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Question Number 208199 by efronzo1 last updated on 07/Jun/24

	Answered by Frix last updated on 07/Jun/24

	$y = \sqrt{18 + 3 x - x^{2}} is a semi circle with r = \frac{9}{2}$ $\sqrt{x + 3} + \sqrt{6 - x} has the maximum (\begin{matrix} \frac{3}{2} \\ 3 \sqrt{2} \end{matrix})$ $We have 2 solutions for 0 ⩽ m < \frac{3 \sqrt{2}}{4} and$ $exactly one solution at m = \frac{3 \sqrt{2}}{4}$
	Commented by efronzo1 last updated on 07/Jun/24

	$how \underset{⏟}{≅}$
	Commented by Frix last updated on 07/Jun/24

	$Draw it x \in [- 3, 6]$ $f (x) = 3 ⩽ \sqrt{x + 3} + \sqrt{6 - x} ⩽ 3 \sqrt{2}$ $g (x) = 0 ⩽ \sqrt{(x + 3) (6 - x)} ⩽ \frac{9}{2}$
	Answered by mr W last updated on 07/Jun/24

	$m (\sqrt{3 + x} + \sqrt{6 - x}) = \sqrt{(3 + x) (6 - x)}$ $l e t u = x - \frac{3}{2}$ $m (\sqrt{\frac{9}{2} + u} + \sqrt{\frac{9}{2} - u}) = \sqrt{(\frac{9}{2} + u) (\frac{9}{2} - u)}$ $b o t h L H S a n d R H S a r e e v e n f u n c t i o n s$ $w i t h r e s p e c t t o u .$ $d u e t o s y m m e t r y t h e r e a r e (i s)$ $1) n o n e s o l u t i o n o r$ $2) o n e s o l u t i o n a t u = 0 o r$ $3) t w o s o l u t i o n s a t u = \pm a (a \neq 0) o r$ $4) t h r e e s o l u t i o n s a t u = 0 \land u = \pm a (a \neq 0) o r$ $5) f o u r s o l u t i o n s a t u = \pm a \land u = \pm b (a \neq b \neq 0) o r$ $e t c .$ $s i n c e t h e r e i s o n l y o n e s o l u t i o n,$ $i t m u s t b e a t u = 0, i . e .$ $m (\sqrt{\frac{9}{2} + 0} + \sqrt{\frac{9}{2} - 0}) = \sqrt{(\frac{9}{2} + 0) (\frac{9}{2} - 0)}$ $2 m \sqrt{\frac{9}{2}} = \frac{9}{2}$ $\Rightarrow m = \frac{1}{2} \sqrt{\frac{9}{2}} = \frac{3 \sqrt{2}}{4} ✓$
	Commented by mr W last updated on 07/Jun/24

	$p o s s i b l e c a s e s f o r t h e i n t e r s e c t i o n$ $f r o m t w o e v e n f u n c t i o n s :$ $f (x) = f (- x)$ $g (x) = g (- x)$
	Commented by mr W last updated on 07/Jun/24

	Commented by Frix last updated on 08/Jun/24

	$But in this special case obviously$ $1 . 3 ⩽ \sqrt{x + 3} + \sqrt{6 - x} ⩽ 3 \sqrt{2}$ $2 . 0 ⩽ \sqrt{(x + 3) (6 - x)} ⩽ \frac{9}{2}$ $\Rightarrow 3 . m ⩾ 0$ $4 . We never get more than 2 real solutions$ $These solutions are the solutions of$ $x^{2} - 3 x + (m^{2} + 6) (2 m^{2} - 3) + 2 m^{3} \sqrt{m^{2} + 9} = 0$ $x \in R \Leftrightarrow 0 ⩽ m ⩽ \frac{3 \sqrt{2}}{4}$
	Commented by mr W last updated on 07/Jun/24

	$i w a s t a l k i n g a b o u t g e n e r a l c a s e . i$ ${d i d n}^{'} t m e a n t h a t e v e r y e q u a t i o n m a y$ $h a v e n o s o l u t i o n o r o n e s o l u t i o n s$ $o r t h r e e s o l u t i o n s e t c .$ $i f f (x) = g (x) h a s o n l y o n e s o l u t i o n$ $a n d f (x) a n d g (x) a r e e v e n, t h e n$ $f (0) = g (0) . t h i s i s t h e e a s i e s t w a y$ $t o s o l v e t h i s q u e s t i o n, i t h i n k .$
	Commented by Frix last updated on 08/Jun/24

	$Yes .$ $I thought ‘ ‘ a {solution}^{″} \neq ‘ ‘ exactly 1 {solution}^{″}$
	Commented by mr W last updated on 08/Jun/24

	$y o u a r e b a s i c a l l y r i g h t . b u t i t h i n k$ $m a n y q u e s t i o n e r s h e r e a r e$ $n o n - n a t i v e e n g l i s h s p e a k e r s . t h e y$ $s a y ‘ ‘ a {s o l u t i o n}^{″} a n d m e a n$ $‘ ‘ o n e {s o l u t i o n}^{″} .$
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