Difficulty: Medium
Correct Answer: x ≥ y
Explanation:
Introduction / Context:When comparing two quadratics, each with two real roots, we must confirm that a proposed inequality holds across all four pairings (each x with each y). A single counterexample invalidates a universal statement like x ≥ y. Here we solve both quadratics and compare the root sets carefully.
Given Data / Assumptions:
Concept / Approach:Compute exact roots via discriminants. Then order the roots numerically and check whether the smallest x is still ≥ the largest y (for x ≥ y), or similar tests for other relations. If equality can occur for some pairing and strict inequality for others, then ≥ (not >) may be the correct consistent relation.
Step-by-Step Solution:
I) D = 38^2 − 4*24*15 = 1444 − 1440 = 4 ⇒ √D = 2x = (−38 ± 2)/(48) ⇒ x ∈ {−36/48, −40/48} = {−3/4, −5/6}II) D = 28^2 − 4*12*15 = 784 − 720 = 64 ⇒ √D = 8y = (−28 ± 8)/(24) ⇒ y ∈ {−20/24, −36/24} = {−5/6, −3/2}Order: x-roots = {−0.75, −0.833…}; y-roots = {−0.833…, −1.5}Check: smallest x = −0.833… ≥ largest y = −0.833… (true, equality), and any x ≥ any y (true)Verification / Alternative check:Test all pairings: (−0.75 ≥ −0.833…), (−0.75 ≥ −1.5), (−0.833… ≥ −0.833…), (−0.833… ≥ −1.5). All hold; at least one pairing yields equality, so x ≥ y is the strongest always-true statement.
Why Other Options Are Wrong:
Common Pitfalls:Overlooking equal roots across equations or assuming strict inequality without testing equality. Always check extremes to decide between > and ≥.
Final Answer:x ≥ y
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