x,y,z are three consecutive positive odd integers that satisfy y^2 - x^2 = 384. Find the value of z.
+128460 Let y = n x = n - 2 and z = n + 2
So
n^2 - ( n - 2)^2 = 384
n^2 - n^2 + 4n - 4 = 384
4n = 384 + 4
4n = 388
n = 388/ 4 = 97
So
z = n + 2 = 99
+514 well, y must be greater than x, so x + 2 = y.
\(y^2-x^2=384\) and \(x + 2 = y\)
so solving that system of equations... u get \(x=95, \: y=97\)
so 97 + 2 = 99
