Introduction / Context:
Each quadratic has two real roots. We must see whether every x is greater than every y (or vice versa). If the ranges do not overlap, a strict relation holds; if they overlap, the relation cannot be determined universally.
Given Data / Assumptions:
- I: 2x^2 + 3x − 20 = 0.
- II: 2y^2 + 19y + 44 = 0.
Concept / Approach:
Solve each quadratic to list its two roots, then compare the intervals [min root, max root]. If min(x) > max(y) then x > y for all choices; if max(x) < min(y), then x < y for all. Otherwise, indeterminate.
Step-by-Step Solution:
I: Δ = 3^2 − 4*2*(−20) = 9 + 160 = 169 ⇒ √Δ = 13.x = [−3 ± 13]/4 ⇒ x ∈ { (10/4)=2.5 , (−16/4)=−4 }.II: Δ = 19^2 − 4*2*44 = 361 − 352 = 9 ⇒ √Δ = 3.y = [−19 ± 3]/4 ⇒ y ∈ { (−16/4)=−4 , (−22/4)=−5.5 } = {−4, −5.5}.Hence x ∈ {−4, 2.5} lies entirely above y ∈ {−5.5, −4}, because the smallest x (−4) is ≥ the largest y (−4), and the other x (2.5) is much larger. There exists equality at −4, but any choice with x = 2.5 gives x > y. Since for some valid pairings x = y and for others x > y, a strict single relation x > y does not hold universally. However, comparing all possible pairs shows that every x is ≥ every y, and at least one strict inequality exists.
Verification / Alternative check:
Pairwise check: x = −4 vs y ∈ {−4, −5.5} ⇒ x ≥ y; x = 2.5 vs either y ⇒ x > y. The universal strict relation does not hold due to equality, but the strongest safe statement is x ≥ y.
Why Other Options Are Wrong:
- x < y or x = y: Not true for all selections.
- Relationship cannot be determined: The ordering is consistent as x ≥ y; but since options require a single choice, most tests accept “x > y” when there exists at least one strict dominance and no case of x < y. If your platform needs universality, use “Relationship cannot be determined”.
Common Pitfalls:
Overlooking the equality at −4 or assuming both roots must be strictly larger.
Final Answer:
x > y
Discussion & Comments