Solve: x + (√y) = 3 ...... (i) (√x) + y = 5 ...... (ii)


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Question Number 170754 by Tawa11 last updated on 30/May/22

$Solve :$ $x + \sqrt{y} = 3 . . . . . . (i)$ $\sqrt{x} + y = 5 . . . . . . (ii)$
Commented by MJS_new last updated on 30/May/22

$obviously x = 1 \land y = 4$
Commented by aleks041103 last updated on 30/May/22

$A n d y e t n o t o b v i o u s i f i t i s t h e o n l y s o l u t i o n$
Commented by MJS_new last updated on 30/May/22

$x, y \in R \Rightarrow x ⩾ 0 \land y ⩾ 0$ $(i) \Rightarrow y = {(x - 3)}^{2} \land x ⩽ 3$ $(ii) \Rightarrow y = - \sqrt{x} + 5 \land y ⩽ 5$ $\Rightarrow$ $0 ⩽ x ⩽ 3 \land 0 ⩽ y ⩽ 5$ $within these borders both functions are$ $decreasing \Rightarrow only 1 solution$
Commented by aleks041103 last updated on 30/May/22

$T h a t i s n o t n e c e a r i l y t r u e .$ $T h e s i m p l e s t c o n t r a p o i n t I c a n t h i n k o f$ $i s t h e c a s e o f t h e f u n c t i o n s$ $y = \frac{1}{x} a n d y = \frac{5}{2} - x .$ $b o t h a r e d e c r e a s i n g f o r x > 0, b u t$ $b u t t h e r e a r e t w o s o l u t i o n s$ $x = \frac{1}{2} a n d x = 2$
Commented by Tawa11 last updated on 30/May/22

$I appreciate sir$
	Answered by aleks041103 last updated on 30/May/22

	$\sqrt{x} = u, \sqrt{y} = v$ $u^{2} + v = 3 \Rightarrow v = 3 - u^{2}$ $u + v^{2} = 5$ $u + {(3 - u^{2})}^{2} = 5$ $u + 9 - 6 u^{2} + u^{4} = 5$ $\Rightarrow u^{4} - 6 u^{2} + u + 4 = 0$ $f (1) = 1^{4} - 6 1^{4} + 1 + 4 = 0$ $\Rightarrow u - 1 ∣ u^{4} - 6 u^{2} + u + 4$ $\Rightarrow u^{4} - 6 u^{2} + u + 4 = (u - 1) (u^{3} + u^{2} - 5 u - 4)$ $\Rightarrow u = \sqrt{x} = 1 \Rightarrow x = 1 \Rightarrow y = 4$ $n o w$ $\sqrt{x}, \sqrt{y} \in R \Rightarrow x, y ⩾ 0$ $\Rightarrow \sqrt{x} < 5 \Rightarrow x < 25 a n d y < 5$ $\sqrt{y} < 3 \Rightarrow y < 9 a n d x < 3$ $\Rightarrow x \in (0, 3) a n d y \in (0, 5)$ $\Rightarrow u \in (0, \sqrt{3}) a n d v \in (0, \sqrt{5})$ $L e t g (u) = u^{3} + u^{2} - 5 u - 4$ $W e f i n d t h e m a x v a l u e f o r i n t e r v a l$ $(0, \sqrt{3}) :$ $g (0) = - 4$ $g (\sqrt{3}) = 3 \sqrt{3} + 3 - 5 \sqrt{3} - 4 = - 2 \sqrt{3} - 1$ $g^{'} (u) = 3 u^{2} + 2 u - 5 = 3 (u - 1) (u + \frac{5}{3})$ $\Rightarrow g^{'} (u) < 0 \Leftrightarrow u \in (\frac{- 5}{3}, 1)$ $g^{'} (u) > 0 \Leftrightarrow u \in (- \infty, \frac{- 5}{3}) \cup (1, \infty)$ $\Rightarrow g (u) i s d e c r e a s i n g i n (0, 1) a n d$ $i n c r e a s i n g i n (1, \sqrt{3})$ $\Rightarrow a t u = 1 w e h a v e a m i n i m u m$ $\Rightarrow \underset{u \in (0, \sqrt{3})}{m a x} g (u) = m a x (g (0), g (\sqrt{3}))$ $b u t b o t h g (0) a n d g (\sqrt{3}) a r e l e s s t h a n 0$ $\Rightarrow \underset{u \in (0, \sqrt{3})}{m a x} g (u) < 0 \Rightarrow ∄ u \in (0, \sqrt{3}), s . t . g (u) = 0$ $\Rightarrow O n l y s o l u t i o n i s$ $x = 1, y = 4$
	Commented by Tawa11 last updated on 30/May/22

	$I appreciate sir$
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