2016 AMC 12A #24 Problem 24 on the 2016 AMC 12A asks the following:

There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$?

On AoPS, Solution 1 insists that the second derivative of the polynomial must be 0 at the x-intercept, and Solution 3 insists that the roots must be identical. How do they know either of these things?  
Thanks.
https://www.artofproblemsolving.com/wiki/index.php/2016_AMC_12A_Problems/Problem_24
EDIT: I understand why the roots must be equal, but it's still not clear to me why the second derivative must be 0 at the x-intercept. 
 A: If the roots of the equation are $x_1$, $x_2$ and $x_3$ :
$$
x_1+x_2+x_3=x_1x_2x_3=a
$$
From the sum of roots and product of roots condition. This is sufficient to show that $x_1$, $x_2$, $x_3$ are positive as :
$$
x_1+x_2+x_3 > 0
\\
x_1x_2x_3>0
\\
x_1x_2+x_2x_3+x_1x_3=b>0
$$
As per Descartes Rule of Signs, either all three roots are positive or only one of them is positive. Let's suppose that $x_3$ is the positive root. 
$$
x_1+x_2+x_3 >0
\\
x_1+x_2>-x_3
\\
\lvert x_1 + x_2 \rvert < x_3
$$
We can see that :
$$
\frac{\lvert x_1+x_2 \rvert}{2} \ge \sqrt{x_1x_2}
$$
As $x_1$ and $x_2$ are both negative. 
Squaring :
$$
\frac{(x_1+x_2)^2}{4} \ge x_1x_2
$$
Now, 
$$
x_1x_2+x_2x_3+x_1x_3=x_1x_2+x_3(x_1+x_2)
$$
If this is positive. Then :
$$
x_1x_2>x_3 \lvert x_1+x_2 \rvert
$$
But
$$
x_1x_2 \le \frac{(x_1+x_2)^2}{4}
$$
Hence :
$$
x_3 \lvert x_1 + x_2 \rvert < \frac{(x_1+x_2)^2}{4}
\\
x_3 < \frac{\lvert x_1 + x_2 \rvert}{ 4}
$$
which is impossible as
$$
\lvert x_1 + x_2 \rvert < x_3
$$
EDIT : It appears that there is a much easier way to prove that all three roots are positive. The credit for this goes to @dxiv. Since the quadratic equation is $x^3-ax^2+bx-a$, if $x<0$, then the expression always evaluates to a negative value since $a,b$ are positive. In this case, there won't be any roots to the equation which proves that if there are roots to the equation, then they have to be positive. 
This has been proved earlier. Hence. this equation has three positive real roots. 
Applying the AM-GM inequality :
$$
\frac{x_1+x_2+x_3}{3} \ge \sqrt[3]{x_1x_2x_3}
\\
\frac{a}{3} \ge \sqrt[3]a
\\
a \ge 3\sqrt{3}
\\
\frac{x_1+x_2+x_3}{3} = \sqrt{3}= \sqrt[3]{x_1x_2x_3}
\implies AM=GM
$$
Hence $x_1$, $x_2$ and $x_3$ are equal as the geometric mean of three numbers is equal to their arithmetic mean only if all three are equal. 
A: the minimum value of $b$ is 9.
let the three roots be $0<x_1\le x_2 \le x_3.$ from the vietas relation you get $$x_1+x_2+x_3 = a \\x_1x_2+x_2x_3+x_3x_1 = b \\x_1x_2x_3 = a $$ using the the am-gm inequality together with (1) and (3)  gives you $$ \frac a 3 \ge a^{1/3} \to a \ge \sqrt{27} \tag 1$$
again use the am-gm $$\frac13\left(x_1x_2 + x_2x_3+x_3x_1\right) \ge (x_1x_2x_3)^{2/3}\to b\ge a^{2/3} \tag 2$$
now we need to use the fact that the existence of three real roots for $$x^3-ax^2 + bx-a = 0 $$ implies two roots for the derivative  $$3x^2 - 2ax+b = 0 $$  this in turn makes the discriminant positive requiring $$ a \ge \sqrt{3b}\tag 3$$
now, (1), (2) and (3) gives you $$ b \ge 9.$$
