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.