Some news stuff


I got three things to say:


  1. "Things that annoy me" post was undrafted a while ago because of flame wars in the comments. People thought that I was targeting specific people, so yeah.
  2. My 400th post was sadly http://cavs4tw.blogspot.com/2019/07/the-los-angeles-pelicans.html. I completely forgot to keep track, and I won't be making a replacement partly due to laziness but also because it doesn't feel right. Plus, the 500th post I will not forget, and I saved it for a special person.
  3. So that writing problem thing. I was wrong but I had a good solution. I took another look at it and I found the correct answer. Here it is(really bad solution though xD hard to explain)



There's the solution

And here's the solution non-rendered:

Since $\frac{x^2+x+3}{2x^2+x-6}$ has to be greater than or equal to $0$, we can consider two cases:

$\frac{x^2 + x + 3}{2x^2 + x - 6} > 0$ and $\frac{x^2 + x + 3}{2x^2 + x - 6} = 0$

If $\frac{x^2 + x + 3}{2x^2 + x - 6} > 0$, we can consider both the numerator and denominator to be negative, meaning we can set an inequality like so:

$$x^2+x+3<0$$

Since we see that $\sqrt{b^2-4ac}=\sqrt{1-12}$ which is imaginary, we know that the numerator will not help us, so we move on to the denominator.

$$2x^2+x-6<0$$

We can use the quadratic formula to help factor it.

$$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$\frac{-(1)\pm \sqrt{1-4(2)(-6)}}{2(2)}$$
$$\frac{-1\pm \sqrt{49}}{4}$$
$$\frac{-1\pm 7}{4}$$

If it's plus, then the result becomes $\frac{6}{4}=\frac{3}{2}$
If it's minus, then the result becomes $\frac{-8}{4}=-2$

Meaning our quadratic for the denominator can be factored out to become:
$$(x+2)(x-\frac32)<0$$

With this type of structure, we have found our treasure piece, so we can now ignore the rest and use a number line. 

If we want the product to be negative, we need to let a factor be negative and the other be positive, so we can let that be the case. 

If $x+2$ is negative, $x<-2$.
If $x-\frac32$ is positive, $x>3/2$

After checking a couple of values we find that these statements hold true, meaning that for values of $\boxed{\boxed{x\in (-\infty, -2) \cup (\frac32, \infty)}}$, $\frac{x^2+x+3}{2x^2+x-6}\geq 0$


Lyric(s) of the Post: Wei pho


Comments

Post a Comment