Solving algebraic equations with radicals: $x(x-\sqrt3)(x+1)+3-\sqrt3=0$ I have several problems requiring assistance.
Solve for $x$:
$x\left( x-\sqrt { 3 }  \right) \left( x+1 \right) +3-\sqrt { 3 } \quad =\quad 0$
I've followed the suggestion to get x^2 - (√3 -1)x + (3-2√3) and shall proceed to the quadratic formula. Thanks.
Number 2:
The equation $2k{ x }^{ 2 }+(8k+1)x+8k\quad =\quad 0$ has 2 distinct real roots for $x$. Find the range of values for $k$.
I've used the discriminant and all that to find $k>\frac { -1 }{ 16 } $. 
Am I missing anything? 
 A: The originally posted equation appeared to be $(x - \sqrt{3}) x (x+1) + 3 - \sqrt{3} = 0$ for which this polynomial can be seen in the form
\begin{align}
& x(x+1)(x - \sqrt{3}) + 3 - \sqrt{3} = 0 \\
& x^{2} (x + 1 - \sqrt{3}) - \sqrt{3} (x + 1 - \sqrt{3}) = 0 \\
& (x - \sqrt[4]{3})(x + \sqrt[4]{3})(x + 1 - \sqrt{3}) = 0
\end{align}
of which the solutions are $x \in \{ \sqrt[4]{3}, - \sqrt[4]{3}, \sqrt{3}-1 \}$. 
As it appears now the equation asked to be solved is $(x+1)(x-\sqrt{3}) + 3 - \sqrt{3} = 0$. Expanded and solved is as follows:
\begin{align}
& x^{2} + (1-\sqrt{3})x + 3 - 2 \sqrt{3} = 0 \\
& x= - \frac{1 - \sqrt{3}}{2} \pm \frac{1}{2} \sqrt{(1-\sqrt{3})^{2} - 4 (3 - 2 \sqrt{3})} \\
& x = \frac{\sqrt{3} -1}{2} \pm \frac{\sqrt{6 \sqrt{3} - 8}}{2}.
\end{align}
As to the additional question:
The equation given is $2 k \, x^{2} + (1+8k) \, x + 8k = 0$ for which
$$x^{2} + \left(4 + \frac{1}{2k}\right) \, x + 4 = 0$$
and 
$$x = - 2 - \frac{1}{4k} \pm \frac{\sqrt{1 + 16 k}}{4 k}$$. 
It is fairly evident that the case $k =0$ leads to an unbounded solution. If $k \to \infty$ then the equation becomes $(x+2)^{2}=0$ and has a double solution of $x = -2$. If $\sqrt{1 + 16 k} \geq 0$ is required then $k > - \frac{1}{16}$. The conclusion is $k \geq - \frac{1}{16}, k \neq 0$. 
