \)
It is good practice to substitute solutions back into the equation, as sometimes extraneous solutions form, as we’ll see below.
Extraneous Solutions #
Try solving the equation:
\[ |1x + 2| = -3 \]You might find:
- Case 1: \( x = -5 \)
- Case 2: \( x = 1 \)
But when you substitute either one, you get:
\[ 3 = -3 \quad \text{ false!} \]Both are extraneous solutions. This is because the equation is set equal to a negative, which is impossible with absolute value. Sometimes this negative can sneak its way into a problem, so it’s good practice to always check your solutions.
Exercises #
- Solve for \( x \): \( |4x+2| = 6 \)
- Solve for \( x \): \( 1|2x+3+4| + 5 = 6 \)
- Solve for \( x \): \( 6|5x+4+3| = 2 + 1 \)
- Solve for \( x \): \( |ax + b| = c \)
- Solve for \( x \): \( K|ax + b| = c \)
- Solve for \( x \): \( -1|ax + b| = c \)
Credits #
All explanations and problems written by Adrian Hernandez Vega, unless otherwise noted.
Prefer video? Watch the full explanation here: