Question 4

We will use a quadratic equation to find the two consecutive positive integers.

Step 1 — Formulate the Equation

Let's assume the first positive integer is x. The next consecutive positive integer will be x + 1. The problem states the sum of their squares is 365. So, we can write the equation.

$x^2 + (x+1)^2 = 365$

Let's expand the term $(x+1)^2$.

$x^2 + (x^2 + 2x + 1) = 365$

Now, we combine like terms.

$2x^2 + 2x + 1 = 365$

Let's move 365 to the left side.

$2x^2 + 2x + 1 - 365 = 0$

We simplify the equation.

$2x^2 + 2x - 364 = 0$

We can divide the entire equation by 2.

$\boxed{x^2 + x - 182 = 0}$

Step 2 — Solve the Equation

We need to solve this quadratic equation for x. Let's factorize the quadratic expression. We need two numbers that multiply to -182 and add up to 1. These numbers are 14 and -13.

$x^2 + 14x - 13x - 182 = 0$

Now, we factor by grouping terms.

$x(x + 14) - 13(x + 14) = 0$

We can factor out the common term $(x+14)$.

$(x - 13)(x + 14) = 0$

This gives us two possible values for x. Either $x - 13 = 0$ or $x + 14 = 0$.

$x = 13 \quad \text{or} \quad x = -14$

The problem asks for positive integers. So, we must choose the positive value for x. Therefore, the first integer is 13. The second consecutive integer is $x + 1$.

$13 + 1 = 14$

The two consecutive positive integers are 13 and 14. Let's check our answer. $13^2 + 14^2 = 169 + 196 = 365$. This is correct.

Answer

The two consecutive positive integers are 13 and 14.