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Question Number 195301 by deleteduser4 last updated on 29/Jul/23

$$\{\begin{array}{l}{x}^2+y=11\\ x+y^2=7\end{array}\Rightarrow x,y=?$$

Answered by Frix last updated on 29/Jul/23

$$y=11-x^2$$$$x^4-22x^2+x+114=0$$$$(x-3)(x^3+3x^2-13x-38)=0$$$$x_1=3$$$$x^3+3x^2-13x-38=0$$$$x=t-1$$$$t^3-16t-23=0$$$$3\mathrm{real}\mathrm{solutions}\Rightarrow \mathrm{use}\mathrm{trigonometric}\mathrm{method}$$$$\mathrm{to}\mathrm{get}\mathrm{exact}\mathrm{values}$$

Commented by deleteduser4 last updated on 29/Jul/23

$$thankyousir;+38or-38?$$

Commented by Frix last updated on 29/Jul/23

$$\mathrm{Typo},-38\mathrm{is}\mathrm{correct}$$

Commented by Frix last updated on 29/Jul/23

$$t_1=\frac{8\sqrt{3}}{3}\sin \frac{\pi +{\sin }^{-1}\frac{69\sqrt{3}}{128}}{3}\Rightarrow x\approx 3.58442834\wedge y\approx -1.84812653$$$$t_2=-\frac{8\sqrt{3}}{3}\cos \frac{\pi +{2\sin }^{-1}\frac{69\sqrt{3}}{128}}{6}\Rightarrow x\approx -3.77931025\wedge y\approx -3.28318599$$$$t_3=-\frac{8\sqrt{3}}{3}\sin \frac{{\sin }^{-1}\frac{69\sqrt{3}}{128}}{3}\Rightarrow x\approx -2.80511809\wedge y\approx 3.13131252$$$$\mathrm{and}\mathrm{of}\mathrm{course}x=3\wedge y=2$$

Commented by deleteduser4 last updated on 29/Jul/23

$$veryverynicesir;andinfinitythankyou$$

Commented by Frix last updated on 30/Jul/23

$$\mathrm{To}\mathrm{clarify}:$$$$t^3+pt+q=0$$$$\mathrm{If}\frac{p^3}{27}+\frac{q^2}{4}<0\mathrm{Cardano}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{work}.\mathrm{This}$$$$\mathrm{is}\mathrm{even}\mathrm{wrong}\mathrm{in}\mathrm{wolframalpha}.\mathrm{We}\mathrm{need}$$$$\mathrm{the}\mathrm{Trigonometric}\mathrm{Method}\mathrm{leading}\mathrm{to}$$$$t_k=\frac{2\sqrt{-3p}}{3}\sin \frac{2k\pi +{\sin }^{-1}\frac{3\sqrt{3}q}{2\sqrt{-p^3}}}{3}\mathrm{with}k=1,2,3$$

Commented by Frix last updated on 30/Jul/23

$$\mathrm{My}\mathrm{solution}\mathrm{proves}\mathrm{there}\mathrm{are}4\mathrm{real}\mathrm{intersections}.$$$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{obviously}\mathrm{a}\mathrm{problem}\mathrm{of}\mathrm{the}\mathrm{approximation}.$$$$\mathrm{An}\mathrm{imaginary}\mathrm{part}\mathrm{of}\approx .0000000000000001\mathrm{i}$$$$\mathrm{is}\mathrm{strange}.\mathrm{But}\mathrm{as}\mathrm{I}\mathrm{wrote}\mathrm{before},\mathrm{your}$$$$\mathrm{picture}\mathrm{also}\mathrm{shows}4\mathrm{real}\mathrm{solutions}.\mathrm{Do}\mathrm{you}$$$$\mathrm{trust}\mathrm{more}\mathrm{in}\mathrm{an}\mathrm{algorithm}?$$

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