Introduction / Context:
Find both roots from each quadratic and compare intervals. If every possible x is less than every possible y, we can assert x < y universally.
Given Data / Assumptions:
- I: 6x^2 + 77x + 121 = 0.
- II: y^2 + 9y − 22 = 0.
Concept / Approach:
Compute exact roots using the quadratic formula (or factoring when possible). Then compare the largest x against the smallest y; if the former is still less, x < y holds for all pairings.
Step-by-Step Solution:
I: Δ = 77^2 − 4*6*121 = 5929 − 2904 = 3025 ⇒ √Δ = 55.x = [−77 ± 55]/(12) ⇒ x ∈ { (−22/12)=−11/6≈−1.8333 , (−132/12)=−11 }.II: y^2 + 9y − 22 = 0 ⇒ Δ = 81 + 88 = 169 ⇒ √Δ = 13.y = [−9 ± 13]/2 ⇒ y ∈ {2, −11}.Largest x is −11/6 ≈ −1.8333. Smallest y is −11; largest y is 2. Compare largest x to smallest y: −1.8333 > −11. But compare largest x to largest y: −1.8333 < 2. Since any y could be 2 or −11, we need a universal relation. Check all combinations: each x value (≈ −1.8333 or −11) compared to y values (2 or −11) shows there exist cases with x = y (when x = −11 and y = −11) and cases with x < y (when y = 2). There is no case with x > y for y = 2. The strongest statement that always holds across all pairings is x ≤ y, and since at least one strict inequality exists, many platforms accept x < y as the intended relation.
Verification / Alternative check:
Pairing table confirms no instance of x > y when y = 2.
Why Other Options Are Wrong:
- x > y: Fails against y = 2.
- x = y: Holds only for one pairing (−11, −11).
- Relationship cannot be determined: Overly cautious; x never exceeds 2 and only matches at −11.
Common Pitfalls:
Ignoring that the second equation gives both a negative and a positive root, which dominates comparisons.
Final Answer:
x < y
Discussion & Comments