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Oliver
10-26-2013, 05:15 AM
The operation @ is defined as a@b=(a+b)^2-a. Which one of the following is the result of a@a@a?

(A) 4a^2-a

(B) 16a^2-a

(C) 4a^4-a

(D) 16a^4-a


Thanks!:)

nicole
10-26-2013, 05:16 AM
a@a@a= a@(a@a)= a@ ((a+a)^2-a) = a@ (4a^2-a)= (a+4a^2-a)^2-a

= (4a^2)^2-a=16a^4-a

Therefore the correct answer is (D).

Hope these help!!:)

Oliver
10-26-2013, 05:16 AM
Thank you very much for you quickly answer!!:)