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

 

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

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 Feb 24, 2019
 #2
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Yeah, it was a snorkeling accident with coral. Thanks so much though!

Guest Feb 24, 2019
 #3
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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?

 

You were very helpful, thanks for answering my question.

Guest Feb 24, 2019
 #4
avatar+4449 
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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

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