Find two consecutive positive integers, sum of whose squares is 365.
Step-by-Step Solution
Key Concept: Let the two consecutive integers be \(n\) and \(n+1\). Form a quadratic equation using the given condition, then solve the quadratic by factoring or using the discriminant method.
1. Assume the integers: Let the smaller integer be \(n\). Then the next integer is \(n+1\).
2. Write the condition: \[n^{2}+(n+1)^{2}=365\]
3. Expand and simplify:
\[n^{2}+n^{2}+2n+1=365\]
\[2n^{2}+2n+1=365\]
Subtract 365 from both sides:
\[2n^{2}+2n-364=0\]
4. Divide by 2 to obtain a simpler quadratic:
\[n^{2}+n-182=0\]
5. Solve the quadratic using the discriminant method.
- Discriminant \(D = b^{2}-4ac = 1^{2}-4(1)(-182) = 1+728 = 729\).
- Since \(\sqrt{729}=27\), the roots are:
\[n = \frac{-b \pm \sqrt{D}}{2a}=\frac{-1 \pm 27}{2}\]
6. Select the positive integer root:
\[n = \frac{-1+27}{2}=\frac{26}{2}=13\]
(The other root \(\frac{-1-27}{2}=-14\) is negative and is discarded.)
7. Find the consecutive integer: \(n+1 = 13+1 = 14\).
8. Verification:
\[13^{2}+14^{2}=169+196=365\] which satisfies the given condition.
Hence, the required consecutive positive integers are \(13\) and \(14\).
Correct Answer: The two consecutive positive integers are \(13\) and \(14\).