Extraneous Solution - Algebra-2

1. Fundamental Concepts

Definition: An extraneous solution is a value that appears to be a solution when solving an equation but does not satisfy the original equation.
Origin: Extraneous solutions often arise from operations that are not reversible, such as squaring both sides of an equation or multiplying by a variable expression that could be zero.
Verification: It is crucial to check all potential solutions in the original equation to identify and discard any extraneous solutions.

Basic Rule: $$|x| = a \implies x = a \text{ or } x = -a$$

Solving Process: Solve the equation inside the absolute value and consider both positive and negative scenarios.

Verification Step: Substitute potential solutions back into the original equation to ensure they are valid.

Problem: Solve $$|2x + 3| = 7$$

Set up two equations: $$2x + 3 = 7$$ and $$2x + 3 = -7$$
Solve each equation:
- For $$2x + 3 = 7$$: $$2x = 4 \implies x = 2$$
- For $$2x + 3 = -7$$: $$2x = -10 \implies x = -5$$

Validation: Substitute $$x = 2$$ and $$x = -5$$ into the original equation.

Problem: Solve $$|3x - 4| + 2 = 6$$

Isolate the absolute value term: $$|3x - 4| = 4$$
Set up two equations: $$3x - 4 = 4$$ and $$3x - 4 = -4$$
Solve each equation:
- For $$3x - 4 = 4$$: $$3x = 8 \implies x = \frac{8}{3}$$
- For $$3x - 4 = -4$$: $$3x = 0 \implies x = 0$$

Validation: Substitute $$x = \frac{8}{3}$$ and $$x = 0$$ into the original equation.

For $$x = \frac{8}{3}$$: $$|3\left(\frac{8}{3}\right) - 4| + 2 = |8 - 4| + 2 = 4 + 2 = 6$$ ✓
For $$x = 0$$: $$|3(0) - 4| + 2 = |-4| + 2 = 4 + 2 = 6$$ ✓

Isolation Strategy: Always isolate the absolute value term before setting up the equations.
Case Analysis: Consider both the positive and negative scenarios for the absolute value expression.
Verification Practice: Always verify potential solutions by substituting them back into the original equation.