imaginary algebraic inequality equation

Question

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:

$$4x - 2x(3x + 5) > 19$$

The intended version:

$$4x - 2(3x + 5) > 19$$

Working the typo:

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

Questions:

1. Is it even valid to evaluate imaginary terms in an inequaltiy equation?

2. If so, should the inequality operator have been brought down to the quadratic evaluation?

3. If question 2 is positive, then should I end up with one answer by selecting only the smaller result?

Thanks!

Answer 1

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 $i^2=-1<0$. 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

Answer 2

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< 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.

Answer 3

Your parabola doesn't cross the $x$-axis as Klaramun says, so it doesn't have any roots. but a better explanation is that the parabola has a vertex of $(-.5,-17.5)$, a $y$-intercept = $-19$, and the parabola opens downward meaning that it never crosses the $x$-axis

There is no real $x$-value that satisfies the inequality, so there is no real solution