SOLUTION: A=square Root S (s-a)(s-b)(s-1) make s the subject of the Formula ?

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Question 1094584: A=square Root S (s-a)(s-b)(s-1)
make s the subject of the Formula ?
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
A=sqrt%28s%28s-a%29%28s-b%29%28s-c%29%29

A%5E2=s%28s-a%29%28s-b%29%28s-c%29

Multiplying the right side out (which is a big chore):

A%5E2=++s%5E4+-+as%5E3+-+bs%5E3+-+cs%5E3+%2B+abs%5E2+%2B+acs%5E2+%2B+bcs%5E2+-+abcs

0+=++s%5E4+-+as%5E3+-+bs%5E3+-+cs%5E3+%2B+abs%5E2+%2B+acs%5E2+%2B+bcs%5E2+-+abcs+-+A%5E2

s%5E4+-+as%5E3+-+bs%5E3+-+cs%5E3+%2B+abs%5E2+%2B+acs%5E2+%2B+bcs%5E2+-+abcs+-+A%5E2+=+0

s%5E4+-+%28a%2Bb%2Bc%29s%5E3+%2B+%28ab+%2B+ac+%2B+bc%29s%5E2+-+abcs+-+A%5E2+=+0

While it is theoretically possible to solve 4th degree (quartic) equations,
the formula is so complicated it takes several pages just to write it, and
that formula is only for quartic equations without a term in s3.  So it
would take several more pages to transform this one into one without a term 
in s3. 

This is entirely too complicated a problem to present to anyone.  It can be
done, but it would take hours to substitute in and calculate, so it would
be totally useless.

Edwin