Answer:
40, 42 and 44
Step-by-step explanation:
Mark these even consecutive integers as (2n-2), 2n and (2n+2)
Note that since they are integers, n > 0 and 2n is always even
Writing the equation
(2n-2) * 2n = 38 * (2n+2) + 8
// divide both sides by 2
(2n-2) * n = 19 * (2n+2) + 4
[tex]2n^{2}[/tex] - 2n = 38n + 38 + 4
[tex]2n^{2}[/tex] - 2n - 38n - 38 - 4 = 0
[tex]2n^{2}[/tex] - 40n - 42 = 0
// divide both sides by 2
[tex]n^{2}[/tex] - 20n - 21 = 0
// rewrite the equation
[tex]n^{2}[/tex] + n - 21n - 21 = 0
n * (n+1) - 21n - 21 = 0
n * (n+1) - 21 * (n+1) = 0
(n+1) * (n-21) = 0
// separate the two possible cases
n + 1 = 0 and n - 21 = 0
n = -1 n = 21
Note that here n has to be an integer, so case n = -1 can't be a solution. Therefore we use n = 21
Calculating the consecutive even integers
2n - 2 = 2 * 21 - 2 = 40
2n = 2 * 21 = 42
2n + 2 = 2 * 21 + 2 = 44
