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

{ (((u^2 /v) + (v^2 /u) = 12)),(((1/u) + (1/v) = (1/3))) :} . find u and v ?

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

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

Commented by bobhans last updated on 05/Jun/20

thank you

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

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

{ ((u^3 +v^3 =12uv)),((3(u+v)=uv)) :} ⇒_(uv=q) ^(u+v=p) { ((p(p^2 −3q)=12q)),((3p=q)) :}⇒ { ((p^3 −3pq=12q)),((3p=q)) :} ⇒p^3 −9p^2 −36p=0⇒p(p^2 −9p−36)=0 ⇒ { ((p=0)),((p^2 −9p−36=0⇒p=((9±(√(81+144)))/2)=12,−3)) :} 1. { ((p=0⇒q=0⇒u=v=0 (not ok))),((p=12,−3⇒q=36,−9)) :} 2. { ((u+v=12)),((uv=36)) :}⇒u=v=6 3. { ((u+v=−3)),((uv=−9)) :}z^2 +3z−9=0⇒u∨v=−(3/2)(1±(√5))

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

Commented by bobhans last updated on 05/Jun/20

thank you

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

Answered by 1549442205 last updated on 05/Jun/20

Condition for the system defined as u,v≠0 ⇒uv≠0,u+v≠0.Then the system of eqs is equivalent to { ((u^3 +v^3 =12uv(1))),((3(u+v)=uv(2))) :}⇔ { (((u+v)^3 −3uv(u+v)=12uv)),((9(u+v)^2 =3uv(u+v))) :} Subtructing two equations we get: (u+v)^3 −9(u+v)^2 =36(u+v)⇔_(x=u+v) x^2 −9x−36=0 ⇔(x+3)(x−12)=0⇒x∈{−3;12} a/if x=−3 then putting (2) we get α/ { ((u+v=−3)),((uv=−9)) :} ⇒(u−v)^2 =(u+v)^2 −4uv=9+36=45 ⇒u−v=±3(√5) .We have: { ((u+v=−3)),((u−v=3(√5))) :} ⇔ { ((u=((−3+(√5))/2))),((v=((−3−(√5))/2))) :} β/ { ((u+v=−3)),((u−v=−3(√5))) :} ⇔ { ((u=((−3−3(√5))/2))),((v=((−3+3(√5))/2))) :} b/if x=u+v=12 then putting (2) we get: uv=36⇒(u−v)^2 =(u+v)^2 −4uv=144−144=0 ⇒u=v⇒u=v=6.Thus,the system has three roots: (u;v)∈{(6;6);(((−3+(√5))/2);((−3−(√5))/2));(((−3−(√5))/2);((−3+(√5))/2))}

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

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