$P_1(1+r)^2 + P_2(1+r) = A$ 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:

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

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

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

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

P_1(1+r)^2 + P_2(1+r) = AP_1(1+r)^2 + P_2(1+r) = A for r?