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

2 Answers
May 7, 2016

Answer:

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

Answer:

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.