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Question Number 212114 by Ismoiljon_008 last updated on 01/Oct/24
(x^2 + 1)(y^2 + 1) + 9 = 6(x+y) Find that: x^2 + y^2 = ?
$$ \\ $$$$\:\:\:\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({y}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:+\:\mathrm{9}\:=\:\mathrm{6}\left({x}+{y}\right) \\ $$$$\:\:\:\mathcal{F}{ind}\:{that}:\:\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:=\:? \\ $$$$ \\ $$
Answered by Frix last updated on 01/Oct/24
x, y ∈C ⇒ x^2 +y^2 ∈C (not unique) x, y ∈R is only possible for xy=1 ⇒ x^2 +y^2 =7
$${x},\:{y}\:\in\mathbb{C}\:\Rightarrow\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \in\mathbb{C}\:\left(\mathrm{not}\:\mathrm{unique}\right) \\ $$$${x},\:{y}\:\in\mathbb{R}\:\mathrm{is}\:\mathrm{only}\:\mathrm{possible}\:\mathrm{for}\:{xy}=\mathrm{1}\:\Rightarrow\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{7} \\ $$
Commented by Frix last updated on 01/Oct/24
(x^2 +1)(y^2 +1)+9−6(x+y)=0 Due to symmetry we can exchange x and y (1) Solve for y: y=(3/(x^2 +1))±((x^2 −3x+1)/(x^2 +1))i x^2 +y^2 =((x^6 +x^4 +6x^3 −10x^2 +6x+8)/((x^2 +1)^2 ))±(((x^2 −3x+1)^2 )/((x^2 +1)^2 ))i For x, y ∈R ⇒ x^2 −3x+1=0 ⇒ x=(3/2)±((√5)/2)∧y=(3/2)∓((√5)/2) x^2 +y^2 =7 (2) Let x=u−(√v)∧y=u+(√v) v^2 −2(u^2 −1)v+(u^4 +2u^2 −12u+10)=0 v=u^2 −1±(2u−3)i x^2 +y^2 =2(2u^2 −1)±2(2u−3)i with u∈C x^2 +y^2 ∈R ⇒ 2u−3=0 ⇔ u=(3/2) ⇒ x^2 +y^2 =7 [v=(5/4); x=(3/2)−((√5)/2)∧y=(3/2)+((√5)/2)] (3) (x^2 +1)(y^2 +1)+9−6(x+y)=0 Let y=(1/x) ((x^4 −6x^3 +11x^2 −6x+1)/x^2 )=0 (((x^2 −3x+1)^2 )/x^2 )=0 x=(3/2)±((√5)/2) ⇒ y=(3/2)∓((√5)/2) x^2 +y^2 =7
$$\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({y}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{9}−\mathrm{6}\left({x}+{y}\right)=\mathrm{0} \\ $$$$\mathrm{Due}\:\mathrm{to}\:\mathrm{symmetry}\:\mathrm{we}\:\mathrm{can}\:\mathrm{exchange}\:{x}\:\mathrm{and}\:{y} \\ $$$$ \\ $$$$\left(\mathrm{1}\right) \\ $$$$\mathrm{Solve}\:\mathrm{for}\:{y}: \\ $$$${y}=\frac{\mathrm{3}}{{x}^{\mathrm{2}} +\mathrm{1}}\pm\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{i} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\frac{{x}^{\mathrm{6}} +{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{3}} −\mathrm{10}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\pm\frac{\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{i} \\ $$$$\mathrm{For}\:{x},\:{y}\:\in\mathbb{R}\:\Rightarrow\:{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0}\:\Rightarrow \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\wedge{y}=\frac{\mathrm{3}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{7} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$$\mathrm{Let}\:{x}={u}−\sqrt{{v}}\wedge{y}={u}+\sqrt{{v}} \\ $$$${v}^{\mathrm{2}} −\mathrm{2}\left({u}^{\mathrm{2}} −\mathrm{1}\right){v}+\left({u}^{\mathrm{4}} +\mathrm{2}{u}^{\mathrm{2}} −\mathrm{12}{u}+\mathrm{10}\right)=\mathrm{0} \\ $$$${v}={u}^{\mathrm{2}} −\mathrm{1}\pm\left(\mathrm{2}{u}−\mathrm{3}\right)\mathrm{i} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2}\left(\mathrm{2}{u}^{\mathrm{2}} −\mathrm{1}\right)\pm\mathrm{2}\left(\mathrm{2}{u}−\mathrm{3}\right)\mathrm{i}\:\mathrm{with}\:{u}\in\mathbb{C} \\ $$$$ \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} \in\mathbb{R}\:\Rightarrow\:\mathrm{2}{u}−\mathrm{3}=\mathrm{0}\:\Leftrightarrow\:{u}=\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{7} \\ $$$$\left[{v}=\frac{\mathrm{5}}{\mathrm{4}};\:{x}=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\wedge{y}=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right] \\ $$$$ \\ $$$$\left(\mathrm{3}\right) \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({y}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{9}−\mathrm{6}\left({x}+{y}\right)=\mathrm{0} \\ $$$$\mathrm{Let}\:{y}=\frac{\mathrm{1}}{{x}} \\ $$$$\frac{{x}^{\mathrm{4}} −\mathrm{6}{x}^{\mathrm{3}} +\mathrm{11}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\frac{\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} }=\mathrm{0} \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\:{y}=\frac{\mathrm{3}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{7} \\ $$
Answered by Rasheed.Sindhi last updated on 01/Oct/24
(x^2 + 1)(y^2 + 1) + 9 = 6(x+y) x^2 y^2 +x^2 +y^2 +10−6(x+y)=0 (xy)^2 +(x+y)^2 −2xy+10−6(x+y)=0 xy=a , x+y=b a^2 +b^2 −2a−6b+10=0 a^2 −2a+1+b^2 −6b+9=0 (a−1)^2 +(b−3)^2 =0 a−1=0 ∧ b−3=0 a=1 ∧ b=3 xy=1 ∧ x+y=3 x(3−x)=1 3x−x^2 −1=0 x^2 −3x+1=0 x=((3±(√(9−4)))/2)=((3±(√5))/2) y=3−x=3−((3±(√5))/2)=((6−3∓(√5) )/2) y=((3∓(√5) )/2) x^2 +y^2 =(((3±(√5))/2))^2 +(((3∓(√5) )/2))^2 =((9+5±6(√5))/4)+((9+5∓6(√5))/4) =((14±6(√5) +14∓6(√5))/4)=7
$$\:\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({y}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:+\:\mathrm{9}\:=\:\mathrm{6}\left({x}+{y}\right) \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} +{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{10}−\mathrm{6}\left({x}+{y}\right)=\mathrm{0} \\ $$$$\left({xy}\right)^{\mathrm{2}} +\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}+\mathrm{10}−\mathrm{6}\left({x}+{y}\right)=\mathrm{0} \\ $$$${xy}={a}\:,\:{x}+{y}={b} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{a}−\mathrm{6}{b}+\mathrm{10}=\mathrm{0} \\ $$$${a}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}+{b}^{\mathrm{2}} −\mathrm{6}{b}+\mathrm{9}=\mathrm{0} \\ $$$$\left({a}−\mathrm{1}\right)^{\mathrm{2}} +\left({b}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${a}−\mathrm{1}=\mathrm{0}\:\wedge\:{b}−\mathrm{3}=\mathrm{0} \\ $$$${a}=\mathrm{1}\:\wedge\:{b}=\mathrm{3} \\ $$$${xy}=\mathrm{1}\:\wedge\:{x}+{y}=\mathrm{3} \\ $$$${x}\left(\mathrm{3}−{x}\right)=\mathrm{1} \\ $$$$\mathrm{3}{x}−{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${y}=\mathrm{3}−{x}=\mathrm{3}−\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}=\frac{\mathrm{6}−\mathrm{3}\mp\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$$${y}=\frac{\mathrm{3}\mp\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{3}\mp\sqrt{\mathrm{5}}\:}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{9}+\mathrm{5}\pm\mathrm{6}\sqrt{\mathrm{5}}}{\mathrm{4}}+\frac{\mathrm{9}+\mathrm{5}\mp\mathrm{6}\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{14}\pm\mathrm{6}\sqrt{\mathrm{5}}\:+\mathrm{14}\mp\mathrm{6}\sqrt{\mathrm{5}}}{\mathrm{4}}=\mathrm{7} \\ $$
Answered by A5T last updated on 01/Oct/24
x^2 y^2 +x^2 +y^2 +1+9=6x+6y x^2 y^2 −2xy+1+(x+y)^2 −6(x+y)+9=0 ⇒(xy−1)^2 +(x+y−3)^2 =0 ⇒xy−1=0 and x+y−3=0 ⇒xy=1...(i);x+y=3...(ii) ⇒x^2 +y^2 =(x+y)^2 −2xy=3^2 −2=7
$${x}^{\mathrm{2}} {y}^{\mathrm{2}} +{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}+\mathrm{9}=\mathrm{6}{x}+\mathrm{6}{y} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{2}{xy}+\mathrm{1}+\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{6}\left({x}+{y}\right)+\mathrm{9}=\mathrm{0} \\ $$$$\Rightarrow\left({xy}−\mathrm{1}\right)^{\mathrm{2}} +\left({x}+{y}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{xy}−\mathrm{1}=\mathrm{0}\:{and}\:{x}+{y}−\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow{xy}=\mathrm{1}…\left({i}\right);{x}+{y}=\mathrm{3}…\left({ii}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}=\mathrm{3}^{\mathrm{2}} −\mathrm{2}=\mathrm{7} \\ $$
Commented by Frix last updated on 01/Oct/24
Great!
$$\mathrm{Great}! \\ $$

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