Yvaries inversely as the square of x. if y = 12 when x = 2, then what is x when y = 3?

Remark
When you start out, you should do this problem by solving for k

Formula
y = k/x^2

Givens
y = 12
x = 2

Solve for k
12 = k /2^2
12 = k / 4  Multiply by 4
12*4 = k
48 = k

Problem
k = 48
y = 3
x = ??
y = k/x^2
3 = 48 / x^2  Multiply both sides by x^2

3x^2 = 48     Divide by 3
x^2 = 48/3    Take the square root of both sides.
x^2 = 16
sqrt(x^2) = sqrt(16)
 x = +/- 4   Both work

If only 1 answer is permitted, use plus 4
  

What class is this? i have no idea and i'm not going to give you the wrong answer. do you have a graph that goes with all the questions