# How do you solve for R in a= 2piR^2 + 2piRh?

Jul 12, 2016

R= -h/2 +- sqrt(h^2/4 + a/(2pi)

#### Explanation:

Notice we have a quadratic in R here:

$2 \pi {R}^{2} + 2 \pi h R - a = 0$

We can use the quadratic formula:

$R = \frac{- 2 \pi h \pm \sqrt{{\left(2 \pi h\right)}^{2} - 4 \left(2 \pi\right) \left(- a\right)}}{4 \pi}$

$R = \frac{- 2 \pi h \pm \sqrt{4 {\pi}^{2} {h}^{2} + 8 \pi a}}{4 \pi}$

$R = - \frac{h}{2} \pm \frac{\sqrt{4 {\pi}^{2} {h}^{2} + 8 \pi a}}{4 \pi}$

In the second fraction, we can rewrite the denominator from $4 \pi$ to $\sqrt{16 {\pi}^{2}}$

Our solution becomes:

R = -h/2 +-sqrt((4pi^2h^2+8pia)/(16pi^2)) = -h/2 +- sqrt(h^2/4 + a/(2pi)

Jul 12, 2016

$R = \frac{- 2 \pi h \pm \sqrt{{\left(2 \pi h\right)}^{2} + 8 \pi a}}{4 \pi}$
$a = 2 \pi {R}^{2} + 2 \pi R h$ is a quadratic equation in $R$ and can be written as $2 \pi {R}^{2} + 2 \pi R h - a = 0$.
We can now solve for $R$ using quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ and hence $R = \frac{- 2 \pi h \pm \sqrt{{\left(2 \pi h\right)}^{2} + 8 \pi a}}{4 \pi}$