{ (((x+1)^2 (y+1)^2 =27xy)),(((x^2 +1)(y^2 +1) =10xy)) :}


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 88811 by jagoll last updated on 13/Apr/20
${ (((x+1)^2 (y+1)^2 =27xy)),(((x^2 +1)(y^2 +1) =10xy)) :}$
${\begin{cases} {(x + 1)}^{2} {(y + 1)}^{2} = 27 x y \\ (x^{2} + 1) (y^{2} + 1) = 10 x y \end{cases}$
	Answered by john santu last updated on 13/Apr/20

	$l e t m e t r y$ ${(\sqrt{x} + \frac{1}{\sqrt{x}})}^{2} {(\sqrt{y} + \frac{1}{\sqrt{y}})}^{2} = 27$ $(x + \frac{1}{x}) (y + \frac{1}{y}) = 10$ $t a k e \sqrt{x} + \frac{1}{\sqrt{x}} = u, \sqrt{y} + \frac{1}{\sqrt{y}} = v$ $\Rightarrow u^{2} v^{2} = 27$ $\Rightarrow (u^{2} - 2) (v^{2} - 2) = 10$ $u^{2} v^{2} - 2 (u^{2} + v^{2}) = 6 \Rightarrow u^{2} + v^{2} = \frac{21}{2}$ $u^{2}, v^{2} a r e r o o t s o f w^{2} - \frac{21}{2} w + 27 = 0$ $w e g e t w_{1} = 6 \land w_{2} = \frac{9}{2}$ $t h e n u^{2} = 6 \land v^{2} = \frac{9}{2}$ $u^{2} - 2 = 4, w^{2} - 2 = \frac{5}{2}$ $(1) x + \frac{1}{x} = 4 \Rightarrow x = 2 \pm \sqrt{3}$ $(2) x + \frac{1}{x} = \frac{5}{2} \Rightarrow x = \frac{1}{2}; 2$ $s i n c e e q n (1) & (2) x a n d y s y m e t r i c,$ $t h e s o l u t i o n (x, y) i s (2 \pm \sqrt{3}, 2),$ $(2 \pm \sqrt{3}, \frac{1}{2}), (\frac{1}{2}, 2 \pm \sqrt{3}),$ $(2, 2 \pm \sqrt{3})$
	Commented by jagoll last updated on 13/Apr/20

	$w a w . . . c o o l l$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum