How do you solve for R in $a = 2 π R^{2} + 2 π R h$ $a= 2piR^2 + 2piRh$ ?

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How do you solve for R in $a = 2 π R^{2} + 2 π R h$ $a= 2piR^2 + 2piRh$ ?

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Answer 1 · 2016-07-12T17:21:13

$R = - \frac{h}{2} \pm \sqrt{\frac{h^{2}}{4} + \frac{a}{2 π}}$ $R= -h/2 +- sqrt(h^2/4 + a/(2pi)$

Explanation:

Notice we have a quadratic in R here:

$2 π R^{2} + 2 π h R - a = 0$ $2piR^2 + 2pihR - a =0$

We can use the quadratic formula:

$R = \frac{- 2 π h \pm \sqrt{{(2 π h)}^{2} - 4 (2 π) (- a)}}{4 π}$ $R = (-2pih +-sqrt((2pih)^2-4(2pi)(-a)))/(4pi)$

$R = \frac{- 2 π h \pm \sqrt{4 π^{2} h^{2} + 8 π a}}{4 π}$ $R = (-2pih +-sqrt(4pi^2h^2+8pia))/(4pi)$

$R = - \frac{h}{2} \pm \frac{\sqrt{4 π^{2} h^{2} + 8 π a}}{4 π}$ $R = -h/2 +- (sqrt(4pi^2h^2+8pia))/(4pi)$

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

Our solution becomes:

$R = - \frac{h}{2} \pm \sqrt{\frac{4 π^{2} h^{2} + 8 π a}{16 π^{2}}} = - \frac{h}{2} \pm \sqrt{\frac{h^{2}}{4} + \frac{a}{2 π}}$ $R = -h/2 +-sqrt((4pi^2h^2+8pia)/(16pi^2)) = -h/2 +- sqrt(h^2/4 + a/(2pi)$

Answer 2 · 2016-07-12T17:30:20

$R = \frac{- 2 π h \pm \sqrt{{(2 π h)}^{2} + 8 π a}}{4 π}$ $R=(-2pih+-sqrt((2pih)^2+8pia))/(4pi)$

Explanation:

$a = 2 π R^{2} + 2 π R h$ $a=2piR^2+2piRh$ is a quadratic equation in $R$ $R$ and can be written as $2 π R^{2} + 2 π R h - a = 0$ $2piR^2+2piRh-a=0$ .

We can now solve for $R$ $R$ using quadratic formula $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b+-sqrt(b^2-4ac))/(2a)$ and hence $R = \frac{- 2 π h \pm \sqrt{{(2 π h)}^{2} + 8 π a}}{4 π}$ $R=(-2pih+-sqrt((2pih)^2+8pia))/(4pi)$

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How do you solve for R in $a = 2 π R^{2} + 2 π R h$ $a= 2piR^2 + 2piRh$ ?

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