Wait a minute... The LaTeX bragger.. is....

Hi

This was a problem that I basically didn't know how to solve, but I kinda like bashed?? I didn't even find the correct answer because it was greater than -1, not 0, so there's probably that one decimal number I'm missing >:(

I did spend a heck lot of time both writing this solution and "solving" this problem.

PS(The image was too big for me to drag so I had to download it, meaning that even if I put it on x-large the image will be very small. Here's the link to the PDF)
PS#2(shoot ok lemme try this again i forgot its pps)
PPS(vote on the poll plz)



Anyway,

As a first observation, we begin by noticing that the numerator of the fraction $\frac{x^2+x+3}{2x^2+x-6}$ looks similar to the denominator when multiplied by two.

Multiplying both sides of the inequality by two will achieve the similarity, but some numbers are still off, and we can separate them to create two separate fractions.

$$\frac{x^2+x+3}{2x^2+x-6}\geq 0$$
$$\frac{2x^2+2x+6}{2x^2+x-6}\geq 0$$
$$\frac{2x^2+2x+6+(-12+12)+(-x+x)}{2x^2+x-6}\geq 0$$
$$\frac{2x^2+x-6+12+x}{2x^2+x-6}\geq 0$$
$$\frac{2x^2+x-6}{2x^2+x-6}+\frac{12+x}{2x^2+x-6}\geq 0$$
$$1+\frac{12+x}{2x^2+x-6}\geq 0$$

We can factor $2x^2+x-6$ into $(x+2)(x-3/4)$

$$\frac{12+x}{(x+2)(x-\frac34)}\geq -1$$


We can use logic to solve the rest.

Letting $x$ become $-12$ would lead to the numerator becoming $0$, which always remains when divided by any number. $0$ is greater than $-1$ so $-12$ is a value for $x$ that holds truth.

Let $A=12+x$, $B=x+2$, and $C=x-\frac34$. Substituting turns our complex inequality into $\frac{A}{B\cdot C}\geq -1$

If we let $x$ be a little bit greater than $-12$,

$A$=positive
$B$=negative
$C$=negative

So $\frac{A}{B\cdot C}=\text{positive}$

So it holds true.


We notice that when we enlarge $-12$ how many times we want, $A$ remains positive, $B$ becomes unstable, and $C$ will always hold the value of being a negative number.

Since $B$ is unstable when we infinitely make -12 greater, we need to find the time it changes.

If $x$ is $-3$, then $B$=negative, but when $x$ is -2, $B$=zero, and when $x$ is -1, $B$=positive. From here on $B$ will always be positive because increasing a positive number will have it still be a positive number.

If $B$ becomes $0$, then the fraction would be unsolvable, as we would be left with a denominator with $0$. If $B$ is positive, $\frac{A}{B\cdot C}$ would be negative, a value that is false. Therefore, we do not want $B$ to be positive.

We can make sure $B$ is not positive if $x$ is less than $-2$ or if $x<-2$.

We also knew that $x\leq -12$, so combining our two statements will lead us to our final answer.



For values of $x$ in the interval of $\boxed{\boxed{[-12, -2)}}$, $\frac{x^2+x+3}{2x^2+x-6}$ will always be greater than or equal to $0$.