Question 59924
I found this problem on an SAT practice test
Given 
a > b > 0
a^2 - b^2 = 12 
list all posible values for 
a - b = 

(a+b)(a-b)=a^2-b^2
a-b = (a^2-b^2)/(a+b)=12/(a+b)
since a>b>0 and a^2 = b^2+12
a^2>12
a>sqrt(12)
hence a-b ...maximum value of a-b is 12/[sqrt(12)+] < sqrt (12)
since a could be any number > sqrt of 12
minimum value of a-b is >0
hence 0<a-b<sqrt(12)