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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 \dots . 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 ?