Question-18798

Question Number 18798 by chernoaguero@gmail.com last updated on 29/Jul/17

Answered by behi.8.3.4.1.7@gmail.com last updated on 30/Jul/17

\frac{x + a}{x + b} = t, \frac{x - a}{x - b} = s \Rightarrow \frac{x^{2} - a^{2}}{x^{2} - b^{2}} = t s, \frac{a^{2} + b^{2}}{a b} = m

\Rightarrow t^{2} + s^{2} - m t s = 0 \Rightarrow t = \frac{m s \pm \sqrt{m^{2} s^{2} - 4 s^{2}}}{2} =

\Rightarrow t = \frac{s}{2} (m \pm \sqrt{m^{2} - 4})

m^{2} - 4 = \frac{{(a^{2} + b^{2})}^{2}}{a^{2} b^{2}} - 4 = \frac{{(a^{2} + b^{2})}^{2} - 4 a^{2} b^{2}}{a^{2} b^{2}} =

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

\Rightarrow m \pm \sqrt{m^{2} - 4} = \frac{a^{2} + b^{2}}{a b} \pm \frac{a^{2} - b^{2}}{a b} = 2 \frac{a}{b}, 2 \frac{b}{a}

\Rightarrow \frac{t}{s} = \frac{a}{b}, \frac{b}{a}

1) \frac{t}{s} = \frac{a}{b} \Rightarrow \frac{\frac{x + a}{x + b}}{\frac{x - a}{x - b}} = \frac{a}{b} \Rightarrow

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

(b - a) x^{2} + (a + b) (a - b) x - a b (a - b) = 0

i f : b \neq a \Rightarrow x^{2} - (a + b) x + a b = 0 \Rightarrow

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

2) \frac{t}{s} = \frac{b}{a} \Rightarrow \frac{\frac{x + a}{x + b}}{\frac{x - a}{x - b}} = \frac{b}{a} \Rightarrow

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

(a - b) x^{2} + (a - b) (a + b) x - a b (a - b) = 0

\Rightarrow i f : a \neq b \Rightarrow x^{2} + (a + b) x - a b = 0

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

i f : a = b \Rightarrow t = s \Rightarrow \frac{x + a}{x + a} = \frac{x - a}{x - a} . t r u e : \forall x .

Commented by chernoaguero@gmail.com last updated on 30/Jul/17

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