Does anyone know about Ramanujan's method of solving the quartic? I have read (probably) in Kanigel's book The Man Who Knew Infinity that S. Ramanujan devised his own method of solving the Quartic Equation after he learnt to solve the Cubic Equation. Does anyone know what exactly was Ramanujan's method of solving the Quartic Equation? 

Update
It is now asked here at HSMSE.
 A: You can find it in Ramanujan's Notebooks IV by B. Berndt, Chap. 22 Elementary Results, Entry 20, p.31.
Ramanujan starts with this problem. Let $a,b,c,d$ be arbitrary. Solve the system,
$$x^2+ay = b\tag{20a}$$
$$y^2+cx = d\tag{20b}$$
Eliminating $y$, we find it is equivalent to,
$$a^2(d-cx) = (b-x^2)^2\tag{20.1}$$
Assume without loss of generality that $a=2$. Expanding this out,
$$x^4-2bx^2+4cx+(b^2-4d)=0\tag{20.1a}$$
By a simple linear substitution, the general quartic equation can be expressed in the depressed form,
$$x^4+px^2+qx+r = 0\tag{20.1b}$$
Equating coefficients of ${20.1a}$ and ${20.1b}$, you have a system of 3 equations in 3 unknowns {$b,c,d$}. Hence every quartic can be expressed in the form ${20.1}$. The problem then is to find $x$. Ramanujan defines,
$$x = \alpha+\beta+\gamma\tag{20.1c}$$
$$y = -(\alpha\beta+\beta\gamma+\gamma\alpha)$$
$$-c/2 = \alpha\beta\gamma$$
(If you are familiar with cubic equations, you'll already see where Ramanujan is going.) 
Substitute $x,y,c$ into $(20a)$ and $(20b)$ keeping in mind $a=2$, then,
$$x^2+ay = \alpha^2+\beta^2+\gamma^2= b$$
$$y^2+cx = (\alpha\beta)^2+(\beta\gamma)^2+(\gamma\alpha)^2 = d$$
$$(-c/2)^2 = (\alpha\beta\gamma)^2$$
By elementary symmetric polynomials, we then conclude that $\alpha^2,\,\beta^2,\,\gamma^2$ are roots of the cubic equation,
$$t^3-bt^2+dt-c^2/4=0\tag{20.1d}$$
Of course, by solving $(20.1d)$, one can then find $\alpha,\,\beta,\,\gamma$. Using $(20.1c)$, we then further conclude that the four roots of the quartic are,
$$x = \alpha+\beta+\gamma, \quad\alpha-\beta-\gamma, \quad-\alpha-\beta+\gamma, \quad-\alpha+\beta-\gamma$$
P.S. This is similar to Euler's method where he solves a quartic as $x_i = \sqrt{y_1}\pm \sqrt{y_2}\pm \sqrt{y_3}$ and the $y_i$ are roots of a cubic. There's actually a generalization to this for solvable 8th deg eqns as $x_i = \sqrt{y_1}\pm \sqrt{y_2}\pm\dots\pm \sqrt{y_7}$ and the $y_i$ are roots of a 7th deg eqn. See this mathoverflow post.
