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Question Number 200025 by cortano12 last updated on 12/Nov/23

	Answered by Frix last updated on 12/Nov/23

	$(1) \sqrt{t + 8} + \sqrt{t} = 4 \Rightarrow t = 1 \Rightarrow y = \frac{1}{x}$ $Transforming (2) to$ $x^{3} (x + 13) \sqrt{x + 1} = 6 x^{4} + 14 x^{3} + 8$ $x^{3} (x + 13) \sqrt{x + 1} = 2 (x + 1) (3 x^{3} + 4 x^{2} - 4 x + 4)$ $x = - 1 \land y = - 1 ★$ $x^{3} (x + 13) = 2 (3 x^{3} + 4 x^{2} - 4 x + 4) \sqrt{x + 1}$ $t = \sqrt{x + 1} ⩾ 0$ $t^{8} - 6 t^{7} + 9 t^{6} + 10 t^{5} - 33 t^{4} + 6 t^{3} + 35 t^{2} - 18 t - 12 = 0$ $(t^{2} - 2 t - 1) (t^{6} - 4 t^{5} + 2 t^{4} + 10 t^{3} - 11 t^{2} - 6 t + 12) = 0$ $The 2^{nd} factor ⩾ 3$ $\Rightarrow t = 1 + \sqrt{2} \Rightarrow$ $x = 2 + 2 \sqrt{2} \land y = - \frac{1}{2} + \frac{\sqrt{2}}{2} ★$
	Answered by Rasheed.Sindhi last updated on 12/Nov/23
	$(√(xy+8)) _(a) +(√(xy)) _(b) =4 { (((√(xy+8)) +(√(xy)) =4)),(((√((1/y)+1)) (x^2 +13x−(6/y)(√(x+1)) =8(x+y^2 ))) :} (√(xy+8)) _(a) +(√(xy)) _(b) =4 a+b=4...........(i) a^2 −b^2 =8 a−b=((a^2 −b^2 )/(a+b))=(8/4)=2....(ii) (i) + (ii): 2a=6⇒a=3 ⇒(√(xy+8)) =3⇒xy=1 OR (i)−(ii): 2b=2⇒b=1⇒(√(xy)) =1⇒xy=1 ⇒y=(1/x) (√((1/y)+1)) (x^2 +13x−(6/y)(√(x+1))) =8(x+y^2 ) (√(x+1)) (x^2 +13x−6x(√(x+1)) )=8(x+(1/x^2 )) (√(x+1)) (x^2 +13x−6x(√(x+1)) )=8x+(8/x^2 ) (√(x+1)) (x^4 +13x^3 −6x^3 (√(x+1)) )=8x^3 +8 Let (√(x+1)) =t⇒x=t^2 −1 t ((t^2 −1)^4 +13(t^2 −1)^3 −6(t^2 −1)^3 t )=8(t^2 −1)^3 +8 Complicated approach.... ......$
	$\underset{a}{\underset{⏟}{\sqrt{x y + 8}}} + \underset{b}{\underset{⏟}{\sqrt{x y}}} = 4$ ${\begin{cases} \underset{⏟}{\sqrt{x y + 8}} + \underset{⏟}{\sqrt{x y}} = 4 \\ \sqrt{\frac{1}{y} + 1} (x^{2} + 13 x - \frac{6}{y} \sqrt{x + 1} = 8 (x + y^{2}) \end{cases}$ $\underset{a}{\underset{⏟}{\sqrt{x y + 8}}} + \underset{b}{\underset{⏟}{\sqrt{x y}}} = 4$ $a + b = 4 . . . . . . . . . . . (i)$ $a^{2} - b^{2} = 8$ $a - b = \frac{a^{2} - b^{2}}{a + b} = \frac{8}{4} = 2 . . . . (i i)$ $(i) + (i i) : 2 a = 6 \Rightarrow a = 3$ $\Rightarrow \sqrt{x y + 8} = 3 \Rightarrow x y = 1$ $OR$ $(i) - (i i) : 2 b = 2 \Rightarrow b = 1 \Rightarrow \sqrt{x y} = 1 \Rightarrow x y = 1$ $\Rightarrow y = \frac{1}{x}$ $\sqrt{\frac{1}{y} + 1} (x^{2} + 13 x - \frac{6}{y} \sqrt{x + 1}) = 8 (x + y^{2})$ $\sqrt{x + 1} (x^{2} + 13 x - 6 x \sqrt{x + 1}) = 8 (x + \frac{1}{x^{2}})$ $\sqrt{x + 1} (x^{2} + 13 x - 6 x \sqrt{x + 1}) = 8 x + \frac{8}{x^{2}}$ $\sqrt{x + 1} (x^{4} + 13 x^{3} - 6 x^{3} \sqrt{x + 1}) = 8 x^{3} + 8$ $L e t \sqrt{x + 1} = t \Rightarrow x = t^{2} - 1$ $t ({(t^{2} - 1)}^{4} + 13 {(t^{2} - 1)}^{3} - 6 {(t^{2} - 1)}^{3} t) = 8 {(t^{2} - 1)}^{3} + 8$ $C o m p l i c a t e d a p p r o a c h . . . .$ $. . . . . .$
	Answered by MathematicalUser2357 last updated on 15/Nov/23

	$Yay, 200 K!$
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