gremather wrote:
a = 5b^2 - 10b + 7
Subtract b from both sides
a - b = 5b^2 - 11b + 7
Since the coefficient of b^2 is positive the RHS is >0.
So a-b>0 --> a>b.
Answer: A
This is a false approach. Let a = 5b^2 - 10b
In that case, a-b = 5b^2 - 11b. Coefficient of b^2 is positive, but the RHS isn't always positive.
Consider b = 1, the RHS equals -6. The only thing you can say for certain when b^2 has a positive coefficient is that the RHS will be positive as b tends to infinity.