If √(29 − 12√5) = a + b√5, then the value of (2a + 3b) is:

Let √(29 − 12√5) = a + b√5.

Square both sides:

a² + 5b² + 2ab√5 = 29 − 12√5

Equate rational and irrational parts:

a² + 5b² = 29

2ab = −12 ⇒ ab = −6

Check integer pairs with product −6 that might satisfy a² + 5b² = 29:

• a = 3, b = −2 → a² + 5b² = 9 + 20 = 29 (works)

• a = −3, b = 2 → a² + 5b² = 9 + 20 = 29 (works)

Both pairs satisfy the squared equation, but the principal square root is nonnegative. Evaluate a + b√5 for both:

• 3 − 2√5 ≈ −1.472 (negative), so this does not match the principal square root.

• −3 + 2√5 ≈ 1.472 (positive), so take a = −3, b = 2.

Now compute 2a + 3b = 2(−3) + 3(2) = −6 + 6 = 0.

Answer: 0