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Solve for r.

 

f/(r ^ 2) + a = n + k

 Dec 16, 2019
 #1
avatar+22114 
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f/(r ^ 2)  = n + k - a     after subtracting  'a' from both sides of the equation

 

f/(n+k-a) = r^2      after re-arranging    now take sqrt of both sides

 

r = +- sqrt ( f/(n+k-a) )       (and as WIJ pointed out   n+k-a cannot equal 0)

 Dec 16, 2019
edited by ElectricPavlov  Dec 16, 2019
edited by ElectricPavlov  Dec 16, 2019
 #2
avatar+30 
0

\(r=\sqrt{\frac{f}{n+k-a}}, r=-\sqrt{\frac{f}{n+k-a}};\quad \:a\ne \:n+k\)

.
 Dec 16, 2019

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