In each of these questions, two equations (I) and (II) are given. Solve both…

Concept

A quadratic ax² + bx + c = 0 has roots given by x = (−b ± √(b² − 4ac)) / (2a). To compare the two variables, find every root of each equation, then check the relation between each x-root and each y-root. A single answer (>, ≥, <, ≤, or no relation) holds only if it is consistent across ALL pairings.

Application

1. Equation I: 2x2 + 13x − 150 = 0. Discriminant = 132 − 4(2)(−150) = 169 + 1200 = 1369 = 372.

2. So x = (−13 ± 37) / 4, giving x = 24/4 = 6 and x = −50/4 = −12.5. Thus x ∈ {6, −12.5}.

3. Equation II: y2 − 17y + 66 = 0 factors as (y − 6)(y − 11) = 0, so y ∈ {6, 11}.

4. Compare each x with each y: 6 vs 6 gives x = y; 6 vs 11 gives x < y; −12.5 vs 6 gives x < y; −12.5 vs 11 gives x < y.

5. Across all four pairings x is either equal to or less than y, and never greater. The consistent relation is therefore x ≤ y.

Cross-check

Plug the roots back: 2(6)² + 13(6) − 150 = 72 + 78 − 150 = 0 and (6)² − 17(6) + 66 = 36 − 102 + 66 = 0, both valid. The largest x (6) only ties the smaller y (6) and is below the larger y (11), confirming x can equal y but never exceed it, so x ≤ y.