Comparison of x and y from two quadratics (use mapping below): I. 16x^2 + 20x + 6 = 0 II. 10y^2 + 38y + 24 = 0 Mapping: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined.

Introduction / Context:
The task is to compare the possible values of x and y when each variable must satisfy its own quadratic. Because a quadratic can have two roots, we must check all admissible x and y values and see if a single consistent relation holds in every case.

Given Data / Assumptions:

• I: 16x^2 + 20x + 6 = 0
• II: 10y^2 + 38y + 24 = 0

Concept / Approach:
Factor (or reduce) each quadratic to find its real roots. Then compare every x-root with every y-root. If for all combinations x is greater than y, we can conclude x > y unequivocally.

Step-by-Step Solution:
I: Divide by 2 → 8x^2 + 10x + 3 = 0 = (4x + 3)(2x + 1)So x ∈ {−3/4 = −0.75, −1/2 = −0.5}II: Divide by 2 → 5y^2 + 19y + 12 = 0 = (5y + 4)(y + 3)So y ∈ {−4/5 = −0.8, −3}Compare: each x (−0.75, −0.5) is greater than each y (−0.8, −3).

Verification / Alternative check:
Numerically: max(y) = −0.8, min(x) = −0.75; since −0.75 > −0.8, even the least x exceeds the greatest y. Hence x > y for all admissible cases.

Why Other Options Are Wrong:
“x < y” and “x = y” are contradicted by direct comparison; “Relationship cannot be determined” is invalid because the relationship is consistent across all root pairs.

Common Pitfalls:
Stopping after one comparison. You must ensure the relation holds for every root combination.

Final Answer:
x > y