If 4b^4 − 41b^2 + 100 = 0, then what is the sum of the two greatest possible values of b ?
Options:
A) 4  B) 9 / 2  C) 7  D) 41 / 4  E) 25

Rationale: just forget about the value of b ^ 4 for the moment and let b ^ 2 = x . ( its better to work with power of 2 than 4 ) now we have equation as 4 x ^ 2 - 41 x + 100 = 0 , factoring them gives us 4 x ^ 2 - 16 x - 25 x + 100 = 0 . which can be solved as ( 4 x - 25 ) ( x - 4 ) , so x = 4 and 25 / 4 . so b ^ 2 = 4 and 25 / 4 so b = + 2 and minus 2 and + 5 / 2 and minus 5 / 2 the two greatest values are + 2 and + 5 / 2 . so their sum 2 + 5 / 2 = 9 / 2 = b
Answer: B