In each question two equations numbered (I) and (II) are given. You should…

Concept

In a two-equation "quadratic comparison" problem, you first find the roots of each quadratic, then compare every x-root against every y-root. A single fixed relation (x>y, x<y, x≥y, x≤y, or x=y) holds only if the comparison is consistent across all root pairs; if different pairs give opposite relations, no relation can be established.

Application

1. Equation I: 2x2 – 9x + 10 = 0. Factor as (2x – 5)(x – 2) = 0, giving x = 5/2 = 2.5 or x = 2.

2. Equation II: y2 – 12y + 35 = 0. Factor as (y – 5)(y – 7) = 0, giving y = 5 or y = 7.

3. So x belongs to {2, 2.5} and y belongs to {5, 7}. The largest x-value is 2.5 and the smallest y-value is 5.

4. Since the largest x (2.5) is still smaller than the smallest y (5), every x-value is strictly less than every y-value.

Cross-check

• Compare each pair: 2 < 5, 2 < 7, 2.5 < 5, 2.5 < 7 — all four comparisons give x smaller than y.

• The relation is strict (the x-values never equal any y-value) and it never reverses, so the result is the strict less-than relation, not the "or equal" variant and not the no-relation case.

Therefore the established relationship is x < y.