# How to solve $\left|\frac{x+4}{ax+2}\right| > \frac1x$

How to solve: $$\left|\frac{x+4}{ax+2}\right| > \frac{1}{x}$$

What I have done:

I) $x < 0$:

Obviously this part of the inequation is $x\in(-\infty, 0), x\neq \frac{-2}{a}$

II) $x > 0$:

$$\left|\frac{x+4}{ax+2}\right| > \frac1x$$ $$\frac{|x+4|}{|ax+2|} > \frac1x$$ because $x > 0$ we can transform it to:

$x^2 + 4x > |ax + 2|$
$(x^2 + 4x)^2 > (ax + 2)^2$
$(x^2 + 4x)^2 - (ax + 2)^2 > 0$
$(x^2 + 4x - ax - 2) (x^2 +4x +ax + 2) > 0$

$(x^2 + (4-a)x - 2)(x^2 + (4+a)x + 2) > 0$

I am kind of stuck here. I also need to discuss the solution for various values of the parameter a. What is the easiest way out of this step? Something with the discriminants, perhaps?

• Are you solving for $x$ or finding values of $a$ that make this inequality true? Commented Nov 4, 2014 at 16:56
• I need to solve it for x, but the solution won't look the same for some different parameters a. Commented Nov 4, 2014 at 16:58
• I guess there'll be subcases a > 0, a < 0 or something like that. Commented Nov 4, 2014 at 17:01

You have already identified and made use of the points where sides of the inequality are undefined. So all that is needed is to solve for equality and then make determinations about the subintervals between the points of equality and points where values are undefined.

Equality happens when $x=\left|\frac{ax+2}{x+4}\right|$. Note that $x$ and $x+4$ are certainly positive in any solution. So either \begin{align} x&=\frac{ax+2}{x+4}&{-x}&=\frac{ax+2}{x+4}\\ \end{align}

Visually, we are intersecting a diagonal line (either of slope $1$ or $-1$) with a simple rational function that has a singularity at $-4$ and a $y$-intercept at $\frac12$. The first of these equations clearly always has precisely one solution for $x$ with $x>0$ and $ax+2>0$ no matter what $a$ is (it's clear if you are considering the graphs). That solution must be the larger of the two solutions from the quadratic equation $x^2+(4-a)x-2=0$, namely $x=\frac{a-4+\sqrt{(a-4)^2+8}}{2}$.

The second of these equations has no positive solutions for $x$ unless $a$ is large enough negative, to make the initial slope of $\frac{ax+2}{x+4}$ steep enough negative to meet $y=-x$. When $a$ is large enough negative, these two curves will meet with multiplicity $2$, and then for larger negative $a$, there will be two points of intersection. The critical $a$ happens when $x^2+(4+a)x+2=0$ has a doubled root and $a$ is large enough negative: when $a=-\sqrt{8}-4$. For $a\leq-\sqrt{8}-4$, the two solutions are $x=\frac{-(a+4)\pm\sqrt{(a+4)^2-8}}{2}$. Using the $+$ sign from the $\pm$, this solution clearly has $ax+2<0$. But with the $-$ sign, for $a$ too negative, $ax+2$ might reach $0$. \begin{align} a\frac{-(a+4)-\sqrt{(a+4)^2-8}}{2}+2&=0\\ -(a^2+4a)-a\sqrt{(a+4)^2-8}+4&=0\\ -a\sqrt{(a+4)^2-8}+8&=(a+2)^2\\ \end{align} A calculus investigation reveals that this equation in fact has no solutions with $a\leq-\sqrt{8}-4$.

So in summary, the list of critical values for $x$ for your relation is $\left\{-{\frac2a},0,\frac{a-4+\sqrt{(a-4)^2+8}}{2}\right\}$ And additionally, if $a\leq-\sqrt{8}-4$, $\left\{\frac{-(a+4)+\sqrt{(a+4)^2-8}}{2},\frac{-(a+4)-\sqrt{(a+4)^2-8}}{2}\right\}$

At these $3,4$, or $5$ $x$-values, the sign of $\left|\frac{x+4}{ax+2}\right| - \frac{1}{x}$ may change. Depending on what $a$ is, the critical values are ordered in different ways. But once you order them, you can test with a single value within each of the subintervals whether the difference is positive or negative.

$D_1= (a-4)^2+8$ and $D_2= (a+4)^2-8$. This shows that you have up to 4 roots. Can you go on from here?
• I think you made a mistake, one discriminant is with -8. And no, I'm completely lost. Getting the roots is easy, but they're complicated, and can be positive and negative depending on a. $(x + (4 + a + sqrt((4+a)^2 -8))/2)(x - (-4 -a + sqrt((4+a)^2 - 8)/2)(x - (a+4-sqrt((4-a)^2 +8)/2) (a+4 + sqrt((4-a)^2 +8)/2) > 0$ is just plain impossible. And now I need consider various cases of a? No way... Commented Nov 4, 2014 at 17:38