Let's denote the two consecutive positive integers as $n$ and $n+1$.

The sum of their squares is given by:

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

Expanding and simplifying the equation:

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

$2n^2 + 2n + 1 = 365$

Subtract 1 from both sides:

$2n^2 + 2n = 364$

Divide the entire equation by 2:

$n^2 + n = 182$

Now, rearrange the equation:

$n^2 + n - 182 = 0$

This is a quadratic equation. We can solve it using the quadratic formula:

$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For our equation $n^2 + n - 182 = 0$, the coefficients are:

• $a = 1$
• $b = 1$
• $c = -182$

Substituting these values into the quadratic formula:

$n = \frac{-1 \pm \sqrt{1^2 + 4 \times 1 \times 182}}{2 \times 1}$

$n = \frac{-1 \pm \sqrt{1 + 728}}{2}$

$n = \frac{-1 \pm \sqrt{729}}{2}$

$n = \frac{-1 \pm 27}{2}$

This gives us two solutions:

$n = \frac{26}{2} = 13$

or

$n = \frac{-28}{2} = -14$

Since we are looking for positive integers, $n = 13$.

Thus, the two consecutive positive integers are 13 and 14.

Let's verify:

$13^2 + 14^2 = 169 + 196 = 365$

The sum of the squares is indeed 365.

Would you like more details or have any questions?

Here are 5 related questions:

1. What is the general approach to solving quadratic equations?
2. Can you find two consecutive even integers whose squares sum up to 340?
3. How would the problem change if the sum of cubes was required instead of squares?
4. Can you determine two consecutive negative integers that satisfy the same condition?
5. How does the discriminant help in determining the nature of the roots?

Tip: When solving quadratic equations, always check if the discriminant is a perfect square for easy factorization.