Solve-the-system-y-x-x-y-x-y-x-y-xy-Find-all-the-real-solutions-other-than-x-0-and-y-0-

Question Number 152178 by mathdanisur last updated on 26/Aug/21

Solve the system

{\begin{cases} y \sqrt{x} + x \sqrt{y} = x + y \\ \sqrt{x} + \sqrt{y} = xy \end{cases}

Find all the real solutions other than

x = 0 and y = 0

Commented by john_santu last updated on 26/Aug/21

{\begin{cases} \sqrt{xy} (\sqrt{y} + \sqrt{x}) = x + y \Rightarrow \sqrt{y} + \sqrt{x} = \frac{x + y}{\sqrt{xy}} \\ \sqrt{y} + \sqrt{x} = xy \end{cases}

Answered by MJS_new last updated on 26/Aug/21

x \sqrt{y} + \sqrt{x} y = x + y

\sqrt{x} + \sqrt{y} = x y

x = p^{2} \land p ⩾ 0

y = q^{2} \land q ⩾ 0

p^{2} q + {p q}^{2} = p^{2} + q^{2}

p + q = p^{2} q^{2}

p^{2} q + {p q}^{2} - p^{2} - q^{2} = 0

p^{2} q^{2} - p - q = 0

p = u - v \land q = u + v

2 (u^{3} - u^{2} - {u v}^{2} - v^{2}) = 0 \Rightarrow v^{2} = \frac{u^{2} (u - 1)}{u + 1}

u^{4} - 2 u^{2} v^{2} + v^{4} - 2 u = 0 \Rightarrow

\Rightarrow

u (u^{3} - \frac{1}{2} u^{2} - u - \frac{1}{2}) = 0

\Rightarrow u = v = 0 \Rightarrow x = y = 0

u^{3} - \frac{1}{2} u^{2} - u - \frac{1}{2} = 0

\Rightarrow u = \frac{1}{6} (1 + \sqrt[3]{73 - 6 \sqrt{87}} + \sqrt[3]{73 + 6 \sqrt{87}})

\Leftrightarrow u \approx 1.43756489708

\Rightarrow v \approx . 609074760404

\Rightarrow p \approx . 828490136678 \land q \approx 2.04663965749

\Rightarrow x \approx . 686395906572 \land y \approx 4.18873388759

and obviously we can exchange x and y

Answered by mr W last updated on 26/Aug/21

u = \sqrt{x} + \sqrt{y}

v = \sqrt{x y}

\Rightarrow u v = u^{2} - 2 v

\Rightarrow u = v^{2}

v^{3} = v^{4} - 2 v

\Rightarrow v = 0 \Rightarrow u = 0 \Rightarrow x = y = 0

o r

v^{3} - v^{2} - 2 = 0

\frac{1}{v^{2}} + \frac{1}{2 v} - \frac{1}{2} = 0

\frac{1}{v} = \sqrt[3]{\frac{\sqrt{87}}{36} + \frac{1}{4}} - \sqrt[3]{\frac{\sqrt{87}}{36} - \frac{1}{4}}

\Rightarrow v = \frac{1}{\sqrt[3]{\frac{\sqrt{87}}{36} + \frac{1}{4}} - \sqrt[3]{\frac{\sqrt{87}}{36} - \frac{1}{4}}}

\sqrt{x} a n d \sqrt{y} a r e r o o t s o f

t^{2} - u t + v = 0

t^{2} - v^{2} t + v = 0

\sqrt{x}, \sqrt{y} = t = \frac{1}{2} (v^{2} \pm \sqrt{v^{4} - 4 v})

\Rightarrow x, y = \frac{1}{4} {(v^{2} \pm \sqrt{v^{4} - 4 v})}^{2} \approx {\begin{cases} 4.188733888 \\ 0.686395906 \end{cases}

Commented by mathdanisur last updated on 26/Aug/21

Thank you Ser cool