If p varies inversely as the square of q and p = 8 when q = 4, find q when p = 32
We can set up the equation for the inverse variation as follows:
p = k/q^2
where k is a constant. To find the value of k, we can plug in the values p = 8 and q = 4:
8 = k/4^2
8 = k/16
To solve for k, we multiply both sides by 16:
128 = k
Now that we have the value of k, we can use it to find q when p = 32:
32 = 128/q^2
Divide both sides by 32:
1 = 4/q^2
To solve for q, we rearrange the equation:
q^2 = 4/1
q^2 = 4
Taking the square root of both sides:
q = ±√4
q = ±2
Therefore, when p = 32, q can be either 2 or -2.
To solve this problem, we can use the formula for inverse variation:
p ∝ 1/q^2
Let's substitute the given values into the formula to find the constant of variation.
When p = 8 and q = 4:
8 = k/4^2
8 = k/16
To solve for k, we multiply both sides of the equation by 16:
8 * 16 = k
k = 128
Now that we have the constant of variation (k = 128), we can find q when p = 32 by rearranging the formula:
p = k/q^2
Substituting p = 32 and k = 128, we get:
32 = 128/q^2
Multiplying both sides by q^2:
32q^2 = 128
Dividing both sides by 32:
q^2 = 4
Taking the square root of both sides:
q = ±2
So, when p = 32, q can be either +2 or -2.