Question-200379

Question Number 200379 by cortano12 last updated on 18/Nov/23

Commented by Frix last updated on 18/Nov/23

No exact solutions possible. Transform to { ((y^2 +((x^2 +x+1)/(2(x+1)))y+((x^2 +x−20)/(4(x+1)))=0)),((y^2 +((x^2 +6x+6)/(2x(x+1)))y+((3x−20)/(2x(x+1)))=0)) :} Subtract and solve for y y=−(((x−2)(x^2 +3x−20))/(2(x^3 −5x−6))) Insert and transform x^6 −35x^5 +34x^4 +500x^3 −784x^2 −680x+560=0 This has got 6 real solutions...

$$\mathrm{No}\:\mathrm{exact}\:\mathrm{solutions}\:\mathrm{possible}.\:\mathrm{Transform}\:\mathrm{to} \\ $$$$\begin{cases}{{y}^{\mathrm{2}} +\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}{y}+\frac{{x}^{\mathrm{2}} +{x}−\mathrm{20}}{\mathrm{4}\left({x}+\mathrm{1}\right)}=\mathrm{0}}\\{{y}^{\mathrm{2}} +\frac{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{6}}{\mathrm{2}{x}\left({x}+\mathrm{1}\right)}{y}+\frac{\mathrm{3}{x}−\mathrm{20}}{\mathrm{2}{x}\left({x}+\mathrm{1}\right)}=\mathrm{0}}\end{cases} \\ $$$$\mathrm{Subtract}\:\mathrm{and}\:\mathrm{solve}\:\mathrm{for}\:{y} \\ $$$${y}=−\frac{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{20}\right)}{\mathrm{2}\left({x}^{\mathrm{3}} −\mathrm{5}{x}−\mathrm{6}\right)} \\ $$$$\mathrm{Insert}\:\mathrm{and}\:\mathrm{transform} \\ $$$${x}^{\mathrm{6}} −\mathrm{35}{x}^{\mathrm{5}} +\mathrm{34}{x}^{\mathrm{4}} +\mathrm{500}{x}^{\mathrm{3}} −\mathrm{784}{x}^{\mathrm{2}} −\mathrm{680}{x}+\mathrm{560}=\mathrm{0} \\ $$$$\mathrm{This}\:\mathrm{has}\:\mathrm{got}\:\mathrm{6}\:\mathrm{real}\:\mathrm{solutions}… \\ $$

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