x-y-11-x-y-7-

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

{\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})

\dots 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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