Bonjour. Aidez moi a resoudre le systeme suivant dans R^2 x−y=2 xy=20


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Question Number 72289 by mathocean1 last updated on 27/Oct/19

$B o n j o u r .$ $A i d e z m o i a r e s o u d r e l e s y s t e m e$ $s u i v a n t d a n s R^{2}$ $x - y = 2$ $x y = 20$
	Answered by mr W last updated on 27/Oct/19

	$x = 2 + y$ $(2 + y) y = 20$ $y^{2} + 2 y - 20 = 0$ $y = \frac{- 2 \pm \sqrt{2^{2} - 4 (- 20)}}{2} = \frac{- 2 \pm 2 \sqrt{21}}{2} = - 1 \pm \sqrt{21}$ $x = 2 + y = 1 \pm \sqrt{21}$
	Answered by Henri Boucatchou last updated on 27/Oct/19
	$poser z = −y on a { (( x+z = 2)),(( xz = −20)) :} ⇒ x et z sont solutions de x^2 − 2x −20 = 0 soit (x−1)^2 −1−20 = (x−1+ (√(21)))(x−1−(√(21)))=0 x=1−(√(21)) ⇒ z =−y = 1+(√(21)) ⇒ y = −1−(√(21)) x=1+(√(21)) ⇒ z = −y= 1−(√(21)) ⇒ y=−+(√(21))$
	$p o s e r z = - y$ $o n a {\begin{cases} x + z = 2 \\ x z = - 20 \end{cases} \Rightarrow x e t z s o n t s o l u t i o n s d e$ $x^{2} - 2 x - 20 = 0$ $s o i t {(x - 1)}^{2} - 1 - 20 = (x - 1 + \sqrt{21}) (x - 1 - \sqrt{21}) = 0$ $x = 1 - \sqrt{21} \Rightarrow z = - y = 1 + \sqrt{21} \Rightarrow y = - 1 - \sqrt{21}$ $x = 1 + \sqrt{21} \Rightarrow z = - y = 1 - \sqrt{21} \Rightarrow y = - + \sqrt{21}$
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