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Compare x and y (define the answer codes): I. 16x^2 + 20x + 6 = 0 II. 10y^2 + 38y + 24 = 0 x and y are real roots (any root from each). Choose: A) x > y B) x < y C) x = y D) Relationship cannot be determined

Difficulty: Medium

Correct Answer: x > y

Explanation:


Introduction / Context:
We compare any root x from the first quadratic with any root y from the second. If every x exceeds every y (or vice versa), then the relationship is determined. Compute all roots and compare the ranges systematically.

Given Data / Assumptions:

  • I: 16x^2 + 20x + 6 = 0.
  • II: 10y^2 + 38y + 24 = 0.
  • Real roots only; any choice per equation.


Concept / Approach:
Find both roots of each quadratic precisely. Determine the interval of possible x values and the interval of possible y values. If the minimum x is greater than the maximum y, then x > y for all combinations.


Step-by-Step Solution:

I: Divide by 2 ⇒ 8x^2 + 10x + 3 = 0. Δ = 10^2 − 4*8*3 = 100 − 96 = 4 ⇒ √Δ = 2.x = [−10 ± 2]/(16) ⇒ x ∈ {−12/16 = −0.75, −8/16 = −0.5}.II: Divide by 2 ⇒ 5y^2 + 19y + 12 = 0. Δ = 19^2 − 4*5*12 = 361 − 240 = 121 ⇒ √Δ = 11.y = [−19 ± 11]/(10) ⇒ y ∈ {−30/10 = −3, −8/10 = −0.8}.Compare ranges: min(x) = −0.75, max(x) = −0.5; min(y) = −3, max(y) = −0.8. Since −0.75 > −0.8 and −0.5 > −0.8, every x is greater than every y.


Verification / Alternative check:
Test all combinations explicitly: {−0.75, −0.5} minus {−3, −0.8} ⇒ all differences are positive.


Why Other Options Are Wrong:

  • x < y or x = y: Contradicted by computed values.
  • Relationship cannot be determined: Here it is determined since intervals do not overlap.


Common Pitfalls:
Arithmetic errors in discriminants or sign mistakes when dividing. Always simplify first to reduce errors.


Final Answer:

x > y
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