Suppose that $a$ varies inversely with $b^2$. If $a=9$ when $b=2$, find the value of $a$ when $b=3$.

What is the second value of b? Is it 2 or 3 ?? Look at the title and the body of the question !

At any rate, I believe it goes like this:

a = k x 1/b^2, where k = constant

9 = k x 1/2^2 

k= 36 

a = 36 x 1/4 =9 when b=2, or

a = 36 x 1/9 =4 when b=3