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Question Number 119795 by bobhans last updated on 27/Oct/20
Solve in real numbers the system of equations { (((3x+y)(x+3y)(√(xy)) =14)),(((x+y)(x^2 +14xy+y^2 )= 36)) :}
$${Solve}\:{in}\:{real}\:{numbers}\:{the}\:{system}\:{of} \\ $$$${equations}\:\begin{cases}{\left(\mathrm{3}{x}+{y}\right)\left({x}+\mathrm{3}{y}\right)\sqrt{{xy}}\:=\mathrm{14}}\\{\left({x}+{y}\right)\left({x}^{\mathrm{2}} +\mathrm{14}{xy}+{y}^{\mathrm{2}} \right)=\:\mathrm{36}}\end{cases}\: \\ $$
Answered by 1549442205PVT last updated on 27/Oct/20
Put x=ky we get (((3k+1)(k+3)(√k) )/((k+1)(k^2 +14k+1)))=(7/(18)) ⇔18(3k+1)(k+3)(√k) =7(k+1)(k^2 +14k+1) { (((3x+y)(x+3y)(√(xy)) =14 (1))),(((x+y)(x^2 +14xy+y^2 )= 36 (2))) :} (2)⇔x^4 +15x^2 y+15xy^2 +y^3 −36=0(3) ⇔324k(3k^2 +10k+3)^2 =49(k^3 +15k^2 +15k+1)^2 ⇔324k(9k^4 +118k^2 +60k^3 +60k+9) =49(k^6 +30k^5 +255k^4 +452k^3 +255k^2 +30k+1) ⇔49k^6 −1446k^5 −6945k^4 −16084k^3 −6945k^2 −1446k+49i ⇔49(k^3 +(1/k^3 ))−1446(k^2 +(1/k^2 ))−6945(k+(1/k))−16084=0(∗) Put k+(1/k)=t⇒k^2 +(1/k^2 )=t^2 −2 k^3 +(1/k^3 )=t^3 −3t.Replace into we get 49(t^3 −3t)−1446(t^2 −2)−6945t−16084=0 ⇔49t^3 −1446t^2 −7092t−13192=0 49(t−34)(49t^2 +220t+338)=0 ⇒t=34⇔k+(1/k)=34⇔k^2 −34k+1=0 Δ′=17^2 −1=288=12^2 .2 k=17±12(√2) i)k=17+12(√2) ⇒x=(17+12(√2))y Replace into (3)we get (3)⇔37/y^3 =(k^3 +15k^2 +15k+1) =k(k^2 +1)+15(k^2 +1)+14k−14 =k(34k)+15.34k+14k−14 =34(k^2 +1)+524k−48=34.34k+524k−48 =1680k−48=1680(17+12(√2))−48=36/y^3 ⇔28512+20160(√2)=36/y^3 y^3 =((36)/(28512+20160(√2))) ⇒y=^3 (√(1/(792+560(√2)))) =^3 (√(1/((6+4(√2))^3 )))=(1/(6+4(√2)))=((3−2(√2))/2) x=(17+12(√2)).((3−2(√2))/2)=((3+2(√2))/2) ii)k=17−12(√2) .Similarly,we get y=((3+2(√2))/2),x=((3−2(√2))/2) Thus the given system has two solutions (x,y)∈{(((3−2(√2))/2),((3+2(√2))/2)),(((3+2(√2))/2),((3−2(√2))/2))}
$$\mathrm{Put}\:\mathrm{x}=\mathrm{ky}\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{\left(\mathrm{3k}+\mathrm{1}\right)\left(\mathrm{k}+\mathrm{3}\right)\sqrt{\mathrm{k}}\:}{\left(\mathrm{k}+\mathrm{1}\right)\left(\mathrm{k}^{\mathrm{2}} +\mathrm{14k}+\mathrm{1}\right)}=\frac{\mathrm{7}}{\mathrm{18}} \\ $$$$\Leftrightarrow\mathrm{18}\left(\mathrm{3k}+\mathrm{1}\right)\left(\mathrm{k}+\mathrm{3}\right)\sqrt{\mathrm{k}}\:=\mathrm{7}\left(\mathrm{k}+\mathrm{1}\right)\left(\mathrm{k}^{\mathrm{2}} +\mathrm{14k}+\mathrm{1}\right) \\ $$$$\:\begin{cases}{\left(\mathrm{3}{x}+{y}\right)\left({x}+\mathrm{3}{y}\right)\sqrt{{xy}}\:=\mathrm{14}\:\left(\mathrm{1}\right)}\\{\left({x}+{y}\right)\left({x}^{\mathrm{2}} +\mathrm{14}{xy}+{y}^{\mathrm{2}} \right)=\:\mathrm{36}\:\left(\mathrm{2}\right)}\end{cases}\: \\ $$$$\left(\mathrm{2}\right)\Leftrightarrow\mathrm{x}^{\mathrm{4}} +\mathrm{15x}^{\mathrm{2}} \mathrm{y}+\mathrm{15xy}^{\mathrm{2}} +\mathrm{y}^{\mathrm{3}} −\mathrm{36}=\mathrm{0}\left(\mathrm{3}\right) \\ $$$$\Leftrightarrow\mathrm{324k}\left(\mathrm{3k}^{\mathrm{2}} +\mathrm{10k}+\mathrm{3}\right)^{\mathrm{2}} =\mathrm{49}\left(\mathrm{k}^{\mathrm{3}} +\mathrm{15k}^{\mathrm{2}} +\mathrm{15k}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{324k}\left(\mathrm{9k}^{\mathrm{4}} +\mathrm{118k}^{\mathrm{2}} +\mathrm{60k}^{\mathrm{3}} +\mathrm{60k}+\mathrm{9}\right) \\ $$$$=\mathrm{49}\left(\mathrm{k}^{\mathrm{6}} +\mathrm{30k}^{\mathrm{5}} +\mathrm{255k}^{\mathrm{4}} +\mathrm{452k}^{\mathrm{3}} +\mathrm{255k}^{\mathrm{2}} +\mathrm{30k}+\mathrm{1}\right) \\ $$$$\Leftrightarrow\mathrm{49k}^{\mathrm{6}} −\mathrm{1446k}^{\mathrm{5}} −\mathrm{6945k}^{\mathrm{4}} −\mathrm{16084k}^{\mathrm{3}} −\mathrm{6945k}^{\mathrm{2}} −\mathrm{1446k}+\mathrm{49i} \\ $$$$\Leftrightarrow\mathrm{49}\left(\mathrm{k}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{3}} }\right)−\mathrm{1446}\left(\mathrm{k}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\right)−\mathrm{6945}\left(\mathrm{k}+\frac{\mathrm{1}}{\mathrm{k}}\right)−\mathrm{16084}=\mathrm{0}\left(\ast\right) \\ $$$$\mathrm{Put}\:\mathrm{k}+\frac{\mathrm{1}}{\mathrm{k}}=\mathrm{t}\Rightarrow\mathrm{k}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }=\mathrm{t}^{\mathrm{2}} −\mathrm{2} \\ $$$$\mathrm{k}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{3}} }=\mathrm{t}^{\mathrm{3}} −\mathrm{3t}.\mathrm{Replace}\:\mathrm{into}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{49}\left(\mathrm{t}^{\mathrm{3}} −\mathrm{3t}\right)−\mathrm{1446}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{2}\right)−\mathrm{6945t}−\mathrm{16084}=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{49t}^{\mathrm{3}} −\mathrm{1446t}^{\mathrm{2}} −\mathrm{7092t}−\mathrm{13192}=\mathrm{0} \\ $$$$\mathrm{49}\left(\mathrm{t}−\mathrm{34}\right)\left(\mathrm{49t}^{\mathrm{2}} +\mathrm{220t}+\mathrm{338}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{t}=\mathrm{34}\Leftrightarrow\mathrm{k}+\frac{\mathrm{1}}{\mathrm{k}}=\mathrm{34}\Leftrightarrow\mathrm{k}^{\mathrm{2}} −\mathrm{34k}+\mathrm{1}=\mathrm{0} \\ $$$$\Delta'=\mathrm{17}^{\mathrm{2}} −\mathrm{1}=\mathrm{288}=\mathrm{12}^{\mathrm{2}} .\mathrm{2} \\ $$$$\mathrm{k}=\mathrm{17}\pm\mathrm{12}\sqrt{\mathrm{2}} \\ $$$$\left.\mathrm{i}\right)\mathrm{k}=\mathrm{17}+\mathrm{12}\sqrt{\mathrm{2}}\:\Rightarrow\mathrm{x}=\left(\mathrm{17}+\mathrm{12}\sqrt{\mathrm{2}}\right)\mathrm{y} \\ $$$$\mathrm{Replace}\:\mathrm{into}\:\left(\mathrm{3}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\left(\mathrm{3}\right)\Leftrightarrow\mathrm{37}/\mathrm{y}^{\mathrm{3}} =\left(\mathrm{k}^{\mathrm{3}} +\mathrm{15k}^{\mathrm{2}} +\mathrm{15k}+\mathrm{1}\right) \\ $$$$=\mathrm{k}\left(\mathrm{k}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{15}\left(\mathrm{k}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{14k}−\mathrm{14} \\ $$$$=\mathrm{k}\left(\mathrm{34k}\right)+\mathrm{15}.\mathrm{34k}+\mathrm{14k}−\mathrm{14} \\ $$$$=\mathrm{34}\left(\mathrm{k}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{524k}−\mathrm{48}=\mathrm{34}.\mathrm{34k}+\mathrm{524k}−\mathrm{48} \\ $$$$=\mathrm{1680k}−\mathrm{48}=\mathrm{1680}\left(\mathrm{17}+\mathrm{12}\sqrt{\mathrm{2}}\right)−\mathrm{48}=\mathrm{36}/\mathrm{y}^{\mathrm{3}} \\ $$$$\Leftrightarrow\mathrm{28512}+\mathrm{20160}\sqrt{\mathrm{2}}=\mathrm{36}/\mathrm{y}^{\mathrm{3}} \\ $$$$\mathrm{y}^{\mathrm{3}} =\frac{\mathrm{36}}{\mathrm{28512}+\mathrm{20160}\sqrt{\mathrm{2}}}\:\Rightarrow\mathrm{y}=\:^{\mathrm{3}} \sqrt{\frac{\mathrm{1}}{\mathrm{792}+\mathrm{560}\sqrt{\mathrm{2}}}} \\ $$$$=\:^{\mathrm{3}} \sqrt{\frac{\mathrm{1}}{\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right)^{\mathrm{3}} }}=\frac{\mathrm{1}}{\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}}=\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{x}=\left(\mathrm{17}+\mathrm{12}\sqrt{\mathrm{2}}\right).\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}=\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\left.\mathrm{ii}\right)\mathrm{k}=\mathrm{17}−\mathrm{12}\sqrt{\mathrm{2}}\:.\mathrm{Similarly},\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{y}=\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}},\mathrm{x}=\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{Thus}\:\mathrm{the}\:\mathrm{given}\:\mathrm{system}\:\mathrm{has}\:\mathrm{two}\:\mathrm{solutions} \\ $$$$\left(\mathrm{x},\mathrm{y}\right)\in\left\{\left(\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}},\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}\right),\left(\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}},\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\right\} \\ $$
Answered by MJS_new last updated on 27/Oct/20
in this case it′s easier to let x=p−q∧y=p+q we get { ((4(4p^2 −q^2 )(√(p^2 −q^2 ))=14)),((8p(4p^2 −3q^2 )=36 ⇒ q^2 =((8p^3 −9)/(6p)))) :} insert in first equation (((16p^3 +9)(√(54−12p^3 )))/(9(√p^3 )))=14 squaring and transforming leads to p^9 −((27)/8)p^6 +((27)/(64))p^3 −((729)/(512))=0 (p^3 −((27)/8))(p^6 +((27)/(64)))=0 ⇒ p=(3/2) ⇒ q=±(√2) ⇒ x=(3/2)∓(√2)∧y=(3/2)±(√2)
$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{it}'\mathrm{s}\:\mathrm{easier}\:\mathrm{to}\:\mathrm{let} \\ $$$${x}={p}−{q}\wedge{y}={p}+{q} \\ $$$$\mathrm{we}\:\mathrm{get} \\ $$$$\begin{cases}{\mathrm{4}\left(\mathrm{4}{p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)\sqrt{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }=\mathrm{14}}\\{\mathrm{8}{p}\left(\mathrm{4}{p}^{\mathrm{2}} −\mathrm{3}{q}^{\mathrm{2}} \right)=\mathrm{36}\:\Rightarrow\:{q}^{\mathrm{2}} =\frac{\mathrm{8}{p}^{\mathrm{3}} −\mathrm{9}}{\mathrm{6}{p}}}\end{cases} \\ $$$$\mathrm{insert}\:\mathrm{in}\:\mathrm{first}\:\mathrm{equation} \\ $$$$\frac{\left(\mathrm{16}{p}^{\mathrm{3}} +\mathrm{9}\right)\sqrt{\mathrm{54}−\mathrm{12}{p}^{\mathrm{3}} }}{\mathrm{9}\sqrt{{p}^{\mathrm{3}} }}=\mathrm{14} \\ $$$$\mathrm{squaring}\:\mathrm{and}\:\mathrm{transforming}\:\mathrm{leads}\:\mathrm{to} \\ $$$${p}^{\mathrm{9}} −\frac{\mathrm{27}}{\mathrm{8}}{p}^{\mathrm{6}} +\frac{\mathrm{27}}{\mathrm{64}}{p}^{\mathrm{3}} −\frac{\mathrm{729}}{\mathrm{512}}=\mathrm{0} \\ $$$$\left({p}^{\mathrm{3}} −\frac{\mathrm{27}}{\mathrm{8}}\right)\left({p}^{\mathrm{6}} +\frac{\mathrm{27}}{\mathrm{64}}\right)=\mathrm{0}\:\Rightarrow\:{p}=\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow\:{q}=\pm\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:{x}=\frac{\mathrm{3}}{\mathrm{2}}\mp\sqrt{\mathrm{2}}\wedge{y}=\frac{\mathrm{3}}{\mathrm{2}}\pm\sqrt{\mathrm{2}} \\ $$
Answered by benjo_mathlover last updated on 27/Oct/20
by substituting { ((u=(√x))),((v=(√y))) :} we obtain equivalent form { ((uv(3u^4 +10u^2 v^2 +3v^4 )=14)),((u^6 +15u^4 v^2 +15u^2 v^4 +v^6 =36)) :} Here we should recognize elements of binomial with exponent equal to 6. { ((36+2.14=u^6 +6u^5 v+15u^4 v^2 +20u^3 v^3 +15u^2 v^4 +6uv^5 +v^6 )),((36−2.14=u^6 −6u^5 v+15u^4 v^2 −20u^3 v^3 +15u^2 v^4 −6uv^5 +v^6 )) :} therefore { (((u+v)^6 =64)),(((u−v)^6 =8)) :} which implies { ((u+v=2)),((u−v=(√2))) :} since u,v have to be +ve so { ((u=1+((√2)/2))),((v=1−((√2)/2))) :} ⇔ { (((x,y)=((3/2)+(√2), (3/2)−(√2)))),(((x,y)=((3/2)−(√2), (3/2)+(√2)) )) :}
$${by}\:{substituting}\:\begin{cases}{{u}=\sqrt{{x}}}\\{{v}=\sqrt{{y}}}\end{cases} \\ $$$${we}\:{obtain}\:{equivalent}\:{form}\: \\ $$$$\begin{cases}{{uv}\left(\mathrm{3}{u}^{\mathrm{4}} +\mathrm{10}{u}^{\mathrm{2}} {v}^{\mathrm{2}} +\mathrm{3}{v}^{\mathrm{4}} \right)=\mathrm{14}}\\{{u}^{\mathrm{6}} +\mathrm{15}{u}^{\mathrm{4}} {v}^{\mathrm{2}} +\mathrm{15}{u}^{\mathrm{2}} {v}^{\mathrm{4}} +{v}^{\mathrm{6}} =\mathrm{36}}\end{cases} \\ $$$${Here}\:{we}\:{should}\:{recognize}\:{elements} \\ $$$${of}\:{binomial}\:{with}\:{exponent}\:{equal}\:{to}\:\mathrm{6}. \\ $$$$\begin{cases}{\mathrm{36}+\mathrm{2}.\mathrm{14}={u}^{\mathrm{6}} +\mathrm{6}{u}^{\mathrm{5}} {v}+\mathrm{15}{u}^{\mathrm{4}} {v}^{\mathrm{2}} +\mathrm{20}{u}^{\mathrm{3}} {v}^{\mathrm{3}} +\mathrm{15}{u}^{\mathrm{2}} {v}^{\mathrm{4}} +\mathrm{6}{uv}^{\mathrm{5}} +{v}^{\mathrm{6}} }\\{\mathrm{36}−\mathrm{2}.\mathrm{14}={u}^{\mathrm{6}} −\mathrm{6}{u}^{\mathrm{5}} {v}+\mathrm{15}{u}^{\mathrm{4}} {v}^{\mathrm{2}} −\mathrm{20}{u}^{\mathrm{3}} {v}^{\mathrm{3}} +\mathrm{15}{u}^{\mathrm{2}} {v}^{\mathrm{4}} −\mathrm{6}{uv}^{\mathrm{5}} +{v}^{\mathrm{6}} }\end{cases} \\ $$$${therefore}\:\begin{cases}{\left({u}+{v}\right)^{\mathrm{6}} =\mathrm{64}}\\{\left({u}−{v}\right)^{\mathrm{6}} =\mathrm{8}}\end{cases} \\ $$$${which}\:{implies}\:\begin{cases}{{u}+{v}=\mathrm{2}}\\{{u}−{v}=\sqrt{\mathrm{2}}}\end{cases}\:{since}\:{u},{v}\:{have}\:{to}\:{be}\:+{ve} \\ $$$${so}\:\begin{cases}{{u}=\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{{v}=\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{cases}\:\Leftrightarrow\:\begin{cases}{\left({x},{y}\right)=\left(\frac{\mathrm{3}}{\mathrm{2}}+\sqrt{\mathrm{2}},\:\frac{\mathrm{3}}{\mathrm{2}}−\sqrt{\mathrm{2}}\right)}\\{\left({x},{y}\right)=\left(\frac{\mathrm{3}}{\mathrm{2}}−\sqrt{\mathrm{2}},\:\frac{\mathrm{3}}{\mathrm{2}}+\sqrt{\mathrm{2}}\right)\:}\end{cases} \\ $$
Answered by behi83417@gmail.com last updated on 27/Oct/20
{ (((3x^2 +3y^2 +10xy)(√(xy))=14)),(((x+y)(x^2 +14xy+y^2 )=36)) :} ⇒_(xy=q^2 ) ^(x+y=p) { (([3(x+y)^2 +4xy](√(xy))=14)),(((x+y)[(x+y)^2 +12xy]=36)) :} ⇒ { ((q.(3p^2 +4q^2 )=14)),((p.(p^2 +12q^2 )=36)) :}⇒(q/p).((3p^2 +4q^2 )/(p^2 +12q^2 ))=(7/(18)) ⇒^((p/q)=t) (1/t).((3t^2 +4)/(t^2 +12))=(7/(18))⇒54t^2 +72=7t^3 +84t ⇒7t^3 −54t^2 +84t−72=0 ⇒(t−6)(7t^2 −12t+12)=0 ⇒ { ((t=(p/q)=6)),((t=((6±(√(36−84)))/7)=((6±4i(√3))/7))) :} (p/q)=6⇒6q^2 (36q^2 +12q^2 )=36⇒q^3 =(1/8) ⇒q=(1/2),p=6q=3⇒ { ((x+y=3)),((xy=q^2 =(1/4))) :} ⇒z^2 −3z+(1/4)=0⇒z=x∨y=((3±(√(9−1)))/2) ⇒x∨y=((3±2(√2))/2)=(3/2)±(√2) .■
$$\begin{cases}{\left(\mathrm{3x}^{\mathrm{2}} +\mathrm{3y}^{\mathrm{2}} +\mathrm{10xy}\right)\sqrt{\mathrm{xy}}=\mathrm{14}}\\{\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{14xy}+\mathrm{y}^{\mathrm{2}} \right)=\mathrm{36}}\end{cases} \\ $$$$\underset{\mathrm{xy}=\mathrm{q}^{\mathrm{2}} } {\overset{\mathrm{x}+\mathrm{y}=\mathrm{p}} {\Rightarrow}}\begin{cases}{\left[\mathrm{3}\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} +\mathrm{4xy}\right]\sqrt{\mathrm{xy}}=\mathrm{14}}\\{\left(\mathrm{x}+\mathrm{y}\right)\left[\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} +\mathrm{12xy}\right]=\mathrm{36}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{q}.\left(\mathrm{3p}^{\mathrm{2}} +\mathrm{4q}^{\mathrm{2}} \right)=\mathrm{14}}\\{\mathrm{p}.\left(\mathrm{p}^{\mathrm{2}} +\mathrm{12q}^{\mathrm{2}} \right)=\mathrm{36}}\end{cases}\Rightarrow\frac{\mathrm{q}}{\mathrm{p}}.\frac{\mathrm{3p}^{\mathrm{2}} +\mathrm{4q}^{\mathrm{2}} }{\mathrm{p}^{\mathrm{2}} +\mathrm{12q}^{\mathrm{2}} }=\frac{\mathrm{7}}{\mathrm{18}} \\ $$$$\overset{\frac{\mathrm{p}}{\mathrm{q}}=\mathrm{t}} {\Rightarrow}\frac{\mathrm{1}}{\mathrm{t}}.\frac{\mathrm{3t}^{\mathrm{2}} +\mathrm{4}}{\mathrm{t}^{\mathrm{2}} +\mathrm{12}}=\frac{\mathrm{7}}{\mathrm{18}}\Rightarrow\mathrm{54t}^{\mathrm{2}} +\mathrm{72}=\mathrm{7t}^{\mathrm{3}} +\mathrm{84t} \\ $$$$\Rightarrow\mathrm{7t}^{\mathrm{3}} −\mathrm{54t}^{\mathrm{2}} +\mathrm{84t}−\mathrm{72}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{t}−\mathrm{6}\right)\left(\mathrm{7t}^{\mathrm{2}} −\mathrm{12t}+\mathrm{12}\right)=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{\mathrm{t}=\frac{\mathrm{p}}{\mathrm{q}}=\mathrm{6}}\\{\mathrm{t}=\frac{\mathrm{6}\pm\sqrt{\mathrm{36}−\mathrm{84}}}{\mathrm{7}}=\frac{\mathrm{6}\pm\mathrm{4i}\sqrt{\mathrm{3}}}{\mathrm{7}}}\end{cases} \\ $$$$\frac{\mathrm{p}}{\mathrm{q}}=\mathrm{6}\Rightarrow\mathrm{6q}^{\mathrm{2}} \left(\mathrm{36q}^{\mathrm{2}} +\mathrm{12q}^{\mathrm{2}} \right)=\mathrm{36}\Rightarrow\mathrm{q}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\Rightarrow\mathrm{q}=\frac{\mathrm{1}}{\mathrm{2}},\mathrm{p}=\mathrm{6q}=\mathrm{3}\Rightarrow\begin{cases}{\mathrm{x}+\mathrm{y}=\mathrm{3}}\\{\mathrm{xy}=\mathrm{q}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}}\end{cases} \\ $$$$\Rightarrow\mathrm{z}^{\mathrm{2}} −\mathrm{3z}+\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0}\Rightarrow\mathrm{z}=\mathrm{x}\vee\mathrm{y}=\frac{\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{1}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x}\vee\mathrm{y}=\frac{\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}\pm\sqrt{\mathrm{2}}\:\:\:\:.\blacksquare \\ $$

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