IMO Solving System Of Equation Problem (Heuristic Solution)

Solve the following system of equations in real $a,\,b,\,c,\,d$:

$a+b=9$

$ab+c+d=29$

$ad+bc=39$

$cd=18$

Heuristic solution:

I solved it, but with the hint given by an Indian math friend of mine, he is very smart and intelligent, and he is capable of giving ingenious and original solutions to the really hard Olympiad mathematics problems.

This is an Olympiad math hard problem that he showed me as well. So I must give credit to him for his generous hint and sharing with me of this greatest algebra Olympiad math problem.

If we form a quartic equation as the product of two quadratic equations as follow, we have:

$\small\begin{align*}(x^2+ax+c)(x^2+bx+d)&=x^4+(a+b)x^3+(ab+c+d)x^2+(ad+bc)x+cd\\&=x^4+9x^3+29x^2+39x+18\\&\overset{I}{=}((x+1)(x+2))((x+3)(x+3))=(x^2+3x+2)(x^2+6x+9)\\&\overset{II}{=}((x+1)(x+3))((x+2)(x+3))=(x^2+4x+3)(x^2+5x+6)\\&\overset{III}{=}((x+2)(x+3))((x+1)(x+3))=(x^2+5x+6)(x^2+4x+3)\\&\overset{IV}{=}((x+3)(x+3))((x+1)(x+2))=(x^2+6x+9)(x^2+3x+2)\end{align*}$

Hence, $(a,\,b,\,c,\,d)=(3,\,6,\,2,\,9),\,(4,\,5,\,3,\,6),\,(5,\,4,\,6,\,3),\,(6,\,3,\,9,\,2)$.