Can you help me with my Precalculus?

Question

the question I’m stuck on is “ solve 2/-6x - 3x/x-4 is greater than or equal to 0”

sorry I had to write it out different, there aren’t many options lol. If you can help me out, that would be great :)

Answer 1

Looks like the parenthesis should go like this:

 

-1/(3x) - 3x/(x-4) = 0    <--- otherwise the x will cancel in the second term,  which is  highly unusual

 

We need to find the zeros anyway, so that we can get the intervals for x that need to be tested.

 

The common denominator is (3x)(x-4).

 

-1(x-4) - 3x(3x) = 0

 

4 - x - 9x^2 = 0

 

9x^2 + x - 4 = 0

 

quadratic formula:

 

 (-1 +or- sqrt( 1^2 - 4(9)(-4))/(2*9)

 

 (-1 +or- sqrt( 1 + 144))/18

 

 (-1 +or- sqrt(145))/18 <--- two zeros, both are irrational

 

 

There are vertical asymtopes at x=0 and x=4

 

The intervals of interest are

 

 x< (-1 - sqrt(145))/18

 

(-1 - sqrt(145))/18 < x < 0

 

0 < x < (-1+sqrt(145))/18

 

(-1 +sqrt(145))/18 < x < 4

 

x>4

 

 

Examining the graph, the intervals in bold face are the solutions