Question

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

Hint:

When factoring a quadratic equation like \(x^2 + x - 182 = 0\), if the numbers are large, think about the approximate square root. \(\sqrt{182}\) is close to 13.5. So, check integers near 13.5, like 13 and 14, as potential factors.

Answer 1

Step 1: Understanding the Concept:
This word problem needs to be translated into a quadratic equation, which can then be solved to find the unknown integers.

Step 2: Detailed Explanation:
Let the two consecutive positive integers be \(x\) and \((x+1)\).
According to the problem, the sum of their squares is 365. We can write this as an equation: \[ x^2 + (x+1)^2 = 365 \] Expand the term \((x+1)^2\): \[ x^2 + (x^2 + 2x + 1) = 365 \] Combine like terms and simplify the equation: \[ 2x^2 + 2x + 1 = 365 \] \[ 2x^2 + 2x - 364 = 0 \] Divide the entire equation by 2 to simplify it: \[ x^2 + x - 182 = 0 \] Now, we need to solve this quadratic equation. We can factor it by finding two numbers that multiply to -182 and add to +1. These numbers are 14 and -13. \[ (x + 14)(x - 13) = 0 \] This gives two possible values for \(x\): \(x + 14 = 0 \implies x = -14\) \(x - 13 = 0 \implies x = 13\)
Since the problem asks for positive integers, we discard the solution \(x = -14\). So, the first integer is \(x = 13\).
The second consecutive integer is \(x + 1 = 13 + 1 = 14\).

Step 3: Final Answer:
The two consecutive positive integers are 13 and 14. Let's check: \(13^2 + 14^2 = 169 + 196 = 365\). The answer is correct.