SOLUTION: Let the function f be defined by F(x)=x^2+28. If f(3y)=2f(y), what is the one possible value of y? A)-1 B)1 C)2 D)-3

Algebra ->  Finance -> SOLUTION: Let the function f be defined by F(x)=x^2+28. If f(3y)=2f(y), what is the one possible value of y? A)-1 B)1 C)2 D)-3      Log On


   

Question 1088766: Let the function f be defined by
F(x)=x^2+28. If f(3y)=2f(y), what is the one possible value of y?
A)-1
B)1
C)2
D)-3
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

f%28x%29=x%5E2%2B28
f%283y%29=2f%28y%29
%283y%29%5E2%2B28=2%28y%5E2%2B28%29
9y%5E2%2B28=2y%5E2%2B56
9y%5E2-2y%5E2=56-28
7y%5E2=28
y%5E2=4
y=2 or y=-2
so, the one possible value of y is: C)2