inequality

Hi Pro35hp and MaxWong,

I think that you will both find this interesting.   It ties in nicely with Alan's graph as well.     

If you do not understand then please ask questions :)

$\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0\\ \frac{x^2 + x + 3}{2x^2 + 4x-3x - 6} \ge 0\\ \frac{x^2 + x + 3}{2x(x+2)-3(x+2)} \ge 0\\ \frac{x^2 + x + 3}{(2x-3)(x+2)} \ge 0\\~\\ \mbox{Now I will consider the numerator and put it equal y}\quad y=x^2+x+3\\ \mbox{The discriminant is }1^2-12=-11 \qquad \mbox{ This means there are no real roots}\\ \mbox{Since the leading coefficiant is positive that means that }\quad y=x^2+x+3 \quad \mbox{is a concave up parabola}\\ \mbox{This means that for all real values of x the numerator will be positive}\ so\\ \mbox{If the denominator is > or = zero the statement will be true.}\\ True\;\;when\;\;\\ (2x-3)(x+2)\ge0\\ \mbox{If I graph }y=(2x-3)(x+2) \mbox{then this will be true when y is on or above the x axis, that is when } y \ge0\\~\\ \mbox{This is a concave up parabola so it will be true at the two ends}\\ \mbox{roots are } x=3/2\;\;\;and\;\;\ x=-2\\ \mbox{So the original statement will be true when }x\le-2\quad and \;\;when\;\; x\ge 1.5$

aops XD

$\frac{x^2+x+3}{2x^2+x-6}\geqslant 0$

$x^2+x+3\geqslant 0$

$x \geqslant {-1 \pm \sqrt{1^2-4(1)(3)} \over 2(1)}$

And the right hand side is imaginary.

yeah , my problem is that I can't express words in english .

so in writings problems I use their sentences lol

my level in english is not good enough

This graph might help:

Melody's answer should be:    x < -2    and  x > 3/2

See the graph, here :  https://www.desmos.com/calculator/afq8ukmpam

Yes you are correct pro35.  You are also extremely rude.

You will not win any friends here by being so incredibly rude!

The logic in my answer wis completely correct.

My oversight was that I was dealing with a denominator which can never be 0 therefore the equal part of my inequality answers are invalid.   

This is a minor oversight.