a must be > 0 so b^2/b-1 must also be above 0. b^2 term is always positive so we dont worry bout that. therefore b-1 > 0 so b>1 so final restriction is 1<b<a .
a must be > 0 so b^2/b-1 must also be above 0. b^2 term is always positive so we dont worry bout that. therefore b-1 > 0 so b>1 so final restriction is 1<b<a .
I got to the stage where I did b^2/b-1 > 0 but I did b^2>0 which obviously did not get me anywhere. So the only other thing I could do was b-1>0 but I was not sure if this was right, and if so why
I got to the stage where I did b^2/b-1 > 0 but I did b^2>0 which obviously did not get me anywhere. So the only other thing I could do was b-1>0 but I was not sure if this was right, and if so why
The idea is to check "what values of b would make your calculation (in)valid" at every step. So read from the very beginning of the question and analyze at every step - including the given!
The idea is to check "what values of b would make your calculation (in)valid" at every step. So read from the very beginning of the question and analyze at every step - including the given! Also b-1>0 is not quite right, I think.