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Question Number 126887 by mnjuly1970 last updated on 25/Dec/20
... calculus... please solve: ( with explanation) {_(x^2 +y=3) ^(x+y^2 =5)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{calculus}… \\ $$$${please}\:\:{solve}:\:\left(\:{with}\:{explanation}\right) \\ $$$$\:\:\:\:\left\{_{{x}^{\mathrm{2}} +{y}=\mathrm{3}} ^{{x}+{y}^{\mathrm{2}} =\mathrm{5}} \right. \\ $$$$ \\ $$
Commented by Lordose last updated on 25/Dec/20
I don′t know a calculus approach though
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{a}\:\mathrm{calculus}\:\mathrm{approach}\:\mathrm{though} \\ $$
Commented by bramlexs22 last updated on 25/Dec/20
x=5−y^2 (5−y^2 )^2 +y=3 y^4 −10y^2 +y+22=0 (y−2)(y^3 +2y^2 −6y−11)=0 y=2 ∧x=1 y^3 +2y^2 −6y−11=0 Cardano
$${x}=\mathrm{5}−{y}^{\mathrm{2}} \\ $$$$\left(\mathrm{5}−{y}^{\mathrm{2}} \right)^{\mathrm{2}} +{y}=\mathrm{3} \\ $$$${y}^{\mathrm{4}} −\mathrm{10}{y}^{\mathrm{2}} +{y}+\mathrm{22}=\mathrm{0} \\ $$$$\left({y}−\mathrm{2}\right)\left({y}^{\mathrm{3}} +\mathrm{2}{y}^{\mathrm{2}} −\mathrm{6}{y}−\mathrm{11}\right)=\mathrm{0} \\ $$$${y}=\mathrm{2}\:\wedge{x}=\mathrm{1} \\ $$$$ \\ $$$${y}^{\mathrm{3}} +\mathrm{2}{y}^{\mathrm{2}} −\mathrm{6}{y}−\mathrm{11}=\mathrm{0} \\ $$$${Cardano} \\ $$
Answered by Lordose last updated on 25/Dec/20
x + y^2 = 5 −−−(1) x^2 + y = 3 −−−(2) From (1), x = 5−y^2 (5−y^2 )^2 + y = 3 25−10y^2 + y^4 +y = 3 y^4 −10y^2 + y + 22 = 0 (y−2)(y^3 +2y^2 −6x−11)=0 Only Integer value for y=2 sub in (1) x = 5−2^2 = 1 (x,y) = (1,2)
$$ \\ $$$$\mathrm{x}\:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{5}\:−−−\left(\mathrm{1}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}\:=\:\mathrm{3}\:−−−\left(\mathrm{2}\right) \\ $$$$\mathrm{From}\:\left(\mathrm{1}\right),\:\mathrm{x}\:=\:\mathrm{5}−\mathrm{y}^{\mathrm{2}} \\ $$$$\left(\mathrm{5}−\mathrm{y}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\mathrm{y}\:=\:\mathrm{3} \\ $$$$\mathrm{25}−\mathrm{10y}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{4}} +\mathrm{y}\:=\:\mathrm{3} \\ $$$$\mathrm{y}^{\mathrm{4}} −\mathrm{10y}^{\mathrm{2}} \:+\:\mathrm{y}\:+\:\mathrm{22}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{y}−\mathrm{2}\right)\left(\mathrm{y}^{\mathrm{3}} +\mathrm{2y}^{\mathrm{2}} −\mathrm{6x}−\mathrm{11}\right)=\mathrm{0} \\ $$$$\mathrm{Only}\:\mathrm{Integer}\:\mathrm{value}\:\mathrm{for}\:\mathrm{y}=\mathrm{2} \\ $$$$\mathrm{sub}\:\mathrm{in}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{x}\:=\:\mathrm{5}−\mathrm{2}^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$$\left(\mathrm{x},\mathrm{y}\right)\:=\:\left(\mathrm{1},\mathrm{2}\right) \\ $$
Answered by mr W last updated on 25/Dec/20
x+(3−x^2 )^2 =5 x^4 −6x^2 +x+4=0 (x−1)(x^3 +x^2 −5x−4)=0 ⇒x−1=0 ⇒x=1 x^3 +x^2 −5x−4=0 (t−(1/3))^3 +(t−(1/3))^2 −5(t−(1/3))−4=0 t^3 −t^2 +(t/3)−(1/(27))+t^2 −((2t)/3)+(1/9)−5t+(5/3)−4=0 t^3 −((16t)/3)−((61)/(27))=0 t=2(√((16)/9)) sin ((1/3)sin^(−1) ((−((61)/(54)))/(((16)/9)(√((16)/9))))+((2kπ)/3)) t=(8/3) sin (−(1/3)sin^(−1) ((61)/(128))+((2kπ)/3)) (k=0,1,2) ⇒x=(8/3) sin (−(1/3)sin^(−1) ((61)/(128))+((2kπ)/3))−(1/3) (k=0,1,2) ≈−0.7729, 2.1642, −2.3914
$${x}+\left(\mathrm{3}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{5} \\ $$$${x}^{\mathrm{4}} −\mathrm{6}{x}^{\mathrm{2}} +{x}+\mathrm{4}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}−\mathrm{1}=\mathrm{0}\:\Rightarrow{x}=\mathrm{1} \\ $$$${x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{4}=\mathrm{0} \\ $$$$\left({t}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} +\left({t}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{5}\left({t}−\frac{\mathrm{1}}{\mathrm{3}}\right)−\mathrm{4}=\mathrm{0} \\ $$$${t}^{\mathrm{3}} −{t}^{\mathrm{2}} +\frac{{t}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{27}}+{t}^{\mathrm{2}} −\frac{\mathrm{2}{t}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{9}}−\mathrm{5}{t}+\frac{\mathrm{5}}{\mathrm{3}}−\mathrm{4}=\mathrm{0} \\ $$$${t}^{\mathrm{3}} −\frac{\mathrm{16}{t}}{\mathrm{3}}−\frac{\mathrm{61}}{\mathrm{27}}=\mathrm{0} \\ $$$${t}=\mathrm{2}\sqrt{\frac{\mathrm{16}}{\mathrm{9}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{−\frac{\mathrm{61}}{\mathrm{54}}}{\frac{\mathrm{16}}{\mathrm{9}}\sqrt{\frac{\mathrm{16}}{\mathrm{9}}}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right) \\ $$$${t}=\frac{\mathrm{8}}{\mathrm{3}}\:\mathrm{sin}\:\left(−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{61}}{\mathrm{128}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)\:\left({k}=\mathrm{0},\mathrm{1},\mathrm{2}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{8}}{\mathrm{3}}\:\mathrm{sin}\:\left(−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{61}}{\mathrm{128}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{3}}\:\left({k}=\mathrm{0},\mathrm{1},\mathrm{2}\right) \\ $$$$\approx−\mathrm{0}.\mathrm{7729},\:\mathrm{2}.\mathrm{1642},\:−\mathrm{2}.\mathrm{3914} \\ $$

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