Compare roots from two quadratics (set-based comparison, all cases): I) 3x^2 − 20x − 32 = 0 II) 2y^2 − 3y − 20 = 0 Based on all possible real roots of each equation, determine the correct relationship between x and y.

Introduction / Context:
This is a quantitative comparison question involving two separate quadratic equations. Each quadratic has two real roots; therefore x can take either of two values from the first equation, and y can take either of two values from the second. The task is to compare every possible x with every possible y and decide whether a single, always-true relationship (such as x < y or x ≥ y) can be asserted.

Given Data / Assumptions:

• I) 3x^2 − 20x − 32 = 0
• II) 2y^2 − 3y − 20 = 0
• We must consider all real roots, not just one root from each equation.

Concept / Approach:
Solve both quadratics exactly using the quadratic formula. Then list the root sets for x and y. A determinate comparison (like x > y) is valid only if every possible x is greater than every possible y. If there is any cross-over (some x bigger than some y, but also some x smaller than some y), then no fixed relationship can be established.

Step-by-Step Solution:

Verification / Alternative check:
To assert x ≥ y, the smallest x (−1.333…) must be ≥ largest y (4), which is false. To assert x ≤ y, the largest x (8) must be ≤ smallest y (−2.5), which is false. Hence, no single relation holds for all pairings.

Why Other Options Are Wrong:

• x < y / x ≤ y: Fails because x = 8 and y = −2.5 give x > y.
• x > y / x ≥ y: Fails because x = −1.333… and y = 4 give x < y.

Common Pitfalls:
Comparing only one root from each equation or comparing averages instead of all combinations. In such “both roots” comparisons, check extreme pairings (min x vs max y, max x vs min y) to quickly decide.

Final Answer:
Relationship between x and y cannot be established