Radical equation (repaired interpretation): If [√(3 + x) + √(3 − x)] / [√(3 + x) − √(3 − x)] = 2, then find the value of x.

Introduction / Context:
The database expression lacked parentheses around radicals, which made the denominator zero. Applying the Recovery-First Policy, we repair it to the standard radical form. The problem now involves manipulating a ratio of sums and differences of square roots and solving for the variable under square roots.

Given Data / Assumptions:

• (√(3 + x) + √(3 − x)) / (√(3 + x) − √(3 − x)) = 2
• 3 ± x ≥ 0, hence −3 ≤ x ≤ 3 for real radicals.

Concept / Approach:
Let a = √(3 + x) and b = √(3 − x). The equation becomes (a + b)/(a − b) = 2. Solve for a in terms of b, then square to eliminate radicals and solve for x, ensuring the solution lies within the domain and satisfies the original equation (no extraneous root).

Step-by-Step Solution:

Verification / Alternative check:
Check domain: x = 12/5 = 2.4 ∈ [−3, 3]. Substitute back numerically to confirm the ratio is very close to 2 (rounding errors may occur if not exact arithmetic). The algebraic derivation guarantees equality.

Why Other Options Are Wrong:
5/12, 5/7, 7/5, 3/2 do not satisfy the transformed condition a = 3b when substituted back through the radical definitions.

Common Pitfalls:
Forgetting parentheses leading to division by zero; squaring without checking the original constraint can also introduce extraneous values—always verify in the given equation and domain.

Final Answer:
12/5