Find two conservative positive even integers whose product i

Find two conservative positive even integers whose product is 168.

Solution

Any positive even integer can be expressed as 2n_k formed where k ranges from 1 to infinity
Now two consecutive positive even integers are going to be 2n and 2(n+1) = 2n and 2n+2 (where n>0)
Their product is = 168 meaning
2n* (2n+2) = 168
4n² + 4n - 168 = 0
n² + n -42=0
n²+7n - 6n - 42=0 ;
n(n+7) - 6 (n+7) = 0
(n-6) (n+7) = 0 meaning n=6 or n=-7 but n cannot be = -7 since numbers have to be positive
Thus n = 6 and 2n= 12 and 2(n+1) = 2(6+1) = 2*7=14

Thus the two consecutive positive even integers are 12 and 14
For verification you can multiply 12 and 14; 12*14= 168 (which is correct)