Step 1: Square both sides: (√(x + 3))² = 4² x + 3 = 16

Step 2: Solve for x: x = 16 - 3 x = 13

Step 3: Check the solution: √(13 + 3) = √16 = 4 ✓

Answer: x = 13

Step 1: Isolate the radical $\sqrt{3x - 2} = 6$

Step 2: Square both sides

Step 1: Square both sides: (√(2x - 1))² = (x - 2)² 2x - 1 = x² - 4x + 4

Step 2: Rearrange to standard form: 0 = x² - 4x - 2x + 4 + 1 0 = x² - 6x + 5

Step 3: Factor: 0 = (x - 5)(x - 1)

Step 4: Solve: x = 5 or x = 1

Step 5: Check x = 5: √(2(5) - 1) = √9 = 3 5 - 2 = 3 ✓

Step 6: Check x = 1: √(2(1) - 1) = √1 = 1 1 - 2 = -1 ✗ x = 1 is EXTRANEOUS

Answer: x = 5

Step 1: Isolate one radical: √(x + 7) = 2 + √(x - 5)

Step 2: Square both sides: x + 7 = (2 + √(x - 5))² x + 7 = 4 + 4√(x - 5) + (x - 5) x + 7 = x - 1 + 4√(x - 5)

Step 3: Simplify: x + 7 - x + 1 = 4√(x - 5) 8 = 4√(x - 5) 2 = √(x - 5)

Step 4: Square again: 4 = x - 5 x = 9

Step 5: Check x = 9: √(9 + 7) - √(9 - 5) = √16 - √4 = 4 - 2 = 2 ✓

Answer: x = 9

Step 1: Square both sides $x + 7 = (x - 5)^2$ $x + 7 = x^2 - 10x + 25$

Step 1: Cube both sides: (∛(x + 1))³ = 2³ x + 1 = 8

Step 2: Solve for x: x = 7

Step 3: Check: ∛(7 + 1) = ∛8 = 2 ✓

Step 4: Note about odd roots: Unlike even roots, odd roots can have negative radicands For example, ∛(-8) = -2

Answer: x = 7