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# I need help again.

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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).

Jun 9, 2020

#1
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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" ...

Jun 9, 2020
#2
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Ah okay, small detail that I missed. Thanks so much!

Jun 9, 2020