x+(1/x)=5 and x∙y=4 find x^2 +((1/y))^2 =?


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Question Number 191451 by mathlove last updated on 24/Apr/23

$x + \frac{1}{x} = 5 a n d x \cdot y = 4 f i n d$ $x^{2} + {(\frac{1}{y})}^{2} = ?$
	Answered by mr W last updated on 24/Apr/23

	$x^{2} - 5 x + 1 = 0$ $x = \frac{5 \pm \sqrt{21}}{2}$ $x^{2} = 5 x - 1$ ${(\frac{1}{y})}^{2} = {(\frac{x}{4})}^{2} = \frac{x^{2}}{16}$ $x^{2} + {(\frac{1}{y})}^{2} = \frac{17 x^{2}}{16} = \frac{17}{16} (5 x - 1)$ $= \frac{17}{16} (5 \times \frac{5 \pm \sqrt{21}}{2} - 1) = \frac{391 \pm 85 \sqrt{21}}{32}$
	Commented by mehdee42 last updated on 24/Apr/23

	$s i r W$ $p l e a s e g u i d e m e h o w c a n i s u b m i t a q u e s t i o n i n t h e g r o u p ?$
	Commented by Tinku Tara last updated on 24/Apr/23

	$Just press + button at top .$
	Commented by mr W last updated on 24/Apr/23

	$t o c o m p o s e a n e w q u e s t i o n j u s t p r e s s$ $‘ ‘ +^{″} b u t t o n a t t h e t o p;$ $t o p o s t a n i m a g e a s a q u e s t i o n j u s t$ $p r e s s ‘ ‘ {i m a g e}^{″} b u t t o n a t t h e t o p .$
	Commented by mr W last updated on 24/Apr/23

	Commented by mehdee42 last updated on 24/Apr/23

	$t h a n k y o u v e r y m u c h f o r y o u r k i d n e s s$
	Commented by mathlove last updated on 25/Apr/23

	$t h a n k s$
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