solving 12x^3 - 16x^2 + 7x - 1 = 0; solve 3x/(x^2 - 4) < -1

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xo-hikari

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Sep 18, 2008
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Hello, i was given a worksheet with many questions and I need help getting the answers for 2 questions!
(because i dont have an answer sheet, and im not sure if i followed the correct steps)

1) Solve 12x^3 - 16x^2 + 7x -1 = 0
Let f(x) = 12x^3 - 16x^2 + 7x -1

2) Solve the inequality:
3x / (x^2 -4) < -1

*note: where the "<" is, its supposed to be less than/equal to. (i just cant find a key for it :p )

Thanks again! :wink: :D
 
Re: URGENT!! solving polynomials + inequalities

the first one I can't solve because it is not a equation but a statement
i will use oo for infinity

pb 2) 3x/[x^2-4] <=-1

let us first look at
3x/[x^2-4]=0
sketch the function
1) a zero at x=0
2) a vertical assymptote at x=-2 and at x=2
3 horizontal assymptote at x=0

from x=-oo to x=-2 f[x] goes from assymptotic to y=0- to y=-oo at x=-2
from x=-2 to x=2 f[x] goes from y=oo at x=-2 to y=0 at x=0 and then to f[x]=-oo at x=2
from x=2 to x=oo f[x] goes from +oo at x=2 to f[x]=0 at x=oo

but we are interested in 3x/[x^2-4] <=-1 , not =0
sketch a horizontal line at x=-1

the solution to the inequality is where the function lies below the line
3x/[x^2-4]=-1
3x=-[x^2-4]
3x+x^2-4=0
[x+4][x-1]=0
x=-4 , x=1
the curve is below the line:
-4<=x<=-2 and 1<=x<=2 answer
 
Re: URGENT!! solving polynomials + inequalities

thanks so much!

and the first one would be equal to zero!
i edited it now lol

:wink:
 
Re: URGENT!! solving polynomials + inequalities

Hello, xo-hikari!

\(\displaystyle \text{1) Solve: }\:12x^3 - 16x^2 + 7x -1 \;=\; 0\)
Click to expand...

\(\displaystyle \text{Using the Rational Roots Theorem, the only rational roots are: }\:\pm1,\:\pm\frac{1}{2},\:\pm\frac{1}{3},\:\pm\frac{1}{4},\:\pm\frac{1}{6},\:\pm\frac{1}{12}\)

\(\displaystyle \text{We find that }\, x = \frac{1}{2}\text{ is a root . . . hence: }\,(2x-1)\text{ is a factor.}\)

\(\displaystyle \text{Then we factor the cubic: }\:12x^3 - 16x^2 + 7x - 1 \;=\;(2x-1)(6x^2 - 5x + 1) \;=\;(2x-1)(2x-1)(3x-1)\)


\(\displaystyle \text{Therefore, the roots are: }\:x \:=\:\frac{1}{2},\;\frac{1}{3}\)

 
Re: URGENT!! solving polynomials + inequalities

thanks! i just needed to see if i did the questions correctly! and i did! :)
thanks soo much for taking the time to help me!
i really appreciate it!
 
Re: URGENT!! solving polynomials + inequalities

xo-hikari said:
... I need help getting the answers for 2 questions ...
Click to expand...

xo-hikari said:
... i just needed to see if i did the questions correctly ...
Click to expand...

The next time you come here to use this site for an answer-checking service, please BEGIN by posting your answers and please make it clear that you are ONLY looking for confirmation of your results.

Cheers,

~ Mark :)
 
Re: URGENT!! solving polynomials + inequalities

this was done previously

The first problem is not a equation. There is nothing thay f[x] is equal to , so you have a statement.
Some one assumed f[x]=0 and solved giving the roots:
[2x-1][3x-1][2x-1]=0
x=1/2, x=1/3, x=1/2

Two has vertical assymptotes at x= +/- 2
horizontal assymptotes at y=0
zero at x=0
-4<x<-2 and 1<x<2 " the signs should be "equal or less than"
Arthur
 
thanks everyone for helping!
i understand how to solve most polynomial equations now and similar inequality problems :wink: :D
 
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