x√x = x+√x. [1] x√x - x - √x = 0 √x(x - √x - 1) = 0 One solution is √x = 0, that is trivial. Put √x = u (& x = u^2) and the above expression becomes u(u^2 - u - 1)= 0. [2] The second solution is that for u^2 - u - 1= 0 that is a quadratic equation. When solved this gives u', u" = (1± sqrt5)/2. [3] {a standard result} x = u^2 = [(1± sqrt5)/2]^2 u^2 = u +1 [4]. x = (6± 2.sqrt5)/4 x', x" = (3± sqrt5)/2
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x√x = x+√x. [1]
x√x - x - √x = 0
√x(x - √x - 1) = 0
One solution is √x = 0, that is trivial.
Put √x = u (& x = u^2) and the above expression becomes
u(u^2 - u - 1)= 0. [2]
The second solution is that for
u^2 - u - 1= 0 that is a quadratic equation.
When solved this gives
u', u" = (1± sqrt5)/2. [3] {a standard result}
x = u^2 = [(1± sqrt5)/2]^2
u^2 = u +1 [4].
x = (6± 2.sqrt5)/4
x', x" = (3± sqrt5)/2