$P_{1} {(1 + r)}^{2} + P_{2} (1 + r) = A$ $P_1(1+r)^2 + P_2(1+r) = A$ for r?

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Answer 1 · 2017-11-14T05:13:27

$r = \frac{- P_{2} \pm \sqrt{P_{2}^{2} + 4 P_{1} A}}{2 P_{1}} - 1$ $r=(-P_2 \pm sqrt(P_2^2+4P_1A)) / (2P_1) -1$

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

$P_{1} R^{2} + P_{2} R - A = 0$ $P_1R^2+P_2R-A=0$

We can now use the quadratic formula:

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

$R = \frac{- P_{2} \pm \sqrt{P_{2}^{2} - 4 P_{1} (- A)}}{2 (P_{1})}$ $R = (-P_2 \pm sqrt(P_2^2-4P_1(-A))) / (2(P_1))$

$R = r + 1 = \frac{- P_{2} \pm \sqrt{P_{2}^{2} + 4 P_{1} A}}{2 P_{1}}$ $R =r+1= (-P_2 \pm sqrt(P_2^2+4P_1A)) / (2P_1)$

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

P_1(1+r)^2 + P_2(1+r) = AP1(1+r)2+P2(1+r)=AP_1(1+r)^2 + P_2(1+r) = A for r?