Consecutive positive odd integers (sum of squares = 290): Find two consecutive positive odd integers whose squares add up to 290.

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Solve this multiple-choice question and choose the correct option.

Accepted Answer

Introduction / Context:Let the odd integers be n and n + 2. Translating the statement “their squares add to 290” yields a quadratic equation in n. Solving it produces the valid positive odd integers. This is a standard modeling step from words to algebra.Given Data / Assumptions: Integers are consecutive and odd: n and n + 2. n > 0. n^2 + (n + 2)^2 = 290. Concept / Approach:Expand, simplify to a quadratic in n, and solve. Discard negative solutions since the integers must be positive. Step-by-Step Solution: n^2 + (n + 2)^2 = 290 ⇒ n^2 + n^2 + 4n + 4 = 290.2n^2 + 4n − 286 = 0 ⇒ divide by 2 ⇒ n^2 + 2n − 143 = 0.Discriminant D = 2^2 + 4*143 = 4 + 572 = 576 ⇒ √D = 24.n = [−2 ± 24]/2 ⇒ n = 11 or n = −13 (reject the negative).Thus the integers are 11 and 13. Verification / Alternative check:11^2 + 13^2 = 121 + 169 = 290 as required. Why Other Options Are Wrong: 13, 15: 169 + 225 = 394, not 290. 9, 11: 81 + 121 = 202, not 290. None of these: Incorrect since 11 and 13 satisfy the condition. Common Pitfalls:Forgetting that consecutive odds differ by 2 (not 1), or arithmetic mistakes while expanding squares. Final Answer: 11, 13

Consecutive positive odd integers (sum of squares = 290): Find two consecutive positive odd integers whose squares add up to 290.

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