the product of two even consecutive integers is 224. find the integers

let the first integer be x

then the next even integer is x+2

x(x+2) = 224
x^2 + 2x - 224 = 0
(x-14)(x+16) = 0
x=14 or x = -16

if x=12, then x+2 = 16
if x = -16 , then x+2 = -14

the two integers are 14 and 16 or -14 and -16

Stop spamming, I already answered you.

<<<if x=12, then x+2 = 16>>>>

You mean if x=14, then x+2=16

To find the integers, let's call them n and n+2.

Since they are even and consecutive, we know that n is the smaller integer and n+2 is the next even number after n.

According to the problem, the product of these two integers is 224. So we can set up the equation:

n * (n+2) = 224

Expanding the equation, we get:

n^2 + 2n = 224

Rearranging the equation to make it a quadratic equation:

n^2 + 2n - 224 = 0

Now, we can solve this quadratic equation by factoring or using the quadratic formula.

To use the quadratic formula, we have:

n = (-b ± √(b^2 - 4ac)) / (2a)

For this equation, a = 1, b = 2, and c = -224.

Plugging in the values, we get:

n = (-2 ± √(2^2 - 4*1*(-224))) / (2*1)

Simplifying further, we have:

n = (-2 ± √(4 + 896)) / 2
n = (-2 ± √900) / 2
n = (-2 ± 30) / 2

Now, we have two possible values for n:

1. n = (-2 + 30) / 2 = 28 / 2 = 14
2. n = (-2 - 30) / 2 = -32 / 2 = -16

Since we are looking for even consecutive integers, we discard the negative value (-16) and consider 14 as the smaller even integer.

Therefore, the two even consecutive integers whose product is 224 are 14 and 16.