Menu Close

Question-192720

Tinku Tara 0 Comments
Pin




Question Number 192720 by pascal889 last updated on 25/May/23
Answered by Frix last updated on 25/May/23
x=p−q∧y=p+q ⇒ 5p^4 +6p^2 q^2 +5q^4 =109 3p^2 +q^2 =13 ⇒ q^2 =13−3p^2 Inserting in 1^(st) equation p^4 −((39p^2 )/4)+23=0 p=±2∨p=±((√(23))/2) The rest is easy
$${x}={p}−{q}\wedge{y}={p}+{q} \\ $$$$\Rightarrow \\ $$$$\mathrm{5}{p}^{\mathrm{4}} +\mathrm{6}{p}^{\mathrm{2}} {q}^{\mathrm{2}} +\mathrm{5}{q}^{\mathrm{4}} =\mathrm{109} \\ $$$$\mathrm{3}{p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\mathrm{13}\:\Rightarrow\:{q}^{\mathrm{2}} =\mathrm{13}−\mathrm{3}{p}^{\mathrm{2}} \\ $$$$\mathrm{Inserting}\:\mathrm{in}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{equation} \\ $$$${p}^{\mathrm{4}} −\frac{\mathrm{39}{p}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{23}=\mathrm{0} \\ $$$${p}=\pm\mathrm{2}\vee{p}=\pm\frac{\sqrt{\mathrm{23}}}{\mathrm{2}} \\ $$$$\mathrm{The}\:\mathrm{rest}\:\mathrm{is}\:\mathrm{easy} \\ $$
Answered by BaliramKumar last updated on 25/May/23
x^4 +3x^2 y^2 +y^4 = 109 .......(i) x^2 +y^2 +xy = 13 ......(ii) x^4 +3x^2 y^2 +y^4 = 109 (x^2 +y^2 )^2 −2x^2 y^2 +3x^2 y^2 = 109 (x^2 +y^2 )^2 +x^2 y^2 = 109 (x^2 +y^2 +xy)^2 −2xy(x^2 +y^2 ) = 109 (13)^2 −2xy(x^2 +y^2 ) = 109 169−2xy(x^2 +y^2 ) = 109 xy(x^2 +y^2 ) = 30 ......(iii) put x^2 +y^2 = α xy = β α+β = 13 αβ = 30 t^2 −13t+30 = 0 t = 10, 3 α = 10 or 3 β = 3 or 10 x^2 +y^2 = 10 x^2 +y^2 = 3 xy = 3 xy = 10 (x, y)= (3, 1) & (−3, −1) or (1, 3), (−1, −3)
$${x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{4}} \:=\:\mathrm{109}\:\:…….\left({i}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}\:=\:\mathrm{13}\:\:\:……\left({ii}\right) \\ $$$${x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{4}} \:=\:\mathrm{109}\:\: \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:=\:\mathrm{109} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:=\:\mathrm{109} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}\right)^{\mathrm{2}} −\mathrm{2}{xy}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\:=\:\mathrm{109} \\ $$$$\left(\mathrm{13}\right)^{\mathrm{2}} −\mathrm{2}{xy}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\:=\:\mathrm{109} \\ $$$$\mathrm{169}−\mathrm{2}{xy}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\:=\:\mathrm{109} \\ $$$${xy}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\:=\:\mathrm{30}\:\:……\left({iii}\right) \\ $$$${put}\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\alpha\:\:\:\:\:\:{xy}\:=\:\beta \\ $$$$\alpha+\beta\:=\:\mathrm{13}\:\:\:\:\:\:\:\:\:\:\:\:\alpha\beta\:=\:\mathrm{30} \\ $$$$\mathrm{t}^{\mathrm{2}} −\mathrm{13t}+\mathrm{30}\:=\:\mathrm{0} \\ $$$$\mathrm{t}\:=\:\mathrm{10},\:\mathrm{3} \\ $$$$\alpha\:=\:\mathrm{10}\:\mathrm{or}\:\mathrm{3}\:\:\:\:\:\:\:\beta\:=\:\mathrm{3}\:\mathrm{or}\:\mathrm{10} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\mathrm{10}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\mathrm{3} \\ $$$${xy}\:=\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{xy}\:=\:\mathrm{10} \\ $$$$\left({x},\:{y}\right)=\:\left(\mathrm{3},\:\mathrm{1}\right)\:\&\:\left(−\mathrm{3},\:−\mathrm{1}\right)\:\mathrm{or}\:\left(\mathrm{1},\:\mathrm{3}\right),\:\left(−\mathrm{1},\:−\mathrm{3}\right) \\ $$
Answered by York12 last updated on 25/May/23
x^2 +y^2 +2xy−2xy=13−xy (x+y)^2 −2xy=13−xy x^4 +3x^2 y^2 +y^4 =((x+y)^2 −2xy_(13−xy) )^2 +x^2 y^2 =109 (13−xy)^2 =109−x^2 y^2 x^2 y^2 −26xy+169=109−x^2 y^2 2x^2 y^2 −26xy+60_(A quadratic equation in (xy)) =0 →(xy−3)(2xy−20)=0 ∴ xy=3 ∨, xy = 10 case (I) for xy = 3 → y = (3/x) substitute (y) in the second equation x^2 +(9/x^2 ) = 10 →x^4 − 10x^2 + 9 = 0 x^2 = 9 → x = +_− ( 3 ) ∨ x^2 =1 → x = +_− (1) then by substituting we get the required values {(3,1) , (−3,−1) ,(1,3) , (−1,−3)}
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}−\mathrm{2}{xy}=\mathrm{13}−{xy}\: \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}=\mathrm{13}−{xy} \\ $$$${x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{4}} =\left(\underset{\mathrm{13}−{xy}} {\underbrace{\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}}}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{109} \\ $$$$\left(\mathrm{13}−{xy}\right)^{\mathrm{2}} =\mathrm{109}−{x}^{\mathrm{2}} {y}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{26}{xy}+\mathrm{169}=\mathrm{109}−{x}^{\mathrm{2}} {y}^{\mathrm{2}} \\ $$$$\underset{{A}\:{quadratic}\:{equation}\:{in}\:\left({xy}\right)} {\underbrace{\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{26}{xy}+\mathrm{60}}}=\mathrm{0}\:\rightarrow\left({xy}−\mathrm{3}\right)\left(\mathrm{2}{xy}−\mathrm{20}\right)=\mathrm{0} \\ $$$$\therefore\:{xy}=\mathrm{3}\:\:\vee,\:{xy}\:=\:\mathrm{10}\: \\ $$$${case}\:\left({I}\right)\:{for}\:{xy}\:=\:\mathrm{3}\:\rightarrow\:{y}\:=\:\frac{\mathrm{3}}{{x}} \\ $$$${substitute}\:\left({y}\right)\:{in}\:{the}\:{second}\:{equation} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{9}}{{x}^{\mathrm{2}} }\:=\:\mathrm{10}\:\rightarrow{x}^{\mathrm{4}} \:−\:\mathrm{10}{x}^{\mathrm{2}} \:+\:\mathrm{9}\:=\:\mathrm{0} \\ $$$${x}^{\mathrm{2}} =\:\mathrm{9}\:\rightarrow\:{x}\:=\:\:\underset{−} {+}\left(\:\mathrm{3}\:\right)\:\vee\:{x}^{\mathrm{2}} =\mathrm{1}\:\rightarrow\:{x}\:=\:\underset{−} {+}\:\left(\mathrm{1}\right) \\ $$$${then}\:{by}\:{substituting}\:{we}\:{get}\:{the}\:{required}\:{values} \\ $$$$\left\{\left(\mathrm{3},\mathrm{1}\right)\:,\:\left(−\mathrm{3},−\mathrm{1}\right)\:,\left(\mathrm{1},\mathrm{3}\right)\:,\:\left(−\mathrm{1},−\mathrm{3}\right)\right\} \\ $$
Answered by witcher3 last updated on 25/May/23
2⇒x^4 +y^4 +x^2 y^2 +2x^2 y^2 +2xy(x^2 +y^2 )=169 ⇔2xy(13−xy)=169−109=60 t=xy⇔2t(13−t)−60=0 t^2 −13t+30=(t−10)(t−3) t=3⇒2⇒(x+y)^2 −xy⇒(x+y)^2 =16 x+y=+_− 4 T^2 −4T+3=0⇒(T−3)(T−1)⇒T∈{1,3} (x,y)∈{(1,3);(3,1)} T^2 +4T+3⇒(x,y)∈{(−1,−3),(−3,−1)} x+y=+_− (√(23)) ⇒T^2 −(√(23))T+10=0⇒23−40=−17 (x,y)∈{+_− ((((√(23))+i(√(17)))/2),(((√(23))−i(√(17)))/2))}∈C−R withe if (x,y) Solution (y,x) also
$$\mathrm{2}\Rightarrow\mathrm{x}^{\mathrm{4}} +\mathrm{y}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} +\mathrm{2x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} +\mathrm{2xy}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)=\mathrm{169} \\ $$$$\Leftrightarrow\mathrm{2xy}\left(\mathrm{13}−\mathrm{xy}\right)=\mathrm{169}−\mathrm{109}=\mathrm{60} \\ $$$$\mathrm{t}=\mathrm{xy}\Leftrightarrow\mathrm{2t}\left(\mathrm{13}−\mathrm{t}\right)−\mathrm{60}=\mathrm{0} \\ $$$$\mathrm{t}^{\mathrm{2}} −\mathrm{13t}+\mathrm{30}=\left(\mathrm{t}−\mathrm{10}\right)\left(\mathrm{t}−\mathrm{3}\right) \\ $$$$\mathrm{t}=\mathrm{3}\Rightarrow\mathrm{2}\Rightarrow\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} −\mathrm{xy}\Rightarrow\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} =\mathrm{16} \\ $$$$\mathrm{x}+\mathrm{y}=\underset{−} {+}\mathrm{4} \\ $$$$\mathrm{T}^{\mathrm{2}} −\mathrm{4T}+\mathrm{3}=\mathrm{0}\Rightarrow\left(\mathrm{T}−\mathrm{3}\right)\left(\mathrm{T}−\mathrm{1}\right)\Rightarrow\mathrm{T}\in\left\{\mathrm{1},\mathrm{3}\right\} \\ $$$$\left(\mathrm{x},\mathrm{y}\right)\in\left\{\left(\mathrm{1},\mathrm{3}\right);\left(\mathrm{3},\mathrm{1}\right)\right\} \\ $$$$\mathrm{T}^{\mathrm{2}} +\mathrm{4T}+\mathrm{3}\Rightarrow\left(\mathrm{x},\mathrm{y}\right)\in\left\{\left(−\mathrm{1},−\mathrm{3}\right),\left(−\mathrm{3},−\mathrm{1}\right)\right\} \\ $$$$\mathrm{x}+\mathrm{y}=\underset{−} {+}\sqrt{\mathrm{23}} \\ $$$$\Rightarrow\mathrm{T}^{\mathrm{2}} −\sqrt{\mathrm{23}}\mathrm{T}+\mathrm{10}=\mathrm{0}\Rightarrow\mathrm{23}−\mathrm{40}=−\mathrm{17} \\ $$$$\left(\mathrm{x},\mathrm{y}\right)\in\left\{\underset{−} {+}\left(\frac{\sqrt{\mathrm{23}}+\mathrm{i}\sqrt{\mathrm{17}}}{\mathrm{2}},\frac{\sqrt{\mathrm{23}}−\mathrm{i}\sqrt{\mathrm{17}}}{\mathrm{2}}\right)\right\}\in\mathbb{C}−\mathbb{R} \\ $$$$\mathrm{withe}\:\mathrm{if}\:\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{Solution}\:\left(\mathrm{y},\mathrm{x}\right)\:\mathrm{also} \\ $$
Commented by senestro last updated on 25/May/23
i don′t understand your genius
$${i}\:{don}'{t}\:{understand}\:{your}\:{genius} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *