SOLUTION: Given that p=re^2/(r+b)^2 make r the subject formula

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Question 1205062: Given that p=re^2/(r+b)^2 make r the subject formula
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
+p=re%5E2%2F%28r%2Bb%29%5E2

Clear the fraction by multiplying both sides by %28r%2Bb%29%5E2

+p%28r%2Bb%29%5E2=re%5E2

Square out %28r%2Bb%29%5E2 as r%5E2%2B2rb%2Bb%5E2

+p%28r%5E2%2B2rb%2Bb%5E2%29=re%5E2

Remove the parentheses by distributing:

+pr%5E2%2B2prb%2Bpb%5E2=re%5E2

Get the term with r%5E2 first, then terms with r then other terms:

pr%5E2%2B2prb-re%5E2%2Bpb%5E2=0

Factor r out of the two terms that contain r:

pr%5E2%2B%282pb-e%5E2%29r%2Bpb%5E2=0

Use the quadratic formula: x+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29+
with x=r, A=p, B=(2pb-e2) and C=pb^2 

r+=+%28-%282pb-e%5E2%29+%2B-+sqrt%28+%282pb-e%5E2%29%5E2-4prb%5E2+%29%29%2F%282p%29+

remove the parentheses of the first term in the numerator
Square the binomial under the square root radical:

r+=+%28-2pb%2Be%5E2+%2B-+sqrt%284b%5E2p%5E2-4bpe%5E2%2Be%5E4-4b%5E2p%5E2+%29%29%2F%282p%29+

The 4b%5E2p%5E2 and -4b%5E2p%5E2 cancel under the square root radical:

r+=+%28-2pb%2Be%5E2+%2B-+sqrt%28-4bpe%5E2%2Be%5E4+%29%29%2F%282p%29+

Factor out e%5E2 under the square root radical

r+=+%28-2pb%2Be%5E2+%2B-+sqrt%28e%5E2%28-4bp%2Be%5E2%29+%29%29%2F%282p%29+

Take out the the square root of e%5E2 as e in front of the square root

r+=+%28-2pb%2Be%5E2+%2B-+e%2Asqrt%28-4bp%2Be%5E2+%29%29%2F%282p%29+

Reverse the terms under the radical so the positive term comes first
which makes the + sign unnecessary:

r+=+%28-2bp%2Be%5E2+%2B-+e%2Asqrt%28e%5E2+-+4bp%29%29%2F%282p%29+

Edwin