the product of two consecutive even numbers is 440. What is the sum of the numbers?

Let's assume the first even number as "x". So, the next consecutive even number would be "x + 2".

According to the given information, the product of these two numbers is 440. We can write this as an equation:

x * (x + 2) = 440

Expanding the equation:

x^2 + 2x = 440

Rearranging the equation:

x^2 + 2x - 440 = 0

Now, we can use the quadratic formula to solve for "x":

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 1, b = 2, and c = -440. Plugging in these values:

x = (-2 ± √(2^2 - 4 * 1 * -440)) / (2 * 1)

Simplifying:

x = (-2 ± √(4 + 1760)) / 2

x = (-2 ± √(1764)) / 2

x = (-2 ± 42) / 2

Now, we can find the possible values of "x":

1. When x = (-2 + 42) / 2 = 40 / 2 = 20
2. When x = (-2 - 42) / 2 = -44 / 2 = -22

However, we are looking for even numbers, so we discard the negative value.

Therefore, the first even number is 20, and the next consecutive even number is 20 + 2 = 22.

To find the sum of these numbers:

Sum = 20 + 22 = 42

So, the sum of the numbers is 42.

To find the sum of two consecutive even numbers, we first need to determine the two numbers.

Let's assume the first even number is x. Since the numbers are consecutive, the next even number can be represented as x + 2.

Now, we can set up the equation to find the product of the two numbers:

x * (x + 2) = 440

Expanding this equation gives us:

x^2 + 2x = 440

Rearranging the equation and setting it equal to zero, we have:

x^2 + 2x - 440 = 0

Now we can solve this quadratic equation to find the possible values of x by factoring, completing the square or using the quadratic formula.

Factoring may not be suitable in this case since it involves larger numbers. Similarly, completing the square can be a bit cumbersome. Therefore, we will use the quadratic formula, which states:

For an equation in the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation x^2 + 2x - 440 = 0:

a = 1, b = 2, c = -440

Using the quadratic formula, we have:

x = (-2 ± √(2^2 - 4 * 1 * -440)) / (2 * 1)

Simplifying this equation gives us:

x = (-2 ± √(4 + 1760)) / 2

x = (-2 ± √1764) / 2

x = (-2 ± 42) / 2

Solving for both possible values of x:

x₁ = (-2 + 42) / 2 = 40 / 2 = 20

x₂ = (-2 - 42) / 2 = -44 / 2 = -22

Since we are looking for even numbers, we can eliminate the negative value (-22). Therefore, the first even number (x) is 20, and the next even number (x + 2) is 22.

Finally, we can calculate the sum of the two consecutive even numbers:

20 + 22 = 42

Therefore, the sum of the two consecutive even numbers is 42.

Here's a start:

√440 = ?

factors of 440: 1,2,4,8,10,20,22,44,55,110,440

20&22 only consecutive evens
20*22=440
20+22=42