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? $$\ce{H2(g) + I2(g) <=> 2HI(g)}\qquad K_c=50.2\text{ at }445\,\mathrm{^\circ C}$$
First, I found the concentrations of H2 and I2: $$ \begin{aligned} \phantom{}[\ce{H2}]=[\ce{I2}]&=\frac{0.150\text{ mol}}{3.25\text{ L}} \\ &=0.0461538\text{ M} \end{aligned} $$
Setting up the ICE table:
$$ \begin{array}{|l|c c c c c|} \hline &\ce{H2}&+&\ce{I2}&\rightleftharpoons&\ce{2HI} \\ \hline \text{Initial}&0.0461538&&0.0461538&&0 \\ \text{Change}&-x&&-x&&+2x \\ \text{Equilibrium}&0.0461538-x&&0.0461538-x&&2x \\ \hline \end{array} $$ 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 $x$: $$ \begin{aligned} \\ 50.2&=\left(\frac{2x}{0.0461538-x}\right)^2 \\ \sqrt{50.2}&=\pm\frac{2x}{0.0461538-x} \end{aligned} \\ \begin{aligned} 7.0852&=\frac{2x}{0.0461538-x}&&\text{or}&7.0852&=-\frac{2x}{0.0461538-x} \\ 0.327009-7.0852x&=2x&&&0.327009-7.0852x&=-2x \\ 0.327009&=9.0852x&&&0.327009&=5.0852x \\ x&=0.0359936&&&x&=0.064306 \end{aligned} $$ As you can see, there are two positive (and therefore, valid) solutions for $x$. However, only one of the solutions results in a positive $[\ce{H2}]_\mathrm{eq}$: $$ \begin{aligned} \phantom{}[\ce{H2}]_\mathrm{eq}=0.0461538-x \end{aligned} \\ \begin{aligned} \phantom{}[\ce{H2}]_\mathrm{eq}&=0.0461538-0.0359936&&\text{or}&[\ce{H2}]_\mathrm{eq}&=0.0461538-0.064306 \\ &=0.0101602\text{ M}&&&&=-0.0181522\text{ M} \end{aligned} $$ 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.