Compare x and y (define the answer codes): I. x^2 − 4 = 0 II. y^2 + 6y + 9 = 0 Roots are real (any root from each). Choose: A) x > y B) x < y C) x = y D) Relationship cannot be determined

Introduction / Context:
The first quadratic yields two symmetric roots ±2, and the second is a perfect square with a repeated root at −3. We must compare any root x from I with any root y from II and select the correct relation that always holds.

Given Data / Assumptions:

• I: x^2 − 4 = 0 ⇒ x = 2 or x = −2.
• II: y^2 + 6y + 9 = 0 ⇒ (y + 3)^2 = 0 ⇒ y = −3.

Concept / Approach:
List all possible x values and the (single) y value. Compare systematically. If the smallest possible x is still greater than y, then x > y regardless of choice.

Step-by-Step Solution:

Verification / Alternative check:
Draw a number line: y fixed at −3; both x values lie to the right (−2 and 2), confirming x > y.

Why Other Options Are Wrong:

• x < y or x = y: Neither occurs for any valid selection.
• Relationship cannot be determined: Determined here; inequality holds for all choices.

Common Pitfalls:
Missing that y is a repeated single value (−3) and assuming variability comparable to x. The certainty arises from the fixed y.

Final Answer: