+0

# HELP NEEDED!!! ASAP!!!!

0
142
5

This is due in a few hours, so I really need quick help. I was undergoing surgery so I couldn't go to a class called art of problem solving online. There are online classes and I was not able to participate in the class. I know nothing about the unit for this week and need help with solving just one problem. I generally don't like doing this, but I also want to understand how to do the problem.

The question:

(x^2 + x + 3)/(2x^2 + x - 6) ≥ 0.

Please also explain how to solve this so that I can understand this unit. Thanks, I will be so happy if somebody answers this.

Feb 24, 2019

#1
0

$$\text{surgery eh...}\\ x^2 + x + 3 = \left(x+\dfrac 1 2\right)^2 +\dfrac{11}{4} >0\\ 2x^2 + x - 6 = (2x-3 )(x+2)\\ 2x^2 + x - 6 \begin{cases}+ &x<2\\0&x=2\\- &2 < x < \dfrac 3 2\\ 0 &x=\dfrac 3 2\\+&\dfrac 3 2 < x\end{cases}$$

$$+ \div + \Rightarrow +\\ +\div - \Rightarrow -\\ \dfrac{x^2+x+3}{2x^2+x-6}\geq 0 \Rightarrow (-\infty,2]\cup \left[\dfrac 3 2,\infty\right)$$

.
Feb 24, 2019
#2
0

Yeah, it was a snorkeling accident with coral. Thanks so much though!

Guest Feb 24, 2019
#3
0

Wait, I have one quick question. I do not get your solution. I want to understand the problem so I am wondering if you could tell me a tiny bit more about the step where you have the equation 2x^2 + x - 6 and the +, -, and 0's.

Why is there 2 differnet 0's and +'s?

Guest Feb 24, 2019
#4
0

that just denotes where the polynomial is positive, zero, and negative

the numerator is always positive.

pos/pos = pos

pos/neg = neg

Rom  Feb 24, 2019