Solving for Consecutive Even Numbers

Let the two consecutive even natural numbers be represented as $2n$ and $2n + 2$, where $n$ is a natural number.

Formulating the Equation

The problem states that the sum of the squares of these two numbers is 3700. We can write this as an equation:

$ (2n)^2 + (2n + 2)^2 = 3700 $

Solving the Quadratic Equation

1. Expand the equation: $4n^2 + (4n^2 + 8n + 4) = 3700$.
2. Combine like terms: $8n^2 + 8n + 4 = 3700$.
3. Simplify the equation: $8n^2 + 8n - 3696 = 0$.
4. Divide the entire equation by 8: $n^2 + n - 462 = 0$.
5. Factor the quadratic equation. We look for two numbers that multiply to -462 and add to 1. These numbers are 22 and -21.
6. The factored form is $(n + 22)(n - 21) = 0$.
7. This gives two possible values for $n$: $n = -22$ or $n = 21$.
8. Since we are looking for natural numbers, $n$ must be positive. Therefore, we choose $n = 21$.

Finding the Numbers and Their Sum

Using $n = 21$:

• The first even number is $2n = 2 \times 21 = 42$.
• The second consecutive even number is $2n + 2 = 44$.

Let's check: $42^2 + 44^2 = 1764 + 1936 = 3700$. The numbers are correct.

The sum of these two numbers is $42 + 44 = 86$.