2 questions: factoring and inequalities

[Big_Blue824]

Hello
My first question is about inequality factoring.
If you look at the attached picture, you'll notice at step two, which is, "write the nonzero side as one fraction," the denominator of the variable fraction has a changed sign, and the fraction we multiply the -2 by has the same changed sign. Why'd the sign change, whats going on?



My second question is about factoring. So, given the inequality
2x²-x>3, my work is as follows
2x²-x-3>0
Now at this point, I've been told that "In simple cases, find two numbers that multiply together to give AC that when added together, also give B" with regards to the equation Ax²+Bx+C=0
However, when I attempt to use this method here, I end up with the two numbers -3 and 2, when the correctly factored form uses -3 and 1, as
(2x-3)(x+1)>0
Did I perform the method wrong? Is there some special circumstance I'm not aware of which must be true in order to use it? Is it more of a random hit or miss type thing? And in that case, is there another method I should be using instead?
I'm really not a fan of guess and check

Thanks for reading

 

[Big_Blue824]

ALRIGHT, so as to the first question, its just a typo.

Still curious about the second one though

 

[OmarMohamedKhallaf]

For the first question I think the problem is that [MATH]\frac{x-4}{x-5}[/MATH] is a typo and it must be [MATH]\frac{x-4}{x+5}[/MATH].
For the second question: Factoring Quadratics

 

[Dr.Peterson]

Yes, the first is clearly a typo, and you should take the problem as being [MATH]\frac{x-4}{x+5} > 2[/MATH].

As for the AC method of factoring, you have to split the middle term using your two numbers and factor by grouping: 2x² - 3x + 2x - 3. You can't just write the two factors immediately.

 

[Deleted member 4993]

For the first question I think the problem is that [MATH]\frac{x-4}{x-5}[/MATH] is a typo and it must be [MATH]\frac{x-4}{x+5}[/MATH].
For the second question: Factoring Quadratics

If the original problem was:

\(\displaystyle \frac{x-4}{x-5} > 2\)

Then you should continue to have (x-5) in the denominator. The typo happens on the line designated step 2

 

[OmarMohamedKhallaf]

If the original problem was:

\(\displaystyle \frac{x-4}{x-5} > 2\)

Then you should continue to have (x-5) in the denominator. The typo happens on the line designated step 2

I tried both:
[MATH]\frac{x-4}{x-5}-2>0 \implies \frac{-x+6}{x-5}>0[/MATH]and
[MATH]\frac{x-4}{x+5}-2>0 \implies \frac{-x-14}{x+5}[/MATH]It seems that the book meant the second one.

 

[Big_Blue824]

It was a typo

Thanks very much for all the replies!

 

[Steven G]

So what is your answer?

 

[OmarMohamedKhallaf]

Me?
I really wanted to see the rest of the answer that was in the book because I was stuck at this point [MATH]\frac{-x-14}{x+5}[/MATH], so I did some researching and here is what I found.
[MATH]\\ \because\frac{-x-14}{x+5}=\frac{-(x+14)}{x+5}>0 \\ \therefore\frac{x+14}{x+5} < 0 \\ \because\text{The result must be less than 0} \\ \therefore\text{the numerator is positive and the denominator is negative or vice versa} \\ \therefore x+14>0 \text{ and } x+5<0 \iff x>-14 \text{ and } x<-5 \\ \text{ or} \\ x+14<0 \text{ and } x+5>0 \iff x<-14 \text{ and } x>-5 \text{ (rejected)} \\ \therefore x\in\left]-14,-5\right[ [/MATH]