First of all, the claim in the video at 0:40 that the graph of y = 1/√x is a hyperbola is _wrong_ because this is actually one branch of the graph of xy² = 1 which is a _cubic_ curve, not a quadratic curve. The graph of y = 1/x or xy = 1 on the other hand _is_ a hyperbola but like all conic sections this is a _quadratic_ curve. The equation to solve in ℝ is x² + 5x + 5 = 1/√x Substituting x = a² and therefore √x = a assuming a > 0 and then multiplying both sides by a we have (1) a⁵ + 5a³ + 5a = 1 This equation is a special type of quintic equation which can be solved by a hyperbolic substitution. If we substitute (2) a = 2·sinh(t) we have (3) 32·sinh⁵(t) + 40·sinh³(t) + 10·sinh(t) = 1 Dividing both sides by 2 this gives (4) 16·sinh⁵(t) + 20·sinh³(t) + 5·sinh(t) = ½ and using the identity (5) sinh(5t) = 16·sinh⁵(t) + 20·sinh³(t) + 5·sinh(t) we can write (4) as (6) sinh(5t) = ½ Since the hyperbolic sine is a periodic function with period 2πi this gives (7) t = ⅕·arsinh(½) + k·⅖πi, k ∈ ℤ Substituting (7) in (2) the roots of the equation (1) are (8) a = 2·sinh(⅕·arsinh(½) + k·⅖πi), k = 0..4 The equation has five roots, four of which are complex. Setting k = 0 gives the only real root, which is (9) a = 2·sinh(⅕·arsinh(½)) To write this root in algebraic form, we may note that 2·sinh(t) = exp(t) − exp(−t) and arsinh(u) = ln(u + √(u² + 1)). So, we have arsinh(½) = ln(½ + ½√5) and since (½√5 + ½)(½√5 − ½) = 1 we also have −arsinh(½) = −ln(½ + ½√5) = ln(½√5 − ½). Therefore, the real root is a = 2·sinh(⅕·arsinh(½)) = exp(⅕·arsinh(½)) − exp(−⅕·arsinh(½)) = exp(⅕·ln(½ + ½√5)) − exp(⅕·ln(½√5 − ½)) = ⁵√(½ + ½√5) − ⁵√(½√5 − ½) = ⁵√(½ + ½√5) + ⁵√(½ − ½√5) and since x = a² this gives x = (⁵√(½ + ½√5) + ⁵√(½ − ½√5))² Noting that (½ + ½√5)² = ³⁄₂ + ¹⁄₂√5, (½ − ½√5)² = ³⁄₂ − ¹⁄₂√5 and (½ + ½√5)(½ − ½√5) = −1 the exact solution can be written as x = −2 + ⁵√(³⁄₂ + ¹⁄₂√5) + ⁵√(³⁄₂ − ¹⁄₂√5) The numerical value of this solution accurate to seven decimal places is 0.0371649.
