Make-R-the-subject-of-P-RE-2-R-b-2-

Question Number 92275 by otchereabdullai@gmail.com last updated on 05/May/20

Make R the subject of :

P = \frac{{RE}^{2}}{{(R + b)}^{2}}

Answered by MJS last updated on 06/May/20

P {(R + b)}^{2} = E^{2} R

{(R + b)}^{2} = \frac{E^{2}}{P} R

R^{2} + 2 b R + b^{2} = \frac{E^{2}}{P} R

R^{2} + (2 b - \frac{E^{2}}{P}) R + b^{2} = 0

R^{2} + \frac{2 b P - E^{2}}{P} R + b^{2} = 0

R = - \frac{2 b P - E^{2}}{2 P} \pm \sqrt{{(\frac{2 b P - E^{2}}{2 P})}^{2} - b^{2}}

R = \frac{E^{2}}{2 P} - b \pm \frac{E}{2 P} \sqrt{E^{2} - 4 b P}

R = - b + \frac{E}{2 P} (E \pm \sqrt{E^{2} - 4 b P})

Commented by otchereabdullai@gmail.com last updated on 06/May/20

Thank you prof mjs