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Question Number 80448 by mind is power last updated on 03/Feb/20
Hello All of You verry Nice Day, God bless You love peace and happiness Solve for (x,y)∈R^2 { ((x^2 +y^2 =2x+3y+1)),((x^4 +y^4 =4x^2 +9y^2 +12xy+2x^2 y^2 +18)) :}
$${Hello}\:{All}\:{of}\:{You}\:{verry}\:{Nice}\:{Day},\:{God}\:{bless}\:{You}\:{love}\:{peace}\:{and}\: \\ $$$${happiness}\: \\ $$$${Solve}\:{for}\:\left({x},{y}\right)\in\mathbb{R}^{\mathrm{2}} \: \\ $$$$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2}{x}+\mathrm{3}{y}+\mathrm{1}}\\{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} =\mathrm{4}{x}^{\mathrm{2}} +\mathrm{9}{y}^{\mathrm{2}} +\mathrm{12}{xy}+\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{18}}\end{cases} \\ $$$$ \\ $$
Commented by mr W last updated on 03/Feb/20
thank you sir! the same to you!
$${thank}\:{you}\:{sir}!\:{the}\:{same}\:{to}\:{you}! \\ $$
Commented by john santu last updated on 03/Feb/20
x^4 +y^4 =(x^2 +y^2 )^2 −2x^2 y^2 (2x+3y+1)^2 −2x^2 y^2 =4x^2 +9y^2 +12xy+2x^2 y^2 +18 4x^2 +9y^2 +1+2(6xy+2x+3y)−2x^2 y^2 = 4x^2 +9y^2 +12xy+2x^2 y^2 +18 1+4x+6y−2x^2 y^2 =2x^2 y^2 +18 4x^2 y^2 −4x−6y+17=0 4x^2 y^2 −2(2x+3y)+17=0 4x^2 y^2 −2(x^2 +y^2 −1)+17=0 2(x^2 +y^2 )−4x^2 y^2 −19=0 let x^2 = t , y^2 =p 2(t+p)−4tp−19=0 next...
$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} =\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}{x}+\mathrm{3}{y}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{4}{x}^{\mathrm{2}} +\mathrm{9}{y}^{\mathrm{2}} +\mathrm{12}{xy}+\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{18} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{9}{y}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}\left(\mathrm{6}{xy}+\mathrm{2}{x}+\mathrm{3}{y}\right)−\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} = \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{9}{y}^{\mathrm{2}} +\mathrm{12}{xy}+\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{18} \\ $$$$\mathrm{1}+\mathrm{4}{x}+\mathrm{6}{y}−\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{18} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{6}{y}+\mathrm{17}=\mathrm{0} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}{x}+\mathrm{3}{y}\right)+\mathrm{17}=\mathrm{0} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{17}=\mathrm{0} \\ $$$$\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\mathrm{4}{x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{19}=\mathrm{0} \\ $$$${let}\:{x}^{\mathrm{2}} \:=\:{t}\:,\:{y}^{\mathrm{2}} ={p}\: \\ $$$$\mathrm{2}\left({t}+{p}\right)−\mathrm{4}{tp}−\mathrm{19}=\mathrm{0} \\ $$$${next}… \\ $$
Commented by jagoll last updated on 03/Feb/20
to be continue
$${to}\:{be}\:{continue} \\ $$
Answered by MJS last updated on 03/Feb/20
let x=u−v∧y=u+v { ((2u^2 +2v^2 =5u+v+1)),((2u^4 +12u^2 v^2 +2v^4 =2u^4 −4u^2 v^2 +2v^4 +25u^2 +10uv+v^2 +18)) :} { ((2u^2 +2v^2 −5u−v−1=0)),((16u^2 v^2 −25u^2 −10uv−v^2 −18=0)) :} { ((v^2 −(v/2)+((2u^2 −5u−1)/2)=0)),((v^2 −((10u)/(16u^2 −1))v−((25u^2 +18)/(16u^2 −1))=0)) :} subtracting (1) − (2) and solving for v leads to v=((32u^4 −80u^3 +32u^2 +5u+37)/(16u^2 −20u−1)) inserting in (1) or (2) leads to a polynome of 8^(th) degree for u I can only solve approximately ⇒ (all rounded to 6 significant digits) x_1 =−.184612 y_1 =3.18722 x_2 =.323632 y_2 =3.44744 x_(3, 4) =−1.37362±.450762i y_(3, 4) =.705266∓1.34628i x_(5, 6) =.618559±1.80074 y_(5, 6) =−1.22229∓.252316i x_(7, 8) =2.68555±.324293i y_(7, 8) =.199692±.420370i these solutions give the folliwing polynomes: x^8 −4x^7 +2x^6 +6x^5 −16x^4 +15x^3 +((109)/2)x^2 −((17)/2)x−((53)/(16))=0 y^8 −6y^7 +7y^6 +9y^5 −((37)/2)y^4 +((45)/2)y^3 +((123)/4)y^2 −((51)/4)y+((137)/(16))=0
$$\mathrm{let}\:{x}={u}−{v}\wedge{y}={u}+{v} \\ $$$$\begin{cases}{\mathrm{2}{u}^{\mathrm{2}} +\mathrm{2}{v}^{\mathrm{2}} =\mathrm{5}{u}+{v}+\mathrm{1}}\\{\mathrm{2}{u}^{\mathrm{4}} +\mathrm{12}{u}^{\mathrm{2}} {v}^{\mathrm{2}} +\mathrm{2}{v}^{\mathrm{4}} =\mathrm{2}{u}^{\mathrm{4}} −\mathrm{4}{u}^{\mathrm{2}} {v}^{\mathrm{2}} +\mathrm{2}{v}^{\mathrm{4}} +\mathrm{25}{u}^{\mathrm{2}} +\mathrm{10}{uv}+{v}^{\mathrm{2}} +\mathrm{18}}\end{cases} \\ $$$$\begin{cases}{\mathrm{2}{u}^{\mathrm{2}} +\mathrm{2}{v}^{\mathrm{2}} −\mathrm{5}{u}−{v}−\mathrm{1}=\mathrm{0}}\\{\mathrm{16}{u}^{\mathrm{2}} {v}^{\mathrm{2}} −\mathrm{25}{u}^{\mathrm{2}} −\mathrm{10}{uv}−{v}^{\mathrm{2}} −\mathrm{18}=\mathrm{0}}\end{cases} \\ $$$$\begin{cases}{{v}^{\mathrm{2}} −\frac{{v}}{\mathrm{2}}+\frac{\mathrm{2}{u}^{\mathrm{2}} −\mathrm{5}{u}−\mathrm{1}}{\mathrm{2}}=\mathrm{0}}\\{{v}^{\mathrm{2}} −\frac{\mathrm{10}{u}}{\mathrm{16}{u}^{\mathrm{2}} −\mathrm{1}}{v}−\frac{\mathrm{25}{u}^{\mathrm{2}} +\mathrm{18}}{\mathrm{16}{u}^{\mathrm{2}} −\mathrm{1}}=\mathrm{0}}\end{cases} \\ $$$$\mathrm{subtracting}\:\left(\mathrm{1}\right)\:−\:\left(\mathrm{2}\right)\:\mathrm{and}\:\mathrm{solving}\:\mathrm{for}\:{v}\:\mathrm{leads}\:\mathrm{to} \\ $$$${v}=\frac{\mathrm{32}{u}^{\mathrm{4}} −\mathrm{80}{u}^{\mathrm{3}} +\mathrm{32}{u}^{\mathrm{2}} +\mathrm{5}{u}+\mathrm{37}}{\mathrm{16}{u}^{\mathrm{2}} −\mathrm{20}{u}−\mathrm{1}} \\ $$$$\mathrm{inserting}\:\mathrm{in}\:\left(\mathrm{1}\right)\:\mathrm{or}\:\left(\mathrm{2}\right)\:\mathrm{leads}\:\mathrm{to}\:\mathrm{a}\:\mathrm{polynome}\:\mathrm{of} \\ $$$$\mathrm{8}^{\mathrm{th}} \:\mathrm{degree}\:\mathrm{for}\:{u}\:\mathrm{I}\:\mathrm{can}\:\mathrm{only}\:\mathrm{solve}\:\mathrm{approximately} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{all}\:\mathrm{rounded}\:\mathrm{to}\:\mathrm{6}\:\mathrm{significant}\:\mathrm{digits}\right) \\ $$$${x}_{\mathrm{1}} =−.\mathrm{184612}\:\:{y}_{\mathrm{1}} =\mathrm{3}.\mathrm{18722} \\ $$$${x}_{\mathrm{2}} =.\mathrm{323632}\:\:{y}_{\mathrm{2}} =\mathrm{3}.\mathrm{44744} \\ $$$${x}_{\mathrm{3},\:\mathrm{4}} =−\mathrm{1}.\mathrm{37362}\pm.\mathrm{450762i}\:\:{y}_{\mathrm{3},\:\mathrm{4}} =.\mathrm{705266}\mp\mathrm{1}.\mathrm{34628i} \\ $$$${x}_{\mathrm{5},\:\mathrm{6}} =.\mathrm{618559}\pm\mathrm{1}.\mathrm{80074}\:\:{y}_{\mathrm{5},\:\mathrm{6}} =−\mathrm{1}.\mathrm{22229}\mp.\mathrm{252316i} \\ $$$${x}_{\mathrm{7},\:\mathrm{8}} =\mathrm{2}.\mathrm{68555}\pm.\mathrm{324293i}\:\:{y}_{\mathrm{7},\:\mathrm{8}} =.\mathrm{199692}\pm.\mathrm{420370i} \\ $$$$\mathrm{these}\:\mathrm{solutions}\:\mathrm{give}\:\mathrm{the}\:\mathrm{folliwing}\:\mathrm{polynomes}: \\ $$$${x}^{\mathrm{8}} −\mathrm{4}{x}^{\mathrm{7}} +\mathrm{2}{x}^{\mathrm{6}} +\mathrm{6}{x}^{\mathrm{5}} −\mathrm{16}{x}^{\mathrm{4}} +\mathrm{15}{x}^{\mathrm{3}} +\frac{\mathrm{109}}{\mathrm{2}}{x}^{\mathrm{2}} −\frac{\mathrm{17}}{\mathrm{2}}{x}−\frac{\mathrm{53}}{\mathrm{16}}=\mathrm{0} \\ $$$${y}^{\mathrm{8}} −\mathrm{6}{y}^{\mathrm{7}} +\mathrm{7}{y}^{\mathrm{6}} +\mathrm{9}{y}^{\mathrm{5}} −\frac{\mathrm{37}}{\mathrm{2}}{y}^{\mathrm{4}} +\frac{\mathrm{45}}{\mathrm{2}}{y}^{\mathrm{3}} +\frac{\mathrm{123}}{\mathrm{4}}{y}^{\mathrm{2}} −\frac{\mathrm{51}}{\mathrm{4}}{y}+\frac{\mathrm{137}}{\mathrm{16}}=\mathrm{0} \\ $$
Commented by mind is power last updated on 04/Feb/20
thank You Sir nice Worck
$${thank}\:{You}\:{Sir}\:{nice}\:{Worck} \\ $$$$ \\ $$

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