If we solve it on $a$ we get: \[a= {-3x-1\pm (x+1)(2x+1)\over 2}\] 1. case: $a= x^2$ so $a>0$. 2. case $x^2+3x+1+a=0$ so $a< {5\over 4}$ Thus for $a\in (0,{5\over 4})$.

That formula for $a$ is completely wrong. And there exist such $a<0$, for example $a=-0.01$.

I get $-0.0891514\le a\le 0.503879$ but only after solving (numerically) a sixth order polynomial. This is the result of solving the following system: $3a=4x^3+9x^2+2x-1$ and $3a^2=-9x^4-22x^3-12x^2-2x+1.$

Actually, there must be a typo - made either by the problem creators (and hopefully corrected on the spot during the competition!) or by crazyfehmy. Namely, the two red expressions have to be interchanged: crazyfehmy wrote: Find all real values of $a$ for which the equation $x(x+1)^3=(2x+{\color{red}a})(x+{\color{red}a+1})$ has four distinct real roots. i.e. the equation must be $x(x+1)^3=(2x+a+1)(x+a)$ and then the procedure of solving it as a quadratic in $a$ makes sense.

Ha, ha. Indeed, the poster at #2 must have had this inside information, since then his formula is right. The equation then writes $(a-x^2)(a+x^2+3x+1)$, forcing $a \in [0, 5/4]$. Since for $a=0$ the factor $a-x^2$ has a double root, while for $a=5/4$ the factor $a+x^2+3x+1$ has a double root, the poster rightfully eliminated these values from the range of $a$. But there is another case when the four real roots are not distinct, namely when the factors $a-x^2$ and $a+x^2+3x+1$ have themselves a common root; this happens for $0 = 2x^2 + 3x + 1 = (x+1)(2x+1)$, thus for $x\in \{-1, -1/2\}$, leading to $a\in \{1/4,1\}$. The correct answer therefore is $\boxed{a \in (0, 1/4) \cup (1/4,1) \cup (1,5/4)}$.

I knew there was something fishy about this... It made no sense that the solution would require a sixth order messy polynomial. But at least we have the solution to it Since I did all that work for the "wrong" expression, I'll offer a different solution based on the same approach for the "correct" expression $x(x+1)^3=(2x+a+1)(x+a).$ The system now is: $3a=4x^3+9x^2+2x$ and $3a^2=-9x^4-22x^3-12x^2-2x.$ Eliminating $a$ we get $x(x+1)^2(2x+1)^2(2x+3)=0.$ The corresponding pairs of values are: $(x,a)=(0,0),(-1/2, 1/4),(-1,1),(-3/2,5/4).$ Therefore the range of values for $a$ is $[0,5/4]$ excluding the values given by the pairs. That is, as mavropnevma mentioned above, $a\in (0,1/4)\cup (1/4,1)\cup (1,5/4).$ The values $a=0,1/4,1,5/4$ are excluded because the corresponding expression has double roots at the corresponding values of $x,$ namely at $x=0,-1/2,-1,-3/2.$ Indeed: (i) for $a=0,$ the expression becomes $x^2(x^2+3x+1)=0;$ (ii) for $a=1/4,$ the expression becomes $(x-1/2)(x+1/2)^2(x+5/2)=0;$ (iii) for $a=1,$ the expression becomes $(x-1)(x+1)^2(x+2)=0;$ and (iv) for $a=5/4,$ the expression becomes $(x+3/2)^2(x^2-5/4)=0.$