anonymous
  • anonymous
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Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[2x^3+ 5x^2 -1 \le 0\]
anonymous
  • anonymous
solve the inequality
anonymous
  • anonymous
x<=-1-√2

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anonymous
  • anonymous
how?
xapproachesinfinity
  • xapproachesinfinity
what course are you taking! now of some factoring thoerems
xapproachesinfinity
  • xapproachesinfinity
i meant do you know of some factoring theorems or not
xapproachesinfinity
  • xapproachesinfinity
\(2x^3+5x^2-1=0\) has one solution we start by factoring this equation first so we can draw conclusion about the inequality
xapproachesinfinity
  • xapproachesinfinity
@Deeezzzz
anonymous
  • anonymous
i found a factor but don't know how to continue.
anonymous
  • anonymous
|dw:1436640267462:dw|
anonymous
  • anonymous
The course I am currently taking is precalc @xapproachesinfinity
xapproachesinfinity
  • xapproachesinfinity
oh good then what does it mean for ab<0
xapproachesinfinity
  • xapproachesinfinity
one has to be negative right? either a<0, b>0 or a>0, b<0
anonymous
  • anonymous
No idea??
anonymous
  • anonymous
When ab multiply they must be less than zero?
xapproachesinfinity
  • xapproachesinfinity
if it was numbers case if we multiply two negative give positive multiplying two positive return a positive so the remaining case is one negative the other positive
anonymous
  • anonymous
Yes that is true so one of the variables has to be a negative
xapproachesinfinity
  • xapproachesinfinity
now we ask which should be positive and which negative so (x+1/2)<0 and the other>0 or (x+1/2)>0 and parabola<0
anonymous
  • anonymous
x<-1/2
xapproachesinfinity
  • xapproachesinfinity
let's simplify first we can write it as (2x+1)(x^2+2x-1)<=0
xapproachesinfinity
  • xapproachesinfinity
just hold on before that
xapproachesinfinity
  • xapproachesinfinity
can we look at that parabola first
anonymous
  • anonymous
yes
xapproachesinfinity
  • xapproachesinfinity
x^2+2x-1 can this be <0?
xapproachesinfinity
  • xapproachesinfinity
if so when does it be <0 if not it must be >0 just the kind of thinking you need here okay
xapproachesinfinity
  • xapproachesinfinity
first that parabola opens up yes
anonymous
  • anonymous
Yes it does with minimum (-1,-2)
xapproachesinfinity
  • xapproachesinfinity
here is a picture of that parabola http://prntscr.com/7rixyx
xapproachesinfinity
  • xapproachesinfinity
so it has part where it is <0 and as well as part where it is >0 good enough
anonymous
  • anonymous
-1
xapproachesinfinity
  • xapproachesinfinity
not really f(x)<0 when x1
xapproachesinfinity
  • xapproachesinfinity
there are irrational roots
xapproachesinfinity
  • xapproachesinfinity
and f(x)>0 for xx2
xapproachesinfinity
  • xapproachesinfinity
you got that part?
anonymous
  • anonymous
Yes
xapproachesinfinity
  • xapproachesinfinity
let's go back with the first step (2x+1)(x^2+2x-1)<=0 implies 2x+1> 0 and x^2+2x-1<0 first case
xapproachesinfinity
  • xapproachesinfinity
we solve for that case by itself!
xapproachesinfinity
  • xapproachesinfinity
second case (2x+1)<0 and (x^2+2x-1)>0
xapproachesinfinity
  • xapproachesinfinity
we already decided when x^2+2x+1<=0 and when x^2+2x-1>=0 just a matter of solving 2x+1<0 2x+1>0 then combine
xapproachesinfinity
  • xapproachesinfinity
or means union
xapproachesinfinity
  • xapproachesinfinity
you got it now?
xapproachesinfinity
  • xapproachesinfinity
let see the first case 2x+1>0 and x^2+2x-1<0 x>-1/2 and x in [x1,x2] well you still need to decide what is x1 and x2 though
anonymous
  • anonymous
so the final answer would be |dw:1436641853588:dw|
xapproachesinfinity
  • xapproachesinfinity
hm where did you [-1.0] i said x1 and x2 are not -1 and 0
xapproachesinfinity
  • xapproachesinfinity
and what -3 there? you have to explain your work
xapproachesinfinity
  • xapproachesinfinity
why*
anonymous
  • anonymous
|dw:1436641991554:dw|
xapproachesinfinity
  • xapproachesinfinity
roots are x1=-1-root2 , x2=root2-1
anonymous
  • anonymous
|dw:1436642079744:dw|
xapproachesinfinity
  • xapproachesinfinity
so first case is x>=-1/2 and -1-root2<=x<=root2-1
xapproachesinfinity
  • xapproachesinfinity
coming back i have a call
anonymous
  • anonymous
yes that is correct so i just need the roots of the second quadratic
anonymous
  • anonymous
|dw:1436642261054:dw|
xapproachesinfinity
  • xapproachesinfinity
intersection here not union
xapproachesinfinity
  • xapproachesinfinity
root2 not root6
xapproachesinfinity
  • xapproachesinfinity
union is the last step
anonymous
  • anonymous
so how would the final answer look like?
xapproachesinfinity
  • xapproachesinfinity
i don;t know yet, i'm doing with you lol
xapproachesinfinity
  • xapproachesinfinity
so we have first case (-oo, -1/2] with [x1, x2] can you find that interval what is ?
anonymous
  • anonymous
x1= -3.44 x2= 1.44
xapproachesinfinity
  • xapproachesinfinity
do number line
anonymous
  • anonymous
(-oo, -3.44) U (-1/2, 1.44]
xapproachesinfinity
  • xapproachesinfinity
oh wait a minute error you would have (-1/2, oo) not -oo, -1/2 first case! look above
xapproachesinfinity
  • xapproachesinfinity
(-1/2,oo) intersection with [-3.4, 1.4]
xapproachesinfinity
  • xapproachesinfinity
so that would be [-1/2, 1.44]
xapproachesinfinity
  • xapproachesinfinity
that's what they share
xapproachesinfinity
  • xapproachesinfinity
now second case x<-1/2 and x<-3.44 and x>1.44
xapproachesinfinity
  • xapproachesinfinity
(-oo, -0.5) intersect (-oo,-3.44) intersect (1.44, oo)
xapproachesinfinity
  • xapproachesinfinity
you there?
xapproachesinfinity
  • xapproachesinfinity
i think we don't need [1.4, oo) there not intersection with that part
xapproachesinfinity
  • xapproachesinfinity
just (-oo, -0.5] and (-oo, -3.4] find the interval of intersection
xapproachesinfinity
  • xapproachesinfinity
you lost ? lol
anonymous
  • anonymous
Just on whether its a union or a intersection
anonymous
  • anonymous

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