a-b-b-a-b-a-2-3-b-a-b-a-b-a-3-2-a-b-R-a-b-

Question Number 65457 by behi83417@gmail.com last updated on 30/Jul/19

{\begin{cases} \frac{a}{b} + \frac{b - a}{b + a} = 2 \sqrt{3} \\ \frac{b}{a} + \frac{b + a}{b - a} = 3 \sqrt{2} \end{cases} [a, b \in R, a \neq b]

Answered by mr W last updated on 30/Jul/19

l e t t = \frac{a}{b}

e q n . 1 \Rightarrow t + \frac{1 - t}{1 + t} = 2 \sqrt{3} \dots (i)

e q n . 2 \Rightarrow \frac{1}{t} + \frac{1 + t}{1 - t} = 3 \sqrt{2} \dots (i i)

(i) :

t^{2} - 2 \sqrt{3} t - (2 \sqrt{3} - 1) = 0

t = \sqrt{3} \pm \sqrt{2 (\sqrt{3} + 1)} \leftarrow r e a l

(i i) :

1 - t + t^{2} + t = 3 \sqrt{2} (t - t^{2})

(1 + 3 \sqrt{2}) t^{2} - 3 \sqrt{2} t + 1 = 0

t = \frac{3 \sqrt{2} \pm \sqrt{2 (7 - 6 \sqrt{2})}}{2 (1 + 3 \sqrt{2})} \leftarrow n o t r e a l

\Rightarrow n o s o l u t i o n!

Commented by behi83417@gmail.com last updated on 30/Jul/19

thank you very much dear master .