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

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Answer 1 · 2016-07-12T17:21:13

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

Notice we have a quadratic in R here:

#2piR^2 + 2pihR - a =0#

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

#R = (-2pih +-sqrt(4pi^2h^2+8pia))/(4pi)#

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

In the second fraction, we can rewrite the denominator from #4pi# to #sqrt(16pi^2)#

Our solution becomes:

#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=(-2pih+-sqrt((2pih)^2+8pia))/(4pi)#

#a=2piR^2+2piRh# is a quadratic equation in #R# and can be written as #2piR^2+2piRh-a=0#.

We can now solve for #R# using quadratic formula #(-b+-sqrt(b^2-4ac))/(2a)# and hence #R=(-2pih+-sqrt((2pih)^2+8pia))/(4pi)#