Compare x and y given two quadratics (use mapping below): I. x^2 − 7x + 12 = 0 II. y^2 − 12y + 32 = 0 Mapping: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined.

Introduction / Context:
Each variable is restricted by a quadratic with two possible roots. To decide a comparison between x and y, we must test all admissible combinations. If different admissible cases lead to different relations, then the overall comparison cannot be fixed.

Given Data / Assumptions:

• I: x^2 − 7x + 12 = 0
• II: y^2 − 12y + 32 = 0

Concept / Approach:
Factor both quadratics. Enumerate all candidate values for x and y and compare. If at least one pair yields equality while another yields strict inequality, then a unique relation does not exist.

Step-by-Step Solution:
I factors: (x − 3)(x − 4) ⇒ x ∈ {3, 4}II factors: (y − 4)(y − 8) ⇒ y ∈ {4, 8}Test combinations:x = 3 with y = 4 or 8 ⇒ x < yx = 4 with y = 4 ⇒ x = yx = 4 with y = 8 ⇒ x < y

Verification / Alternative check:
Because both x = 4, y = 4 (equality) and x = 3, y = 4 (strict inequality x < y) are valid, there is no single definitive comparison valid for all admissible cases.

Why Other Options Are Wrong:
“x > y”, “x < y”, and “x = y” each fail for at least one admissible pair.

Common Pitfalls:
Choosing “x < y” based on one pair (e.g., 3 vs 8) and forgetting that equality also occurs (4 vs 4). All valid cases must be considered.

Final Answer:
Relationship cannot be determined