Please help with this nonlinear inequality

Question

masterchief · Answer

\[(4x-28)/[(x-5)(x+2)]≥2

masterchief · Answer

I got (-2x^2+12x-18)/[(x-5)(x+1)] but it was a little off

masterchief · Answer

Wow I can't believe this, I accidentally wrote down a 2 instead of a 1. I really appreciate if you tried helping me though, I see my mistake.

tkhunny · Answer

First of all, do some algebra.

$\dfrac{4x-28}{(x-5)(x+2)} \ge 2$ is the same as $\dfrac{2x-14}{(x-5)(x+2)} \ge 1$

Couple of ways to proceed.

#1 - Note that the Rational Function on the left has an horizontal asymptote at y = 0.  This can be an important thing to note.  If we are approaching the asymptote from the top, it will eventually get below y = 1 and never come back.  Also, if we are approaching from the bottom, it will not ever get all the way up to y = 1.

As that may not be enough information, we might need a little more algebra.

$\dfrac{2x-14}{(x-5)(x+2)} \ge 1$

$\dfrac{2x-14}{(x-5)(x+2)} - 1 \ge 0$

$\dfrac{2x-14}{(x-5)(x+2)} - \dfrac{(x-5)(x+2)}{(x-5)(x+2)} \ge 0$

$\dfrac{-(x-1)(x-4)}{(x-5)(x+2)} \ge 0$

$\dfrac{(x-1)(x-4)}{(x-5)(x+2)} \lt 0$

This provides FIVE regions of the Real Number line

$(-\infty,-2)$
$(-2,1)$
$(1,4)$
$(4,5)$
$(5,\infty)$

We just need to examine each on it's own merits, ignoring the magnitude of each:

For $(-\infty,-2)$, we have $\dfrac{(-)(-)}{(-)(-)}$

For $(-2,1)$, we have $\dfrac{(-)(-)}{(-)(+)}$

For $(1,4)$, we have $\dfrac{(+)(-)}{(-)(+)}$

For $(4,5)$, we have $\dfrac{(+)(+)}{(-)(+)}$

For $(5,\infty)$, we have $\dfrac{(+)(+)}{(+)(+)}$

Okay, so which ones are negative?

Note: You could have observed that the result alternates.  That  might be a little faster if you are pressed for time.

tkhunny · Answer

It is a thing of beauty!