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

May 7, 2016

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.

May 7, 2016

Use the quadratic formula to get $r = \frac{- h}{2} \pm \frac{\sqrt{{\pi}^{2} {h}^{2} - 2 \pi A}}{2 \pi}$.

#### 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:
$0 = 2 \pi {r}^{2} + 2 \pi r h - A$

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

We know that $2 \pi$, $h$, and $A$ are constants, so what we have here is a quadratic equation of the form $0 = a {r}^{2} + b r + c$. In our case, $a = 2 \pi$, $b = 2 \pi h$, and $c = A$. We can solve for $r$ using the quadratic formula:
$r = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Making the substitutions $a = 2 \pi$, $b = 2 \pi h$, and $c = A$, we have:
$r = \frac{- \left(2 \pi h\right) \pm \sqrt{{\left(2 \pi h\right)}^{2} - 4 \left(2 \pi\right) \left(A\right)}}{2 \left(2 \pi\right)}$
$= \frac{- 2 \pi h \pm \sqrt{4 {\pi}^{2} {h}^{2} - 4 \left(2 \pi\right) \left(A\right)}}{4 \pi}$
$= \frac{- 2 \pi h \pm \sqrt{4 \left({\pi}^{2} {h}^{2} - 2 \pi A\right)}}{4 \pi}$
$= \frac{- 2 \pi h}{4 \pi} \pm \frac{2 \sqrt{{\pi}^{2} {h}^{2} - 2 \pi A}}{4 \pi}$
$= \frac{- h}{2} \pm \frac{\sqrt{{\pi}^{2} {h}^{2} - 2 \pi A}}{2 \pi}$

Certainly not the cleanest expression, but it will do.