Help me with precalculus! ;_;

[Alex26]

I have a problem, I can't solve a problem ;_;

Inequalities

x^3 -2x^2 -9x - 2 => -20

and

4x^3 - 12x^2 > 0

and

3x-5/x-5 > 4 (i suck a lot at anything related with fractions)


wtf do i do with all those x's, fractions and exponents >.<
I was happy doing everything until i reached that part -_-


I would really appreciate the help, once i know how to do those I think i'll be able to survive the rest of the homework. So please, help!

[Jay3205]

The ideal way to do it would be to use algebra to get the equation to be in the form x > ## or x < ## where "##" is a number. This will work for the 2nd and 3rd equations you posted. For the first equation, I'd find the roots of the function (after you add 20 to each side) and then test a value from each interval between the roots.

An easier way would be to plug the stuff into a graphing calculator and see where the inequality holds.

[Tai Dai]

http://i304.photobucket.com/albums/nn168/Elihor/stuff/Vaizardscomplete.jpg

hopefully that can help you if you're solving for x
double check cause i kinda rushed

it's also kinda upside down >>

[Jay3205]

^^ I would try working that out again. Overall comment is that since these are inequalities, x doesn't equal a specific value... it equals one (or multiple) intervals.

Problem 1: The jump from the 3rd line to the 4th line isn't algebraically correct. If re-multiplied back out, it'll produce a constant of -36, which wasn't in the original equation.

Problem 2: The inequality will hold only for values greater than 3. Plugging zero in results in "0 > 0".

Problem 3: Looks good.

[Alex26]

woot

thanks a lot both of you

I tried the process myself following the tips. The only one that still gives me problem is the first one, I think because im messing up with the factorization.

I go ok until the factor out part. What I have so far is (x-2) (x^2-9), gives me the original equation after adding 20 to both sides of the equation, but I don't know how to take out the results from there. I had this idea of taking the root to end up with (x-3) but then the left side of the equations gets messed up =\


My other 2 results seems good, second turned out to be x>3 or (3,infinite) and third (5,15), corresponds with Panther's results resolving with 0

/happy

surviving the homework so far :p

[smach]

http://i304.photobucket.com/albums/nn168/Elihor/stuff/Vaizardscomplete.jpg

hopefully that can help you if you're solving for x
double check cause i kinda rushed

it's also kinda upside down >>here's an alternative to doing the first one.

once you add 20 to both sides you end up with this:

x^3 - 2x^2 - 9x + 18 >= 0
factor out x-2
(x - 2) (x^2 - 9) >= 0
set both equations greater than or equal to zero1) x - 2 >= 0
add the constant to both sides
x >= 2
2) x^2 - 9 >= 0
add the constant to both sides
x^2 >= 9
solve for x
x >= 3,-3 (positive/negative 3)plug in your results and test them to see which ones work.

in that last part, we look for the square root of 9, which can be both positive three and negative three



EDIT: you could alternatively incorporate step 2 into the second line.

in a nutshell, this is how it'd go:
(x-2)(x-3)(x+3) >= 0
x >= 2,3,-3

test each value on the original equation to find the right one(s).

[ghey b]

this is the start of my mind child :3

[Rorschach]

Well... question was answered =P

/closed