If p varies inverse as the square of q and p = 8 when q =4, find q when p = 32

If p varies inverse as the square of q and p = 8 when q =4, find q when p = 32

Answer Details

P $\alpha$ $\frac{1}{q^2}$
p = $\frac{k}{q^2}$
k = pq2
k = 8 x 42
k = 8 x 16 = 128
find k = pq2
q2 = $\frac{k}{p}$
q = $\frac{k}{p}$
q = $\frac{128}{32}$
q = √4
q = $\pm$2