Solve the system { ((y(√x) + x(√y) = x + y)),(((√x) + (√y) = xy)) :} Find all the real solutions other than x = 0 and y = 0


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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
	$u=(√x)+(√y) v=(√(xy)) ⇒uv=u^2 −2v ⇒u=v^2 v^3 =v^4 −2v ⇒v=0 ⇒u=0 ⇒x=y=0 or v^3 −v^2 −2=0 (1/v^2 )+(1/(2v))−(1/2)=0 (1/v)=((((√(87))/(36))+(1/4)))^(1/3) −((((√(87))/(36))−(1/4)))^(1/3) ⇒v=(1/( ((((√(87))/(36))+(1/4)))^(1/3) −((((√(87))/(36))−(1/4)))^(1/3) )) (√x) and (√y) are roots of t^2 −ut+v=0 t^2 −v^2 t+v=0 (√x), (√y)=t=(1/2)(v^2 ±(√(v^4 −4v))) ⇒x, y=(1/4)(v^2 ±(√(v^4 −4v)))^2 ≈ { ((4.188733888)),((0.686395906)) :}$
	$u = \sqrt{x} + \sqrt{y}$ $v = \sqrt{x y}$ $\Rightarrow u v = u^{2} - 2 v$ $\Rightarrow u = v^{2}$ $v^{3} = v^{4} - 2 v$ $\Rightarrow v = 0 \Rightarrow u = 0 \Rightarrow x = y = 0$ $o r$ $v^{3} - v^{2} - 2 = 0$ $\frac{1}{v^{2}} + \frac{1}{2 v} - \frac{1}{2} = 0$ $\frac{1}{v} = \sqrt[3]{\frac{\sqrt{87}}{36} + \frac{1}{4}} - \sqrt[3]{\frac{\sqrt{87}}{36} - \frac{1}{4}}$ $\Rightarrow v = \frac{1}{\sqrt[3]{\frac{\sqrt{87}}{36} + \frac{1}{4}} - \sqrt[3]{\frac{\sqrt{87}}{36} - \frac{1}{4}}}$ $\sqrt{x} a n d \sqrt{y} a r e r o o t s o f$ $t^{2} - u t + v = 0$ $t^{2} - v^{2} t + v = 0$ $\sqrt{x}, \sqrt{y} = t = \frac{1}{2} (v^{2} \pm \sqrt{v^{4} - 4 v})$ $\Rightarrow x, y = \frac{1}{4} {(v^{2} \pm \sqrt{v^{4} - 4 v})}^{2} \approx {\begin{cases} 4.188733888 \\ 0.686395906 \end{cases}$
	Commented by mathdanisur last updated on 26/Aug/21

	$Thank you Ser cool$
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