Question

Given `A=2(pi)r^2 + 2(pi)rh`, how do you solve for r?

Answer 1

This cannot be done.

Explanation:

My thinking here says that it cannot be done, because there are two different powers of r involved. If you try to factor out the r as a common factor, there will still be an r in the first term.

Answer 2

Use the quadratic formula to get `r=(-h)/(2)+-(sqrt(pi^2h^2-2piA))/(2pi)`.

Explanation:

Although at first it seems like we can't solve, for `r`, we actually can.

Note that if subtract `A` from both sides, we get:
`0=2pir^2+2pirh-A`

If we take out the `2pi` and `h`, we get this simple equation:
`0=r^2+r-A`

We know that `2pi`, `h`, and `A` are constants, so what we have here is a quadratic equation of the form `0=ar^2+br+c`. In our case, `a=2pi`, `b=2pih`, and `c=A`. We can solve for `r` using the quadratic formula:
`r=(-b+-sqrt(b^2-4ac))/(2a)`

Making the substitutions `a=2pi`, `b=2pih`, and `c=A`, we have:
`r=(-(2pih)+-sqrt((2pih)^2-4(2pi)(A)))/(2(2pi))`
`=(-2pih+-sqrt(4pi^2h^2-4(2pi)(A)))/(4pi)`
`=(-2pih+-sqrt(4(pi^2h^2-2piA)))/(4pi)`
`=(-2pih)/(4pi)+-(2sqrt(pi^2h^2-2piA))/(4pi)`
`=(-h)/(2)+-(sqrt(pi^2h^2-2piA))/(2pi)`

Certainly not the cleanest expression, but it will do.