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x-2-1-y-2-1-9-6-x-y-Find-that-x-2-y-2-




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
$${x},\:{y}\:\in\mathbb{C}\:\Rightarrow\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \in\mathbb{C}\:\left(\mathrm{not}\:\mathrm{unique}\right) \\ $$$${x},\:{y}\:\in\mathbb{R}\:\mathrm{is}\:\mathrm{only}\:\mathrm{possible}\:\mathrm{for}\:{xy}=\mathrm{1}\:\Rightarrow\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{7} \\ $$
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 Thyagaraju last updated on 02/Oct/24
(x^2 + 1)(y^2  + 1)+ 9 = 6(x+y)  x^2 (y^2 + 1)+ 1(y^2 + 1) + 9 =6(x+y)  x^2 y^2 + x^2  + y^2  + 1 + 9 = 6(x+y)  (xy)^2  + x^2  + y^2  + 10 = 6(x+y)  x^2  + y^2  = 6(x+y) − (xy)^2  − 10
$$\left({x}^{\mathrm{2}} +\:\mathrm{1}\right)\left({y}^{\mathrm{2}} \:+\:\mathrm{1}\right)+\:\mathrm{9}\:=\:\mathrm{6}\left({x}+{y}\right) \\ $$$${x}^{\mathrm{2}} \left({y}^{\mathrm{2}} +\:\mathrm{1}\right)+\:\mathrm{1}\left({y}^{\mathrm{2}} +\:\mathrm{1}\right)\:+\:\mathrm{9}\:=\mathrm{6}\left({x}+{y}\right) \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} +\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:\mathrm{1}\:+\:\mathrm{9}\:=\:\mathrm{6}\left({x}+{y}\right) \\ $$$$\left({xy}\right)^{\mathrm{2}} \:+\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:\mathrm{10}\:=\:\mathrm{6}\left({x}+{y}\right) \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:=\:\mathrm{6}\left({x}+{y}\right)\:−\:\left({xy}\right)^{\mathrm{2}} \:−\:\mathrm{10} \\ $$
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 02/Oct/24
Great!  We can also use it for x, y ∈C:  (xy−1)^2 +(x+y−3)^2 =0  xy−1=±(x+y−3)i  ⇒ x\{±i}∈C∧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 ))±((6(x^2 −3x+1))/((x^2 +1)^2 ))i  [x=±i ⇒ y=(3/2)∓i; x^2 +y^2 =(1/4)∓3i]
$$\mathrm{Great}! \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{also}\:\mathrm{use}\:\mathrm{it}\:\mathrm{for}\:{x},\:{y}\:\in\mathbb{C}: \\ $$$$\left({xy}−\mathrm{1}\right)^{\mathrm{2}} +\left({x}+{y}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${xy}−\mathrm{1}=\pm\left({x}+{y}−\mathrm{3}\right)\mathrm{i} \\ $$$$\Rightarrow\:{x}\backslash\left\{\pm\mathrm{i}\right\}\in\mathbb{C}\wedge{y}=\frac{\mathrm{3}}{{x}^{\mathrm{2}} +\mathrm{1}}\pm\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{i} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\frac{{x}^{\mathrm{6}} +{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{3}} −\mathrm{10}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\pm\frac{\mathrm{6}\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{i} \\ $$$$\left[{x}=\pm\mathrm{i}\:\Rightarrow\:{y}=\frac{\mathrm{3}}{\mathrm{2}}\mp\mathrm{i};\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\mp\mathrm{3i}\right] \\ $$

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