SOLUTION: Hi! **P=pi (3.14) A=2Pr^2 + 2Prh Solve for r>0. Thank you soooo much in advance. This was in a Calc BC summer packet. Due Tuesday!

Algebra ->  Expressions-with-variables -> SOLUTION: Hi! **P=pi (3.14) A=2Pr^2 + 2Prh Solve for r>0. Thank you soooo much in advance. This was in a Calc BC summer packet. Due Tuesday!      Log On


   

Question 987650: Hi! **P=pi (3.14)
A=2Pr^2 + 2Prh
Solve for r>0.
Thank you soooo much in advance. This was in a Calc BC summer packet. Due Tuesday!
Found 2 solutions by josgarithmetic, jim_thompson5910:
Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
Formula for general solution of a quadratic equation.

2pi%2Ar%5E2%2B2h%2Api%2Ar-A=0

r=%28-2h%2Api%2Bsqrt%28%282h%2Api%29%5E2%2B4%2A2pi%2AA%29%29%2F%282pi%29-------simplify if you can.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
A+=+2%2Api%2Ar%5E2+%2B+2%2Api%2Ar%2Ah


A-A+=+2%2Api%2Ar%5E2+%2B+2%2Api%2Ar%2Ah-A


0+=+2%2Api%2Ar%5E2+%2B+2%2Api%2Ar%2Ah-A


2%2Api%2Ar%5E2+%2B+2%2Api%2Ar%2Ah-A+=+0


2%2Api%2Ar%5E2+%2B+2%2Api%2Ah%2Ar-A+=+0


The last equation is in the form ax%5E2%2Bbx%2Bc=0 where a+=+2pi, b+=+2pi%2Ah and c+=+-A


Use the quadratic formula to solve for r


r+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29


r+=+%28-2pi%2Ah%2B-sqrt%28%282pi%2Ah%29%5E2-4%2A2pi%2A%28-A%29%29%29%2F%282%282pi%29%29


r+=+%28-2pi%2Ah%2B-sqrt%284pi%5E2%2Ah%5E2%2B8Api%29%29%2F%284pi%29


Since r+%3E+0, this means we only focus on the "plus" portion of the "plus/minus". So we end up with


r+=+%28-2pi%2Ah%2Bsqrt%284pi%5E2%2Ah%5E2%2B8Api%29%29%2F%284pi%29