Let \(f(n) = \left\{ \begin{array}{cl} n^2-2 & \text{ if }n<0, \\ 2n-20 & \text{ if }n \geq 0. \end{array} \right.\) . What is the positive difference between the two values of a that satisfy the equation f(-2)+f(2)+f(a)=0?
I solved for f(-2) first. Since -2 is less than 0(n<0), then I plugged it into n^2-2, which results in f(-2)=2. Then I solved for f(2). 2 is greater than 0(n>=0) so I plugged it into 2n-20 and got f(2)= -16.
Adding f(-2)+f(2)+f(a)=0, I have: -14 +f(a) =0.
f(a) must equal to 14 then.
So then I set n^2-2 and 2n-20 to equal to 14.
n^2-2=14
n^2=16
n=4, -4
2n-20=14
2n=34
n=17
But since n has two possible values for n^2-2=14, then there are actually 3 possible values for a in f(a)? So which one am I supposed to subtract from 17(the n value for 2n=20=14).
+23246 The problem arose when you set n2 - 2 = 14 ---> n2 = 16
Normally, you would get two answers, -4 and 4.
However, in this case there is a restriction upon using n2 - 2
--- you are only supposed to use n2 - 2 when n < 0
--- so, for this case, the answer is only -4 (you need to exclude the answer 4)
Everything else that you did was "textbook perfect" ...
