Let's solve the problem step by step.

Given:

• The product of two consecutive even numbers is 125.

Let the two consecutive even numbers be $x$ and $x + 2$.

The product of these two numbers is: $x(x + 2) = 125$

We can simplify this to: $x^2 + 2x = 125$

Rearrange it to form a standard quadratic equation: $x^2 + 2x - 125 = 0$

Next, we solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = 2$, and $c = -125$. Substituting these values in, we get: $x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-125)}}{2 \cdot 1}$ $x = \frac{-2 \pm \sqrt{4 + 500}}{2}$ $x = \frac{-2 \pm \sqrt{504}}{2}$ $x = \frac{-2 \pm 2\sqrt{126}}{2}$ $x = -1 \pm \sqrt{126}$

Since $\sqrt{126}$ is not an integer, the equation $x^2 + 2x - 125 = 0$ does not yield integer solutions for $x$. Therefore, it appears there is an error in the problem statement as there are no two consecutive even integers whose product is 125.

If we assume the problem was meant to state the product of two consecutive odd integers, we can try solving for odd integers instead. Would you like to explore this?

Here are 8 relative questions to further your understanding:

1. What is the general form of a quadratic equation?
2. How do you solve a quadratic equation using the quadratic formula?
3. What are consecutive integers?
4. What are consecutive even integers?
5. What are consecutive odd integers?
6. How can you check if a number is even or odd?
7. What is the importance of the discriminant in the quadratic formula?
8. Can the product of two numbers be negative? How?

Tip: When solving quadratic equations, always check the discriminant ($b^2 - 4ac$) first to determine the nature of the roots (real and distinct, real and repeated, or complex).