Compare x and y from two quadratic statements (use mapping below): I. 2x^2 + 11x + 14 = 0 II. 4y^2 + 12y + 9 = 0 Mapping: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined.

Introduction / Context:
We are asked to compare x and y when each variable is constrained by a quadratic equation. The key is to find all possible real values of x and y from their respective equations and check whether a single consistent inequality holds in all admissible cases.

Given Data / Assumptions:

• I: 2x^2 + 11x + 14 = 0
• II: 4y^2 + 12y + 9 = 0
• Real roots are intended (these quadratics have real roots).

Concept / Approach:
Factor both quadratics to get explicit roots. Then compare every possible x from I with the unique y from II. If for all admissible x we have the same relation to y, the comparison is decided. Otherwise, it is indeterminate.

Step-by-Step Solution:
I: 2x^2 + 11x + 14 = (2x + 7)(x + 2) ⇒ x ∈ {−7/2, −2}II: 4y^2 + 12y + 9 = (2y + 3)^2 ⇒ y = −3/2Compare: −7/2 = −3.5 < −1.5, and −2 < −1.5Thus, in all cases, x < y.

Verification / Alternative check:
Compute numeric values precisely and reconfirm: x candidates are −3.5 and −2; y is −1.5. Both comparisons show x is strictly less than y.

Why Other Options Are Wrong:
“x > y” and “x = y” are contradicted by the computed values. “Relationship cannot be determined” is wrong because the relationship is the same in all admissible cases.

Common Pitfalls:
Missing the perfect square in II, or overlooking that I has two roots and both must be checked. Failing to test all cases can lead to incorrect indeterminacy.

Final Answer:
x < y