A Nice Math radical problem| what is the value of a | Math Olympiad | a=?

[@AfsarSaid-f8s]

Good question waheeb Sir

[@RyanLewis-Johnson-wq6xs]

(a+Sqrt[a^2-1])/(a-Sqrt[a^2-1])+(a-Sqrt[a^2-1])/(a+Sqrt[a^2-1])=34 a=±3

[@RyanLewis-Johnson-wq6xs]

It’s in my head.

[@yiqiwang]

So easy, I can solve it without a pen.

[@ManojkantSamal]

4a^2-2=34
4a^2=34+2=36
a^2=36/4=9
a=±3

[@key_board_x]

{ [a + √(a² - 1)] / [a - √(a² - 1)] } + { [a - √(a² - 1)] / [a + √(a² - 1)] } = 34 → let: b = √(a² - 1)
[(a + b)/(a - b)] + [(a - b)/(a + b)] = 34
[(a + b)/(a - b)] + [1/{(a + b)/(a - b)}] = 34 → let: x = (a + b)/(a - b)
x + (1x) = 34
(x² + 1)/x = 34
x² + 1 = 34x
x² - 34x + 1 = 0
Δ = (- 34)² - (4 * 1) = 1152 = 2 * 24²
x = (34 ± 24√2)/2
x = 17 ± 12√2
First case: x = 17 + 12√2
(a + b)/(a - b) = 17 + 12√2
a + b = (17 + 12√2).(a - b)
a + b = 17a - 17b + 12a√2 - 12b√2
16a - 18b + 12a√2 - 12b√2 = 0
8a - 9b + 6a√2 - 6b√2 = 0
8a + 6a√2 = 9b + 6b√2
2a.(4 + 3√2) = 3.(3 + 2√2).b → recall: b = √(a² - 1)
2a.(4 + 3√2) = 3.(3 + 2√2).√(a² - 1)
√(a² - 1) = 2a.(4 + 3√2)/[3.(3 + 2√2)] → by squaring both sides
a² - 1 = 4a².(4 + 3√2)²/[3.(3 + 2√2)]²
a² - 1 = 4a².(16 + 24√2 + 18)/[9.(9 + 12√2 + 8)]
a² - 1 = 4a².(34 + 24√2)/[9.(17 + 12√2)]
a² - 1 = 8a².(17 + 12√2)/[9.(17 + 12√2)]
a² - 1 = 8a²/9
a² - (8a²/9) = 1
(9a² - 8a²)/9 = 1
a² = 9
a = ± 3
Second case: x = 17 - 12√2
(a + b)/(a - b) = 17 - 12√2 → we can easily deduce that the result will be the same as above
a = ± 3 → recall: b = √(a² - 1)
b = √8
b = 2√2
First solution: a = 3 → b = 2√2
Second solution: a = - 3 → b = 2√2