F.5 Quadratic equation(2)

(1) Let sqrt ( x + 1) = y
y + 6/y = 5
y^2 - 5y + 6 = 0
(y - 3)(y - 2) = 0, y = 3 or 2.
(a) sqrt ( x + 1) = 3
x + 1 = 9, x = 8.
(b) sqrt (x + 1) = 2
x + 1 = 4, x = 3.
(2)
x ( x^4 - 3x^2 - 54) = 0
x [(x^2 - 9)(x^2 - 6)] = 0
so x = 0, +/- 3 or +/- sqrt 6.
(3)
x^3(4x^4 - 17x^2 + 4) = 0
x^3 [(4x^2 - 1)(x^2 - 4)] = 0
so x = 0, +/- sqrt (1/2) or +/- 2.