let x and y such that 2x^2 +4x−2y=0 y^2 −(x+6)^2 =0 find the possibles value of x+y


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Question Number 34211 by candre last updated on 02/May/18

$l e t x a n d y s u c h t h a t$ $2 x^{2} + 4 x - 2 y = 0$ $y^{2} - {(x + 6)}^{2} = 0$ $f i n d t h e p o s s i b l e s v a l u e o f x + y$
	Answered by MJS last updated on 02/May/18
	$I. 2x^2 +4x−2y=0 II. y^2 −(x+6)^2 =0 ⇒ y=±(x+6) case 1 y=−x−6 ⇒ x+y=−6 2x^2 +4x−2(−x−6)=0 2x^2 +6x+12=0 x_1 =−(3/2)−((√7)/2)i; y_1 =−(9/2)+((√7)/2)i x_2 =−(3/2)+((√7)/2)i; y_2 =−(9/2)−((√7)/2)i case 2 y=x+6 2x^2 +4x−2(x+6)=0 2x^2 +2x−12=0 x_3 =−3; y_3 =3 x_4 =2; y_4 =8 x, y∈C ⇒ (x+y)∈{−6; 0; 10} x, y∈R ⇒ (x+y)∈{0; 10}$
	$I . 2 x^{2} + 4 x - 2 y = 0$ $I I . y^{2} - {(x + 6)}^{2} = 0 \Rightarrow y = \pm (x + 6)$ $case 1$ $y = - x - 6 \Rightarrow x + y = - 6$ $2 x^{2} + 4 x - 2 (- x - 6) = 0$ $2 x^{2} + 6 x + 12 = 0$ $x_{1} = - \frac{3}{2} - \frac{\sqrt{7}}{2} i; y_{1} = - \frac{9}{2} + \frac{\sqrt{7}}{2} i$ $x_{2} = - \frac{3}{2} + \frac{\sqrt{7}}{2} i; y_{2} = - \frac{9}{2} - \frac{\sqrt{7}}{2} i$ $case 2$ $y = x + 6$ $2 x^{2} + 4 x - 2 (x + 6) = 0$ $2 x^{2} + 2 x - 12 = 0$ $x_{3} = - 3; y_{3} = 3$ $x_{4} = 2; y_{4} = 8$ $x, y \in C \Rightarrow (x + y) \in {- 6; 0; 10}$ $x, y \in R \Rightarrow (x + y) \in {0; 10}$
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