solve for : x 1.(√((x−a)(x−b)))+(√((x−b)(x−c)))+(√((x−c)(x−a)))=d [a,b,c,d∈R try for: a=4,b=3,c=2,d=1] 2. (x−a^2 )(√(x−a))+(x−a)(√(x−a^2 ))=a^2 +a+1 3. (x−a^2 )(√(x^2 −a))+(x^2 −a)(√(x−a^2 ))=a^2 +a+1 [a∈R try for: a=(1/2) ]


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Question Number 77885 by behi83417@gmail.com last updated on 11/Jan/20

$solve for : x$ $1 . \sqrt{(x - a) (x - b)} + \sqrt{(x - b) (x - c)} + \sqrt{(x - c) (x - a)} = d$ $[a, b, c, d \in R$ $try for : a = 4, b = 3, c = 2, d = 1]$ $2 . (x - a^{2}) \sqrt{x - a} + (x - a) \sqrt{x - a^{2}} = a^{2} + a + 1$ $3 . (x - a^{2}) \sqrt{x^{2} - a} + (x^{2} - a) \sqrt{x - a^{2}} = a^{2} + a + 1$ $[a \in R$ $try for : a = \frac{1}{2}]$
	Answered by jagoll last updated on 12/Jan/20

	$1 . \sqrt{(x - 4) (x - 3)} + \sqrt{(x - 3) (x - 2)} = 1 - \sqrt{(x - 2) (x - 4)}$ $s q u a r i n g$ $(x - 3) {x - 4 + x - 2} + 2 (x - 3) \sqrt{(x - 4) (x - 2)} = 1 + (x - 4) (x - 2) - 2 \sqrt{(x - 4) (x - 2)}$ $(x - 3) {2 x - 4 + \sqrt{(x - 4) (x - 2)}} = 1 + (x - 4) (x - 2) - 2 \sqrt{(x - 4) (x - 2)}$ $l e t u = x - 4$ $\Leftrightarrow (u - 1) {2 u + 4 + \sqrt{u (u + 2)}} = 1 + u (u + 2) - 2 \sqrt{u (u + 2)}$ $t o b e c o n t i n i u$
	Commented by mr W last updated on 12/Jan/20

	$d e a d - e n d s t r e e t!$ $i t l e a d s t o a n 8 - t h o r d e r e q u a t i o n w h i c h$ $i s n o t s o l v a b l e .$
	Commented by jagoll last updated on 12/Jan/20

	$h a h a h a . . . . t h a n k s s i r$
	Commented by behi83417@gmail.com last updated on 12/Jan/20

	$dear master : mrW! thanks .$ $I hope and wait for any try by you, sir .$
	Commented by mr W last updated on 12/Jan/20

	$i {d o n}^{'} t t h i n k a g e n e r a l s o l u t i o n (f o r m u l a)$ $i s p o s s i b l e .$ $a s s u m e a ⩾ b ⩾ c, w e c a n s a y :$ $i f d < \sqrt{(a - c) \times m i n (a - b, b - c)} \Rightarrow n o s o l u t i o n .$ $i f d ⩾ \sqrt{(a - c) \times m i n (a - b, b - c)}$ $o n e o r t w o s o l u t i o n s e x i s t w h i c h l i e$ $i n t h e r a n g e x ⩽ c o r x ⩾ a . b u t a f o r m u l a$ $f o r t h e r o o t s i s n o t p o s s i b l e .$ $a s f o r t h e e x a m p l e a = 4, b = 3, c = 2,$ $\sqrt{(4 - 2) \times m i n (1, 1)} = \sqrt{2}$ $t h e r e f o r e f o r d = 1 < \sqrt{2} t h e r e i s n o s o l u t i o n .$ $s e e d i a g r a m .$ $M J S s i r h a s g i v e n t h e s a m e s t a t e m e n t .$
	Commented by mr W last updated on 12/Jan/20

	Commented by mr W last updated on 12/Jan/20

	$c u r v e f (x) s t a n d s f o r$ $f (x) = \sqrt{(x - a) (x - b)} + \sqrt{(x - b) (x - c)} + \sqrt{(x - c) (x - a)}$ $a n d$ $d_{m i n} = \sqrt{(a - c) \times m i n (a - b, b - c)}$ $s o l u t i o n f o r f (x) = d e x i s t s w h e n$ $d ⩾ d_{m i n} .$
	Commented by mr W last updated on 12/Jan/20

	$f o l l o w i n g i s a n e x a m p l e w i t h$ $a = 5, b = 3, c = 2 .$ $f o r d = 4 : t w o s o l u t i o n s$ $f o r d = 2.5 : o n e s o l u t i o n .$
	Commented by mr W last updated on 12/Jan/20

	Commented by mr W last updated on 12/Jan/20

	Commented by behi83417@gmail.com last updated on 12/Jan/20

	$thank you very much dear master .$ $it is perfect sir .$ $great solution by great man!$
	Answered by MJS last updated on 12/Jan/20

	$1 .$ $let a ⩽ b ⩽ c; p ⩽ q$ $(x - p) (x - q) ⩾ 0 \Rightarrow x ⩽ p \lor x ⩾ q$ $\Rightarrow$ $(x ⩽ a \lor x ⩾ b) \land (x ⩽ b \lor x ⩾ c) \land (x ⩽ a \lor x ⩾ c)$ $\Rightarrow$ $x ⩽ a \lor x ⩾ c$ $\Rightarrow$ $the minimum of$ $\sqrt{(x - a) (x - b)} + \sqrt{(x - b) (x - c)} + \sqrt{(x - c) (x - a)}$ $is either at x = a or x = c and thus its value$ $is either \sqrt{(a - b) (a - c)} or \sqrt{(c - a) (c - b)}$ $is \sqrt{(a - b) (a - c)} if a + c ⩾ 2 b$ $or \sqrt{(c - a) (c - b)} if a + c ⩽ 2 b$ $with a = 4, b = 3, c = 2 the minimum is \sqrt{2}$ $\Rightarrow no solution for d = 1$
	Commented by behi83417@gmail.com last updated on 12/Jan/20

	$dear proph : MJS! thank you very much$ $sir . is there any chance to find : x in$ $terms of : a, b, c, d ?$
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