Precalc help!?

Question

please help me solve this and please add steps

solve for R

P(r+R)^2 = (E^2)R


Please help

Answer 1

P(r+R)^2 = (E^2)R

P(r^2 + 2rR + R^2) = E^2 * R
Pr^2 + 2PrR + PR^2 = E^2 * R
PR^2 + (2Pr - E^2)*R + Pr^2 = 0
This is a quadratic in R, so use the quadratic formula:

R = { E^2 - 2Pr ± √[ (2Pr-E^2)^2 - 4*P*Pr^2 ] } / 2P

= [ E^2 - 2Pr ± E√(E^2 - 4Pr) ] / 2P

Answer 2

P(r + R)^2 = ( E^2 ) R
P (r^2 + 2 r R + R^2 ) = E^2 R
Pr^2 + 2 r R P + R^2 P = E^2 R
P R^2 + ( 2 r P - E^2 ) R + P r^2 = 0
by quadratic equation
R = [ - ( 2 r P - E^2 ) +- sqr( ( 2 r P - E^2 )^2 - 2 P^2 ] / 2P

Answer 3

P(r^2+2rR+R^2) = E^2*R
PR^2 + (2Pr-E^2)R + Pr^2 = 0
Thus, you can use quadratic formula to solve for R.
R = {E^2-2Pr ±√[2Pr-E^2)^2-4P^2*r^2]} / (2P)
You can further simplify R,
R = [E^2-2Pr ±E√(E^2-4Pr)]/ (2P)
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Remember P is the coefficient of R^2, (2Pr-E^2) is the coefficient of R, and Pr^2 is the constant term.

Answer 4

P(r^2 +2Rr +R^2 = E^2R
Pr^2 +2PRr +PR^2 =E^2R
PR^2 +R(2Pr - E^2) +Pr^2 = 0
R = [E^2-2Pr +/- sqrt((2Pr-E^2)^2 -4P^r^2)]/(2P)