imaginary algebraic inequality equation This problem was actually given to me as a typo. I decided to work it despite it being a typo and it presented a couple of questions regarding applying imaginary results to an inequality equation.

Typo version:

$$4x - 2x(3x + 5) &gt; 19$$

The intended version:

$$4x - 2(3x + 5) &gt; 19$$

Working the typo:

\begin{align*}
4x - 2x(3x + 5) &amp;&gt; 19 \\[1em]
4x - 6x^2 - 10x &amp;&gt; 19 \\[1em]
-6x^2 - 10x + 4x &amp;&gt; 19 \\[1em]
-6x^2 - 6x &amp;&gt; 19 \\[1em]
-6x^2 - 6x - 19 &amp;&gt; 0 \\[1em]
\end{align*}
Setup for quadratic:
\begin{align*}
x &amp;= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[1em]
a &amp;= -6 \\[1em]
b &amp;= -6 \\[1em]
c &amp;= -19 \\[1em]
x &amp;= \frac{--6 \pm \sqrt{(-6)^2 - 4*-6*-19}}{2*-6} \\[1em]
  &amp;= \frac{--6 \pm \sqrt{(-6)^2 - 4*-6*-19}}{-12} \\[1em]
  &amp;= \frac{--6 \pm \sqrt{(-6)^2 - 4*114}}{-12} \\[1em]
  &amp;= \frac{--6 \pm \sqrt{(-6)^2 - 456}}{-12} \\[1em]
  &amp;= \frac{--6 \pm \sqrt{36 - 456}}{-12} \\[1em]
  &amp;= \frac{--6 \pm \sqrt{-420}}{-12}\hspace{1em}:\hspace{1em}NRS! \\[1em]
  &amp;= \frac{--6 \pm \sqrt{420}*\sqrt{-1}}{-12} \\[1em]
  &amp;= \frac{--6 \pm \sqrt{420}i}{-12} \\[1em]
  &amp;= \frac{--6 \pm 2*\sqrt{105}i}{-12} \\[1em]
  &amp;= \frac{6 \pm 2*\sqrt{105}i}{-12} \\[1em]
  &amp;= -\frac{1}{12}*\left(6 \pm 2*\sqrt{105}i \right) \\[1em]
  &amp;= -\frac{1}{12}*2*\left(3 \pm \sqrt{105}i \right) \\[1em]
  &amp;= -\frac{2}{12}*\left(3 \pm \sqrt{105}i \right) \\[1em]
x &amp;= \left[-\frac{1}{6}*\left(3 \pm \sqrt{105}i \right) \right] \\[1em]
+:\hspace{1em} x &amp;= -\frac{1}{6}*\left(3 + \sqrt{105}i \right) \\[1em]
+:\hspace{1em} x &amp;= -\frac{3}{6} - \frac{\sqrt{105}i}{6} \\[1em]
+:\hspace{1em} x &amp;= \left[-\frac{1}{2} - \frac{\sqrt{105}i}{6} \right] \approx \left[-0.5 - 1.708i \right] \\[1em]
-:\hspace{1em} x &amp;= -\frac{1}{6}*\left(3 - \sqrt{105}i \right) \\[1em]
-:\hspace{1em} x &amp;= -\frac{3}{6} --\frac{\sqrt{105}i}{6} \\[1em]
-:\hspace{1em} x &amp;= -\frac{3}{6} + \frac{\sqrt{105}i}{6} \\[1em]
-:\hspace{1em} x &amp;= \left[-\frac{1}{2} + \frac{\sqrt{105}i}{6} \right] \approx \left[-0.5 + 1.708i \right] \\[1em]
\end{align*}

Questions:


Is it even valid to evaluate imaginary terms in an inequaltiy equation?
If so, should the inequality operator have been brought down to the quadratic evaluation?
If question 2 is positive, then should I end up with one answer by selecting only the smaller result?


Thanks!
Lets try plotting the following numbers on a number line: $-3, 0,e,\pi,-0.5+1.708i$.


It's clear that $-3\lt0\lt e\lt \pi$, but what to do with $-0.5+1.708i$ isn't so obvious.

Of course the reason for my confusion is that Complex numbers are not ordered. For, let's assume the opposite, that a complex number (e.g. $i$) is either zero, positive, or negative.
We know that $i\ne0$ because $0^2=0$ and $i^2=-1$. We know that $i$ is not negative because a negative time a negative is a positive and $i^2=-1&lt;0$. Additionally we know that $i$ is not positive because a positive times a positive is a positive and $i^2=-1\lt 0$, thus contradicting our assumption that $i$ is either zero, positive, or negative. I know that this isn't a formal proof, but it does indicate that there are difficulties in ordering complex numbers.

This doesn't mean that Complex numbers don't have magnitude though. It should be obvious that $2i$ is twice as massive as $i$ and that $-1.0+3.416i$ is twice as massive as $-0.5+1.708i$. For more on that, see http://en.wikipedia.org/wiki/Complex_number#Absolute_value_and_argument
You have found that the polynomial $y=-6x^2 - 6x-19=0$ has no real roots, hence it cannot pass through the $y=0$ axis, so for all values of $x$ it always has the same sign, which indeed is (taking for example $x=0$)

\begin{equation*}
-6x^2-6x-19 = -(6x^2 + 6x + 19)  = -19&lt; 0
\end{equation*}

hence, for all values of $x$, it is always a negative quantity (as said, it does not cross the $y=0$ line), so the inequality does not hold for any $x$ real.