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Question Number 194899 by kapoorshah last updated on 19/Jul/23

Answered by Frix last updated on 19/Jul/23

$$y=px\wedge z=qx$$$$\Rightarrow$$$$x^2=\frac{5}{1-pq}=\frac{8}{p^2-q}=\frac{10}{q^2-p}$$$$\frac{5}{1-pq}=\frac{8}{p^2-q}\Rightarrow q=\frac{8-5p^2}{8p-5}$$$$\mathrm{Inserting}\mathrm{in}\frac{5}{1-pq}=\frac{10}{q^2-p}\mathrm{and}\mathrm{transforming}$$$$p^4+\frac{14p^3}{55}=p+\frac{14}{55}$$$$\Rightarrow p=-\frac{14}{55}[\vee p=1\vee p=-\frac{1}{2}\pm \frac{\sqrt{3}}{2}\mathrm{i}{}_{\mathrm{the}\mathrm{given}\mathrm{eqs}}^{{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{satisfy}}]$$$$\mathrm{Testing}\mathrm{all}\mathrm{solutions}\mathrm{we}\mathrm{get}$$$$q=-\frac{12}{11}$$$$x=\pm \frac{55}{\sqrt{437}}\wedge y=\mp \frac{14}{\sqrt{437}}\wedge z=\mp \frac{60}{\sqrt{437}}$$

Commented by kapoorshah last updated on 19/Jul/23

$$nice$$$$thankyou$$

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