Solving for H+ given dissociation of weak acid, c and Ka Can the quadratic equation yield two nontrivial solutions? For example, if a question asks for $[\ce{H+}]$ of a $0.100~\mathrm{M}$ solution of picric acid (monoprotic), given that its $K_\mathrm{a}$ at the system temperature is $4.2 \times 10^{-1},$ following my teacher's instructions I would use the ICE table, which looks like this (for a different example).

$$\begin{array}{cccccccc}
 &amp; \ce{CO (g)} &amp; + &amp; \ce{H2O (g)} &amp; \ce{&lt;=&gt;} &amp; \ce{CO2 (g)} &amp; + &amp; \ce{H2 (g)} \\ 
\text{Initial} &amp; 0.100~\mathrm{M} &amp; &amp; 0.100~\mathrm{M} &amp; &amp; 0 &amp; &amp; 0 \\
\text{Change} &amp; -x~\mathrm{M} &amp; &amp; -x~\mathrm{M} &amp; &amp; x~\mathrm{M} &amp; &amp; x~\mathrm{M} \\
\text{Equilibrium} &amp; (0.100-x)~\mathrm{M} &amp; &amp; (0.100-x)~\mathrm{M} &amp; &amp; x~\mathrm{M} &amp; &amp; x~\mathrm{M} \\
\end{array}$$

I would then solve for $x$ by using the expression for $K_\mathrm{a}$. However, since $0.100/K_\mathrm{a} &lt; 100$ (lowest initial concentration divided by $K_\mathrm{a}$ is less than 100), $x$ is significant and cannot be omitted from the $0.100-x$. Thus, you need to use the quadratic equation, which yields two solutions for my example above: $x=0.0834, 0.503$. The latter is obviously incorrect because it is greater than the initial concentration of the undissociated compound.

However, my question is: Is it possible that both solutions to a given reaction will be nontrivial, that is, is it possible that both answers appear to work at a glance? (No conflicts with the initial concentration, etc.) This question refers not specifically to the example in question, but rather looks at if it is possible at all (is there a reaction and starting concentration for which both answers would seem correct). In addition, if it is possible, how would one determine which answer is nontrivially impossible?

Whenever I get into a new subject I'm not able to "do" it until I work out some small details that stick with me.  It would be best if the answer had some logical explanation, or even a proof as to why the two answers will never be both possible at a glance, or if the answer shows a way to determine without fail which answer is correct. I don't see an obvious reason why both solutions resulting from the quadratic equation will always have one blatantly incorrect result.

Please ask for clarifications if necessary.
What you have got here is a strong acid ($\mathrm{p}K_\mathrm{a} \approx 0.38$) that is not diluted, whereof the latter tells you that you can safely ignore the autoprotolysis of water.

The pH approximation to choose is therefor:
$$[\ce{H+}] = \frac{[\ce{HA}]_0 K_\mathrm{a}}{[\ce{H+}] + K_\mathrm{a}}$$

or in a quadratic form:
$$[\ce{H+}]^2 + [\ce{H+}] K_\mathrm{a} - [\ce{HA}]_0 K_\mathrm{a} = 0$$

which has two solutions:
$$[\ce{H+}] = -\frac{K_\mathrm{a}}{2} \pm \sqrt{\left(\frac{K_\mathrm{a}}{2}\right)^2 + [\ce{HA}]_0 K_\mathrm{a}}$$

From those two solutions, only one is chemically meaningful, which is the positive combination which yields one of your solutions ... $[\ce{H+}] = 0.0834$.