#P_1(1+r)^2 + P_2(1+r) = A# for r?

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Answer 1 · 2017-11-14T05:13:27

#r=(-P_2 \pm sqrt(P_2^2+4P_1A)) / (2P_1) -1#

Notice that if we subtract #A# from both sides and substitute in #R# for #(1+r)#, ew get something familiar:

#P_1R^2+P_2R-A=0#

We can now use the quadratic formula:

# R = (-b \pm sqrt(b^2-4ac)) / (2a) #

# R = (-P_2 \pm sqrt(P_2^2-4P_1(-A))) / (2(P_1)) #

# R =r+1= (-P_2 \pm sqrt(P_2^2+4P_1A)) / (2P_1) #

#:.r=(-P_2 \pm sqrt(P_2^2+4P_1A)) / (2P_1) -1#