Candidate's Name: \( \qquad \) 2. The sum of the squares of two consecutive odd positive integers is 74 . Find the numbers.

**Solution:** Let's find two consecutive odd positive integers whose squares add up to 74. 1. **Define the Variables:** Let the first odd integer be \( n \). Since the integers are consecutive and odd, the next odd integer will be \( n + 2 \). 2. **Set Up the Equation:** According to the problem, the sum of their squares is 74. Therefore: \[ n^2 + (n + 2)^2 = 74 \] 3. **Expand and Simplify the Equation:** \[ n^2 + (n^2 + 4n + 4) = 74 \] \[ 2n^2 + 4n + 4 = 74 \] Subtract 74 from both sides to set the equation to zero: \[ 2n^2 + 4n + 4 - 74 = 0 \] \[ 2n^2 + 4n - 70 = 0 \] 4. **Simplify the Equation:** Divide the entire equation by 2 to make it simpler: \[ n^2 + 2n - 35 = 0 \] 5. **Solve the Quadratic Equation:** You can solve this quadratic equation using the quadratic formula: \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = 2 \), and \( c = -35 \). Plugging in the values: \[ n = \frac{-2 \pm \sqrt{(2)^2 - 4 \times 1 \times (-35)}}{2 \times 1} \] \[ n = \frac{-2 \pm \sqrt{4 + 140}}{2} \] \[ n = \frac{-2 \pm \sqrt{144}}{2} \] \[ n = \frac{-2 \pm 12}{2} \] This gives two possible solutions: \[ n = \frac{-2 + 12}{2} = \frac{10}{2} = 5 \] \[ n = \frac{-2 - 12}{2} = \frac{-14}{2} = -7 \] 6. **Determine the Positive Integer Solution:** Since the problem asks for positive integers, we take \( n = 5 \). 7. **Find the Consecutive Odd Integer:** The next consecutive odd integer is: \[ n + 2 = 5 + 2 = 7 \] **Answer:** The two consecutive odd positive integers are **5** and **7**.