Two alternate odd integers (differing by 4) have a product that exceeds three times the smaller by 12. Find the larger integer.

Difficulty: Medium

Correct Answer: 7

Explanation:

Introduction / Context:
This algebra word problem focuses on modeling consecutive odd integers (skipping one: difference 4) and forming a quadratic equation from the given relationship.

Given Data / Assumptions:

Let the smaller odd integer be n; then the larger is n + 4.
Condition: n(n + 4) = 3n + 12.

Concept / Approach:
Translate the statement into an algebraic equation and solve the quadratic. Choose the odd, positive solution consistent with the context.

Step-by-Step Solution:

n(n + 4) = 3n + 12n^2 + 4n − 3n − 12 = 0 ⇒ n^2 + n − 12 = 0Factor: (n + 4)(n − 3) = 0 ⇒ n = 3 or n = −4Valid smaller odd integer: n = 3 ⇒ larger = n + 4 = 7

Verification / Alternative check:
Product = 3 * 7 = 21; three times smaller = 3 * 3 = 9; 21 exceeds 9 by 12, condition satisfied.

Why Other Options Are Wrong:
9 and 5 do not pair with a smaller odd number 4 less that satisfies the condition; 3 is the smaller, not the larger.

Common Pitfalls:
Using difference 2 instead of 4 for alternate odds; sign mistakes when arranging the quadratic; ignoring the requirement that the numbers are odd.

Two alternate odd integers (differing by 4) have a product that exceeds three times the smaller by 12. Find the larger integer.

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