The sum of the squares of two consecutive positive odd integers is (394). What is the smaller integer?
Answer and explanation
Correct answer: (13)
Let the smaller positive odd integer be x. Consecutive odd integers differ by 2, so the other integer is x+2. Their squared sum gives x²+(x+2)²=394. Expanding and simplifying produces 2x²+4x+4=394, hence x²+2x−195=0. Factoring gives (x+15)(x−13)=0, so x=13 or x=−15. Since the integers are positive, x=13 is the valid value.
Thus the smaller integer is 13, and option B is correct. Checking confirms the result: 13²+15²=169+225=394. The negative algebraic root is rejected because the question asks for positive integers. The supplied explanation reaches the correct answer, although it skips the intermediate division by 2 when displaying the simplified quadratic equation.
Frequently asked questions
What is the correct answer to this question?
(13)
Why is this the correct answer?
Let the smaller positive odd integer be x. Consecutive odd integers differ by 2, so the other integer is x+2. Their squared sum gives x²+(x+2)²=394. Expanding and simplifying produces 2x²+4x+4=394, hence x²+2x−195=0. Factoring gives (x+15)(x−13)=0, so x=13 or x=−15. Since the integers are positive, x=13 is the valid value.
Thus the smaller integer is 13, and option B is correct. Checking confirms the result: 13²+15²=169+225=394. The negative algebraic root is rejected because the question asks for positive integers. The supplied explanation reaches the correct answer, although it skips the intermediate division by 2 when displaying the simplified quadratic equation.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Word Problems and Applications.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
