Can ICE tables give two solutions? What does this mean physically? The following question comes from General Chemistry: Principles and Modern Applications, 11th ed. by Pettruci.


  Exercise 15-33. A mixture consisting of 0.150 mol H2 and 0.150 mol I2 is brought to equilibrium at 445 °C, in a 3.25 L flask. What are the equilibrium amounts of H2, I2, and HI?
  $$\ce{H2(g) + I2(g) &lt;=&gt; 2HI(g)}\qquad K_c=50.2\text{ at }445\,\mathrm{^\circ C}$$


First, I found the concentrations of H2 and I2:
$$
\begin{aligned}
\phantom{}[\ce{H2}]=[\ce{I2}]&amp;=\frac{0.150\text{ mol}}{3.25\text{ L}} \\
&amp;=0.0461538\text{ M}
\end{aligned}
$$

Setting up the ICE table:

$$
\begin{array}{|l|c c c c c|}
\hline
&amp;\ce{H2}&amp;+&amp;\ce{I2}&amp;\rightleftharpoons&amp;\ce{2HI} \\
\hline
\text{Initial}&amp;0.0461538&amp;&amp;0.0461538&amp;&amp;0 \\
\text{Change}&amp;-x&amp;&amp;-x&amp;&amp;+2x \\
\text{Equilibrium}&amp;0.0461538-x&amp;&amp;0.0461538-x&amp;&amp;2x \\
\hline
\end{array}
$$
Then,
$$
K_c=50.2=\frac{(2x)^2}{(0.0461538-x)(0.0461538-x)}.
$$
Most solutions I've seen for similar problems involve taking the square root of both sides of this equation. However, they only take the principal root, effectively ignoring a second valid solution. Here, I'll do the proper solution for finding $x$:
$$
\begin{aligned}
 \\
50.2&amp;=\left(\frac{2x}{0.0461538-x}\right)^2 \\
\sqrt{50.2}&amp;=\pm\frac{2x}{0.0461538-x}
\end{aligned} \\
\begin{aligned}
7.0852&amp;=\frac{2x}{0.0461538-x}&amp;&amp;\text{or}&amp;7.0852&amp;=-\frac{2x}{0.0461538-x} \\
0.327009-7.0852x&amp;=2x&amp;&amp;&amp;0.327009-7.0852x&amp;=-2x \\
0.327009&amp;=9.0852x&amp;&amp;&amp;0.327009&amp;=5.0852x \\
x&amp;=0.0359936&amp;&amp;&amp;x&amp;=0.064306
\end{aligned}
$$
As you can see, there are two positive (and therefore, valid) solutions for $x$. However, only one of the solutions results in a positive $[\ce{H2}]_\mathrm{eq}$:
$$
\begin{aligned}
\phantom{}[\ce{H2}]_\mathrm{eq}=0.0461538-x
\end{aligned}
\\
\begin{aligned}
\phantom{}[\ce{H2}]_\mathrm{eq}&amp;=0.0461538-0.0359936&amp;&amp;\text{or}&amp;[\ce{H2}]_\mathrm{eq}&amp;=0.0461538-0.064306 \\
&amp;=0.0101602\text{ M}&amp;&amp;&amp;&amp;=-0.0181522\text{ M}
\end{aligned}
$$
And $x=0.0359936$ does indeed lead to the correct answer according to the textbook.

My question is: was it just a coincidence that I was able to eliminate one of the possible solutions for $x$? In other words, is there a scenario where I could end up with two possible $[\ce{H2}]_\mathrm{eq}$s? If not, what forbids this?

Edit: To reiterate, I am aware that one of the solutions is to be eliminated for being "non-physical". My question is whether it is possible for both solutions to be physical.
In other words, is there a scenario where I could end up with two possible [solutions]?


No, there is always a single solution. The reaction quotient Q assumes values from zero (no products) to infinity (no reactants). It varies monotonically with the extent of reaction (the x in the ICE table). So there is always a solution (because Q covers the full range of possible equilibrium constant), and it is always unique (because for a given set of starting values, every value of Q is unique - follows from the monotonic variation).

The only exception is a process where none of the concentration vary with x (when all species are pure solids or liquids). In this case, Q takes a single value (except for running out of reactant or product). If that value is equal to K, x can take a range of values. If Q is different from K, there is no solution, and there won't be any equilibrium (i.e. the reaction would go to completion).