Réitigh operatorname{IRR}=22000/(1+r)^2-18000

$$\operatorname { IRR } = \frac { 22000 } { ( 1 + r ) ^ { 2 } } - 18000$$

Réitigh do I. (complex solution)

$$\left\{\begin{matrix}I=-\frac{2000\left(9r^{2}+18r-2\right)}{\left(R\left(r+1\right)\right)^{2}}\text{, }&r\neq -1\text{ and }R\neq 0\\I\in \mathrm{C}\text{, }&\left(r=-\frac{\sqrt{11}}{3}-1\text{ or }r=\frac{\sqrt{11}}{3}-1\right)\text{ and }R=0\end{matrix}\right.$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Méadaigh an dá thaobh den chothromóid faoi \left(r+1\right)^{2}.

$$IRR\left(r+1\right)^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)$$

Méadaigh R agus R chun R^{2} a fháil.

$$IR^{2}\left(r+1\right)^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)$$

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(r+1\right)^{2} a leathnú.

$$IR^{2}\left(r^{2}+2r+1\right)=22000+\left(r+1\right)^{2}\left(-18000\right)$$

Úsáid an t-airí dáileach chun IR^{2} a mhéadú faoi r^{2}+2r+1.

$$IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)$$

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(r+1\right)^{2} a leathnú.

$$IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000+\left(r^{2}+2r+1\right)\left(-18000\right)$$

Úsáid an t-airí dáileach chun r^{2}+2r+1 a mhéadú faoi -18000.

$$IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000-18000r^{2}-36000r-18000$$

Dealaigh 18000 ó 22000 chun 4000 a fháil.

$$IR^{2}r^{2}+2IR^{2}r+IR^{2}=4000-18000r^{2}-36000r$$

Comhcheangail na téarmaí ar fad ina bhfuil I.

$$\left(R^{2}r^{2}+2R^{2}r+R^{2}\right)I=4000-18000r^{2}-36000r$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\left(R^{2}r^{2}+2rR^{2}+R^{2}\right)I=4000-36000r-18000r^{2}$$

Roinn an dá thaobh faoi R^{2}r^{2}+2rR^{2}+R^{2}.

$$\frac{\left(R^{2}r^{2}+2rR^{2}+R^{2}\right)I}{R^{2}r^{2}+2rR^{2}+R^{2}}=\frac{4000-36000r-18000r^{2}}{R^{2}r^{2}+2rR^{2}+R^{2}}$$

Má roinntear é faoi R^{2}r^{2}+2rR^{2}+R^{2} cuirtear an iolrúchán faoi R^{2}r^{2}+2rR^{2}+R^{2} ar ceal.

$$I=\frac{4000-36000r-18000r^{2}}{R^{2}r^{2}+2rR^{2}+R^{2}}$$

Roinn 4000-36000r-18000r^{2} faoi R^{2}r^{2}+2rR^{2}+R^{2}.

$$I=\frac{2000\left(2-18r-9r^{2}\right)}{R^{2}\left(r+1\right)^{2}}$$

Réitigh do I.

$$\left\{\begin{matrix}I=-\frac{2000\left(9r^{2}+18r-2\right)}{\left(R\left(r+1\right)\right)^{2}}\text{, }&r\neq -1\text{ and }R\neq 0\\I\in \mathrm{R}\text{, }&\left(r=-\frac{\sqrt{11}}{3}-1\text{ or }r=\frac{\sqrt{11}}{3}-1\right)\text{ and }R=0\end{matrix}\right.$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Méadaigh an dá thaobh den chothromóid faoi \left(r+1\right)^{2}.

$$IRR\left(r+1\right)^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)$$

Méadaigh R agus R chun R^{2} a fháil.

$$IR^{2}\left(r+1\right)^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)$$

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(r+1\right)^{2} a leathnú.

$$IR^{2}\left(r^{2}+2r+1\right)=22000+\left(r+1\right)^{2}\left(-18000\right)$$

Úsáid an t-airí dáileach chun IR^{2} a mhéadú faoi r^{2}+2r+1.

$$IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000+\left(r+1\right)^{2}\left(-18000\right)$$

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(r+1\right)^{2} a leathnú.

$$IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000+\left(r^{2}+2r+1\right)\left(-18000\right)$$

Úsáid an t-airí dáileach chun r^{2}+2r+1 a mhéadú faoi -18000.

$$IR^{2}r^{2}+2IR^{2}r+IR^{2}=22000-18000r^{2}-36000r-18000$$

Dealaigh 18000 ó 22000 chun 4000 a fháil.

$$IR^{2}r^{2}+2IR^{2}r+IR^{2}=4000-18000r^{2}-36000r$$

Comhcheangail na téarmaí ar fad ina bhfuil I.

$$\left(R^{2}r^{2}+2R^{2}r+R^{2}\right)I=4000-18000r^{2}-36000r$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\left(R^{2}r^{2}+2rR^{2}+R^{2}\right)I=4000-36000r-18000r^{2}$$

Roinn an dá thaobh faoi R^{2}r^{2}+2rR^{2}+R^{2}.

$$\frac{\left(R^{2}r^{2}+2rR^{2}+R^{2}\right)I}{R^{2}r^{2}+2rR^{2}+R^{2}}=\frac{4000-36000r-18000r^{2}}{R^{2}r^{2}+2rR^{2}+R^{2}}$$

Má roinntear é faoi R^{2}r^{2}+2rR^{2}+R^{2} cuirtear an iolrúchán faoi R^{2}r^{2}+2rR^{2}+R^{2} ar ceal.

$$I=\frac{4000-36000r-18000r^{2}}{R^{2}r^{2}+2rR^{2}+R^{2}}$$

Roinn 4000-18000r^{2}-36000r faoi R^{2}r^{2}+2rR^{2}+R^{2}.

$$I=\frac{2000\left(2-18r-9r^{2}\right)}{\left(R\left(r+1\right)\right)^{2}}$$

Réitigh do R. (complex solution)

$$\left\{\begin{matrix}R=-\frac{20iI^{-\frac{1}{2}}\sqrt{45r^{2}+90r-10}}{r+1}\text{; }R=\frac{20iI^{-\frac{1}{2}}\sqrt{45r^{2}+90r-10}}{r+1}\text{, }&r\neq -1\text{ and }I\neq 0\\R\in \mathrm{C}\text{, }&\left(r=-\frac{\sqrt{11}}{3}-1\text{ or }r=\frac{\sqrt{11}}{3}-1\right)\text{ and }I=0\end{matrix}\right.$$

Réitigh do R.

$$\left\{\begin{matrix}R=\frac{20\sqrt{-\frac{5\left(9r^{2}+18r-2\right)}{I}}}{|r+1|}\text{; }R=-\frac{20\sqrt{-\frac{5\left(9r^{2}+18r-2\right)}{I}}}{|r+1|}\text{, }&\left(I<0\text{ or }r\leq \frac{\sqrt{11}}{3}-1\right)\text{ and }\left(I<0\text{ or }r\geq -\frac{\sqrt{11}}{3}-1\right)\text{ and }r\neq -1\text{ and }\left(I>0\text{ or }r\geq \frac{\sqrt{11}}{3}-1\text{ or }r\leq -\frac{\sqrt{11}}{3}-1\right)\text{ and }I\neq 0\\R\in \mathrm{R}\text{, }&\left(r=-\frac{\sqrt{11}}{3}-1\text{ or }r=\frac{\sqrt{11}}{3}-1\right)\text{ and }I=0\end{matrix}\right.$$