{ (((√x) + y = 11)),((x + (√y) = 7 )) :}


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Question Number 125185 by bemath last updated on 08/Dec/20

${\begin{cases} \sqrt{x} + y = 11 \\ x + \sqrt{y} = 7 \end{cases}$
	Answered by Olaf last updated on 09/Dec/20
	${ (((√x)+y = 11 (1))),((x+(√y) = 7 (2))) :} (4,9) is a quite evident solution. (2) : (√y) = 7−x ⇒ y = x^2 −14x+49 (1) : (√x)+x^2 −14x+49 = 11 x^2 −14x+ (√x)+38 = 0 (3) Let X = (√x) x = 4 is a root ⇒ X = 2 is a root (3) : X^4 −14X^2 +X+38 = 0 (X−2)(X^3 +2X^2 −10X−19) = 0 Let f(X) = X^3 +2X^2 −10X−19 f′(X) = 3X^2 +4X−10 f′(X) = 3(X^2 +(4/3)X−((10)/3)) f′(X) = 3(X+((2+(√(34)))/3))(X+((2−(√(34)))/3)) ...etc f admit one real positive root : X = (1/3)(14+((3194+3(√(334731))))^(1/3) +((193(4)^(1/3) )/( ((6388+6(√(334731))))^(1/3) ))] X ≈ 14,119789502 ⇒ x > 196 ⇒ y < 0 : impossible ! Finally, only one solution (x,y) = (4,9)$
	${\begin{cases} \sqrt{x} + y = 11 (1) \\ x + \sqrt{y} = 7 (2) \end{cases}$ $(4, 9) is a quite evident solution .$ $(2) : \sqrt{y} = 7 - x$ $\Rightarrow y = x^{2} - 14 x + 49$ $(1) : \sqrt{x} + x^{2} - 14 x + 49 = 11$ $x^{2} - 14 x + \sqrt{x} + 38 = 0 (3)$ $Let X = \sqrt{x}$ $x = 4 is a root \Rightarrow X = 2 is a root$ $(3) : X^{4} - {14 X}^{2} + X + 38 = 0$ $(X - 2) (X^{3} + 2 X^{2} - 10 X - 19) = 0$ $Let f (X) = X^{3} + {2 X}^{2} - 10 X - 19$ $f^{'} (X) = {3 X}^{2} + 4 X - 10$ $f^{'} (X) = 3 (X^{2} + \frac{4}{3} X - \frac{10}{3})$ $f^{'} (X) = 3 (X + \frac{2 + \sqrt{34}}{3}) (X + \frac{2 - \sqrt{34}}{3})$ $. . . etc$ $f admit one real positive root :$ $X = \frac{1}{3} (14 + \sqrt[3]{3194 + 3 \sqrt{334731}} + \frac{193 \sqrt[3]{4}}{\sqrt[3]{6388 + 6 \sqrt{334731}}}]$ $X \approx 14, 119789502$ $\Rightarrow x > 196 \Rightarrow y < 0 : impossible!$ $Finally, only one solution (x, y) = (4, 9)$
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