Solve-for-real-numbers-x-a-x-b-a-x-x-b-x-a-b-x-a-b-R-and-a-b-

Question Number 155650 by mathdanisur last updated on 03/Oct/21

Solve for real numbers :

\sqrt{\frac{x - a}{x - b}} + \frac{a}{x} = \sqrt{\frac{x - b}{x - a}} + \frac{b}{x}

a; b \in R and a \neq b

Answered by som(math1967) last updated on 03/Oct/21

\sqrt{\frac{x - a}{x - b}} - \sqrt{\frac{x - b}{x - a}} = \frac{b - a}{x}

\frac{x - a - x + b}{\sqrt{x^{2} - (a + b) x + a b}} = \frac{b - a}{x}

\frac{1}{\sqrt{x^{2} - (a + b) x + a b}} = \frac{1}{x} [∵ a \neq b]

x^{2} - (a + b) x + a b = x^{2} [s q u a r i n g b o t h s i d e]

x = \frac{a b}{a + b} a n s

Commented by Rasheed.Sindhi last updated on 03/Oct/21

Y o u a t t a c k a f e w q u e s t i o n s b u t

y o u r a n s w e r s a r e a l w a y s p e r f e c t! :)

Commented by som(math1967) last updated on 03/Oct/21

T h a n k y o u s i r

Answered by ajfour last updated on 03/Oct/21

\frac{x - a}{x - b} = t^{2} \Rightarrow x = \frac{{b t}^{2} - a}{t^{2} - 1}

t - \frac{1}{t} = \frac{(b - a) (t^{2} - 1)}{{b t}^{2} - a}

a s t \neq 1 \Rightarrow {b t}^{2} - a = (b - a) t

x = b + \frac{b - a}{t^{2} - 1} = b + \frac{b (b - a)}{(b - a) t - (b - a)}

x = b + \frac{b}{t - 1}

t = \frac{(b - a)}{2 b} \pm \sqrt{\frac{{(b - a)}^{2}}{4 b^{2}} + \frac{a}{b}}

t = \frac{b - a \pm ∣ a + b ∣}{2 b}

x = b + \frac{2 b^{2}}{- a - b \pm ∣ a + b ∣}

r e j e c t i n g + v e s i g n

\Rightarrow x = b - \frac{b^{2}}{a + b}

x = \frac{a b}{a + b} .

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