You can put this solution on YOUR website! Solve for c. a = 1/2 bc2
a = 1/(2bc^2) ----(1)
Multiplying by (2bc^2)
a(2bc^2) = 1
c^2= [1/2ab] (dividing by 2ab which means a cannot be zero and b cannot be zero)
c=+or minus sqrt[1/2ab]
c= +[1/sqrt(2ab)] or c= -[1/sqrt(2ab)]
Note: sqrt(1/t) = sqrt(1)/sqrt(t) = 1/[(sqrt(t)]
Verification: c= +[1/sqrt(2ab)] in (1)
RHS = 1/(2bc^2)=1/{2b[1/sqrt(2ab)]^2} = 1/{[2b/2ab]} = 1/[(1/a)] = a =LHS
c= -[1/sqrt(2ab)] in (1) gives the same result.
Hence our value is right
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