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Question Number 39218 by behi83417@gmail.com last updated on 03/Jul/18

Answered by MJS last updated on 12/Aug/18

$$x_1=1;y_1=\sqrt{2}(\mathrm{plain}\mathrm{to}\mathrm{see}\mathrm{but}\mathrm{there}\mathrm{are}3$$$$\mathrm{more}\mathrm{real}\mathrm{solutions})$$$$y=-x^2+1+\sqrt{2}$$$${(-x^2+1+\sqrt{2})}^2+x-3=0$$$$x^4-2(1+\sqrt{2})x^2+x+2\sqrt{2}=0$$$$(x-1)(x^3+x^2-(1+2\sqrt{2})x-2\sqrt{2})=0$$$$x_1=1$$$$x^3+x^2-(1+2\sqrt{2})x-2\sqrt{2}=0$$$$\mathrm{solved}\mathrm{by}\mathrm{using}\mathrm{trigonometric}\mathrm{method}$$$$(\mathrm{Cardano}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{work}\mathrm{in}\mathrm{case}\mathrm{of}3\mathrm{real}$$$$\mathrm{solutions})$$$$x_2\approx -2.16492;y_2\approx -2.27265$$$$x_3\approx -.700407;y_3\approx 1.92364$$$$x_4\approx 1.86532;y_4\approx -1.06521$$

Commented by behi83417@gmail.com last updated on 04/Jul/18

$$dearMJS!thanksinadvance.$$

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