Radical, Rational, and Absolute Value Equations: Foundations

[Video: Solving Special Equation Types]

Learn techniques for solving equations involving radicals, rational expressions, and absolute values - common question types on the SAT.

Radical Equations

Solving Method

Isolate the radical expression
Square both sides (or raise to appropriate power)
Solve the resulting equation
Check for extraneous solutions

Example: Solve √(2x + 3) = 5

1. Square both sides: 2x + 3 = 25

2. Solve: 2x = 22 → x = 11

3. Check: √(2(11)+3) = √25 = 5 ✓

Important Note

Always check solutions to radical equations, as squaring both sides can introduce extraneous solutions that don't satisfy the original equation.

Rational Equations

Solving Method

Find the Least Common Denominator (LCD)
Multiply both sides by LCD to eliminate denominators
Solve the resulting equation
Check for excluded values

Example: Solve (3/(x-2)) = (5/(x+1))

1. LCD: (x-2)(x+1)

2. Multiply: 3(x+1) = 5(x-2)

3. Solve: 3x + 3 = 5x - 10 → 13 = 2x → x = 6.5

4. Check: x ≠ 2, -1 (valid)

Absolute Value Equations

Solving Method

Isolate the absolute value expression
Create two equations (positive and negative cases)
Solve both equations
Check solutions if needed

Example: Solve |2x - 3| = 7

1. Two cases: 2x - 3 = 7 or 2x - 3 = -7

2. Solve both: x = 5 or x = -2

3. Check: |2(5)-3| = 7 and |2(-2)-3| = 7 ✓

Key Takeaways

For radical equations: isolate, square, solve, check
For rational equations: find LCD, eliminate denominators, solve
For absolute value: consider both positive and negative cases
Always check for extraneous solutions or excluded values
On SAT, these often appear in word problem contexts

Practice Question

What is the solution set for the equation |3x + 2| = x + 4?

A) {-1.5, 1}

B) {1} only

C) {-3, 1}

D) {-1.5} only

Follow these steps:

Set up two cases (without absolute value)
Solve both resulting equations
Check each potential solution in the original equation
Eliminate any extraneous solutions

Step-by-Step Solution:

1. Case 1: 3x + 2 = x + 4 → 2x = 2 → x = 1

Check: |3(1)+2| = 5 and 1+4 = 5 ✓

2. Case 2: 3x + 2 = -(x + 4) → 4x = -6 → x = -1.5

Check: |3(-1.5)+2| = 2.5 and -1.5+4 = 2.5 ✓

3. Both solutions work, so solution set is {-1.5, 1}

The correct answer is A) {-1.5, 1}.