Question Number 125185 by bemath last updated on 08/Dec/20
$$\:\begin{cases}{\sqrt{{x}}\:+\:{y}\:=\:\mathrm{11}}\\{{x}\:+\:\sqrt{{y}}\:=\:\mathrm{7}\:}\end{cases} \\ $$
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)](https://www.tinkutara.com/question/Q125198.png)
$$ \\ $$$$\begin{cases}{\sqrt{{x}}+{y}\:=\:\mathrm{11}\:\left(\mathrm{1}\right)}\\{{x}+\sqrt{{y}}\:=\:\mathrm{7}\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\left(\mathrm{4},\mathrm{9}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{quite}\:\mathrm{evident}\:\mathrm{solution}. \\ $$$$\left(\mathrm{2}\right)\::\:\sqrt{{y}}\:=\:\mathrm{7}−{x} \\ $$$$\Rightarrow\:{y}\:=\:{x}^{\mathrm{2}} −\mathrm{14}{x}+\mathrm{49} \\ $$$$\left(\mathrm{1}\right)\::\:\sqrt{{x}}+{x}^{\mathrm{2}} −\mathrm{14}{x}+\mathrm{49}\:=\:\mathrm{11} \\ $$$${x}^{\mathrm{2}} −\mathrm{14}{x}+\:\sqrt{{x}}+\mathrm{38}\:=\:\mathrm{0}\:\left(\mathrm{3}\right) \\ $$$$\mathrm{Let}\:\mathrm{X}\:=\:\sqrt{{x}} \\ $$$${x}\:=\:\mathrm{4}\:\mathrm{is}\:\mathrm{a}\:\mathrm{root}\:\Rightarrow\:\mathrm{X}\:=\:\mathrm{2}\:\mathrm{is}\:\mathrm{a}\:\mathrm{root} \\ $$$$\left(\mathrm{3}\right)\::\:\mathrm{X}^{\mathrm{4}} −\mathrm{14X}^{\mathrm{2}} +\mathrm{X}+\mathrm{38}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{X}−\mathrm{2}\right)\left(\mathrm{X}^{\mathrm{3}} +\mathrm{2}{X}^{\mathrm{2}} −\mathrm{10}{X}−\mathrm{19}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{Let}\:{f}\left(\mathrm{X}\right)\:=\:\mathrm{X}^{\mathrm{3}} +\mathrm{2X}^{\mathrm{2}} −\mathrm{10X}−\mathrm{19} \\ $$$${f}'\left(\mathrm{X}\right)\:=\:\mathrm{3X}^{\mathrm{2}} +\mathrm{4X}−\mathrm{10} \\ $$$${f}'\left(\mathrm{X}\right)\:=\:\mathrm{3}\left(\mathrm{X}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{3}}\mathrm{X}−\frac{\mathrm{10}}{\mathrm{3}}\right) \\ $$$${f}'\left(\mathrm{X}\right)\:=\:\mathrm{3}\left(\mathrm{X}+\frac{\mathrm{2}+\sqrt{\mathrm{34}}}{\mathrm{3}}\right)\left(\mathrm{X}+\frac{\mathrm{2}−\sqrt{\mathrm{34}}}{\mathrm{3}}\right) \\ $$$$…\mathrm{etc} \\ $$$${f}\:\mathrm{admit}\:\mathrm{one}\:\mathrm{real}\:\mathrm{positive}\:\mathrm{root}\:: \\ $$$$\mathrm{X}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{14}+\sqrt[{\mathrm{3}}]{\mathrm{3194}+\mathrm{3}\sqrt{\mathrm{334731}}}+\frac{\mathrm{193}\sqrt[{\mathrm{3}}]{\mathrm{4}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{6388}+\mathrm{6}\sqrt{\mathrm{334731}}}}\right] \\ $$$$\mathrm{X}\:\approx\:\mathrm{14},\mathrm{119789502} \\ $$$$\Rightarrow\:{x}\:>\:\mathrm{196}\:\Rightarrow\:{y}\:<\:\mathrm{0}\::\:\mathrm{impossible}\:! \\ $$$$ \\ $$$$\mathrm{Finally},\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution}\:\left({x},{y}\right)\:=\:\left(\mathrm{4},\mathrm{9}\right) \\ $$