Question

Find two consecutive positive integers whose sum of squares is 365.

Hint:

Always read the question carefully for constraints like "positive," "negative," "integer," or "even/odd." These constraints are crucial for selecting the correct solution from the roots of your equation.

Answer 1

Step 1: Understanding the Concept:
We need to translate this word problem into a single quadratic equation with one variable and then solve it to find the integers.
Step 2: Key Formula or Approach:
Let the two consecutive positive integers be \(x\) and \(x+1\).
The problem states that the sum of their squares is 365. This gives the equation:
\[ x^2 + (x+1)^2 = 365 \] Step 3: Detailed Explanation:
Expand the equation:
\[ x^2 + (x^2 + 2x + 1) = 365 \] Combine like terms:
\[ 2x^2 + 2x + 1 = 365 \] Move all terms to one side to form a standard quadratic equation:
\[ 2x^2 + 2x - 364 = 0 \] Divide the entire equation by 2 to simplify it:
\[ x^2 + x - 182 = 0 \] Now, we factor this quadratic equation. We need to find two numbers that multiply to -182 and add to 1. The factors of 182 are (1, 182), (2, 91), (7, 26), (13, 14). The pair 14 and -13 fits our requirements (14 \(\times\) -13 = -182 and 14 + (-13) = 1).
\[ (x + 14)(x - 13) = 0 \] This gives two possible solutions for \(x\): \(x = -14\) or \(x = 13\).
The problem specifies that we are looking for positive integers, so we discard \(x = -14\).
Therefore, the first integer is \(x = 13\).
The second consecutive integer is \(x + 1 = 13 + 1 = 14\).
Step 4: Final Answer:
The two consecutive positive integers are 13 and 14. We can check the answer: \(13^2 + 14^2 = 169 + 196 = 365\).