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Question Number 96821 by bobhans last updated on 05/Jun/20

$$\{\begin{array}{l}\frac{{\mathrm{u}}^2}{\mathrm{v}}+\frac{{\mathrm{v}}^2}{\mathrm{u}}=12\\ \frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}=\frac{1}{3}\end{array}.\mathrm{find}\mathrm{u}\mathrm{and}\mathrm{v}?$$

Answered by john santu last updated on 05/Jun/20

Commented by bobhans last updated on 05/Jun/20

$$\mathrm{thank}\mathrm{you}$$

Answered by behi83417@gmail.com last updated on 05/Jun/20

$$\{\begin{array}{l}{\mathrm{u}}^3+{\mathrm{v}}^3=12\mathrm{uv}\\ 3(\mathrm{u}+\mathrm{v})=\mathrm{uv}\end{array}$$$$\underset{\mathrm{uv}=\mathrm{q}}{\stackrel{\mathrm{u}+\mathrm{v}=\mathrm{p}}{\Rightarrow }}\{\begin{array}{l}\mathrm{p}({\mathrm{p}}^2-3\mathrm{q})=12\mathrm{q}\\ 3\mathrm{p}=\mathrm{q}\end{array}\Rightarrow \{\begin{array}{l}{\mathrm{p}}^3-3\mathrm{pq}=12\mathrm{q}\\ 3\mathrm{p}=\mathrm{q}\end{array}$$$$\Rightarrow {\mathrm{p}}^3-{9\mathrm{p}}^2-36\mathrm{p}=0\Rightarrow \mathrm{p}({\mathrm{p}}^2-9\mathrm{p}-36)=0$$$$\Rightarrow \{\begin{array}{l}\mathrm{p}=0\\ {\mathrm{p}}^2-9\mathrm{p}-36=0\Rightarrow \mathrm{p}=\frac{9\pm \sqrt{81+144}}{2}=12,-3\end{array}$$$$1.\{\begin{array}{l}\mathrm{p}=0\Rightarrow \mathrm{q}=0\Rightarrow \mathrm{u}=\mathrm{v}=0\left(\mathrm{not}\mathrm{ok}\right)\\ \mathrm{p}=12,-3\Rightarrow \mathrm{q}=36,-9\end{array}$$$$2.\{\begin{array}{l}\mathrm{u}+\mathrm{v}=12\\ \mathrm{uv}=36\end{array}\Rightarrow \mathrm{u}=\mathrm{v}=6$$$$3.\{\begin{array}{l}\mathrm{u}+\mathrm{v}=-3\\ \mathrm{uv}=-9\end{array}{\mathrm{z}}^2+3\mathrm{z}-9=0\Rightarrow \mathrm{u}\vee \mathrm{v}=-\frac{3}{2}(1\pm \sqrt{5})$$

Commented by bobhans last updated on 05/Jun/20

$$\mathrm{thank}\mathrm{you}$$

Answered by 1549442205 last updated on 05/Jun/20

$$\mathrm{Condition}\mathrm{for}\mathrm{the}\mathrm{system}\mathrm{defined}\mathrm{as}\mathrm{u},\mathrm{v}\ne 0$$$$\Rightarrow \mathrm{uv}\ne 0,\mathrm{u}+\mathrm{v}\ne 0.\mathrm{Then}$$$$\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{eqs}\mathrm{is}\mathrm{equivalent}\mathrm{to}$$$$\{\begin{array}{l}{\mathrm{u}}^3+{\mathrm{v}}^3=12\mathrm{uv}\left(1\right)\\ 3(\mathrm{u}+\mathrm{v})=\mathrm{uv}\left(2\right)\end{array}\iff \{\begin{array}{l}{(\mathrm{u}+\mathrm{v})}^3-3\mathrm{uv}(\mathrm{u}+\mathrm{v})=12\mathrm{uv}\\ 9{(\mathrm{u}+\mathrm{v})}^2=3\mathrm{uv}(\mathrm{u}+\mathrm{v})\end{array}$$$$\mathrm{Subtructing}\mathrm{two}\mathrm{equations}\mathrm{we}\mathrm{get}:$$$$\text{Missing left or extra right}$$$$\iff (\mathrm{x}+3)(\mathrm{x}-12)=0\Rightarrow \mathrm{x}\in \{-3;12\}$$$$\mathrm{a}/\mathrm{if}\mathrm{x}=-3\mathrm{then}\mathrm{putting}\left(2\right)\mathrm{we}\mathrm{get}$$$$\alpha /\{\begin{array}{l}\mathrm{u}+\mathrm{v}=-3\\ \mathrm{uv}=-9\end{array}\Rightarrow {(\mathrm{u}-\mathrm{v})}^2={(\mathrm{u}+\mathrm{v})}^2-4\mathrm{uv}=9+36=45$$$$\Rightarrow \mathrm{u}-\mathrm{v}=\pm 3\sqrt{5}.\mathrm{We}\mathrm{have}:$$$$\{\begin{array}{l}\mathrm{u}+\mathrm{v}=-3\\ \mathrm{u}-\mathrm{v}=3\sqrt{5}\end{array}\iff \{\begin{array}{l}\mathrm{u}=\frac{-3+\sqrt{5}}{2}\\ \mathrm{v}=\frac{-3-\sqrt{5}}{2}\end{array}$$$$\beta /\{\begin{array}{l}\mathrm{u}+\mathrm{v}=-3\\ \mathrm{u}-\mathrm{v}=-3\sqrt{5}\end{array}\iff \{\begin{array}{l}\mathrm{u}=\frac{-3-3\sqrt{5}}{2}\\ \mathrm{v}=\frac{-3+3\sqrt{5}}{2}\end{array}$$$$\mathrm{b}/\mathrm{if}\mathrm{x}=\mathrm{u}+\mathrm{v}=12\mathrm{then}\mathrm{putting}\left(2\right)\mathrm{we}\mathrm{get}:$$$$\mathrm{uv}=36\Rightarrow {(\mathrm{u}-\mathrm{v})}^2={(\mathrm{u}+\mathrm{v})}^2-4\mathrm{uv}=144-144=0$$$$\Rightarrow \mathrm{u}=\mathrm{v}\Rightarrow \mathrm{u}=\mathrm{v}=6.\mathrm{Thus},\mathrm{the}\mathrm{system}\mathrm{has}\mathrm{three}\mathrm{roots}:$$$$(\mathrm{u};\mathrm{v})\in \{(6;6);(\frac{-3+\sqrt{5}}{2};\frac{-3-\sqrt{5}}{2});(\frac{-3-\sqrt{5}}{2};\frac{-3+\sqrt{5}}{2})\}$$

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