Math Olympiad | A Nice Radical Problem | VIJAY Maths

[@Quest3669]

X>0 for real solns

[@raghvendrasingh1289]

Here x should be greater or equal to zero
Let y = √( 1+x) hence y > 1
(1+y)^3 = (y^2-1)^2
Now 1+y is not equal to zero hence we can cancel it both sides
1+ y = ( y-1)^2
y^2-3 y = 0
y = 3 because y > 1
x+1 = 9 or x = 8

[@key_board_x]

√[1 + √(1 + x)] = ³√x
{√[1 + √(1 + x)]}² = (³√x)² → recall: ³√x = x^(1/3) → recall: (³√x)² = x^(2/3)
1 + √(1 + x) = x^(2/3)
√(1 + x) = x^(2/3) - 1
[√(1 + x)]² = [x^(2/3) - 1]²
1 + x = [x^(2/3)]² - 2x^(2/3) + [1]²
x = [x^(2/3)]² - 2x^(2/3)
[x^(2/3)]² - 2x^(2/3) - x = 0
x^(4/3) - 2x^(2/3) - x = 0
x^[(4/3) + 1 - 1] - 2x^[(2/3) + 1 - 1] - x^(1) = 0
x^[1 + (4/3) - 1] - 2x^[1 + (2/3) - 1] - x^(1) = 0
x^[1 + (1/3)] - 2x^[1 - (1/3)] - x^(1) = 0
[x^(1) * x^(1/3)] - 2.[x^(1) * x^(- 1/3)] - x^(1) = 0
x^(1) * [x^(1/3) - 2x^(- 1/3) - 1] = 0
x * [x^(1/3) - 2x^(- 1/3) - 1] = 0
x * [x^(1/3) - 2.{1/x^(1/3)} - 1] = 0
First case: x = 0
Rejected because this value does not satisfy the initial equation
√[1 + √(1 + x)] = ³√x → if x = 0
√[1 + √(1)] = ³√0
√2 = 0 ← false
Second case: [x^(1/3) - 2.{1/x^(1/3)} - 1] = 0
x^(1/3) - 2.{1/x^(1/3)} - 1 = 0 → let: a = x^(1/3) → where: a > 0
a - (2/a) - 1 = 0
(a² - 2 - a)/a = 0 → ecall: a ≠ 0 because exponential
a² - a - 2 = 0
Δ = (- 1)² - (4 * - 2) = 1 + 8 = 9
a = (1 ± 3)/2 → recall: a > 0
a = (1 + 3)/2
a = 2 → recall: a = x^(1/3)
x^(1/3) = 2
[x^(1/3)]³ = 2³
x = 8

[@vijaymaths5483]

Nice solution