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- anonymous

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- anonymous

\[2x^3+ 5x^2 -1 \le 0\]

- anonymous

solve the inequality

- anonymous

x<=-1-√2

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## More answers

- anonymous

how?

- xapproachesinfinity

what course are you taking!
now of some factoring thoerems

- xapproachesinfinity

i meant do you know of some factoring theorems or not

- 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

@Deeezzzz

- anonymous

i found a factor but don't know how to continue.

- anonymous

|dw:1436640267462:dw|

- anonymous

The course I am currently taking is precalc @xapproachesinfinity

- xapproachesinfinity

oh good then
what does it mean for ab<0

- xapproachesinfinity

one has to be negative right? either a<0, b>0 or a>0, b<0

- anonymous

No idea??

- anonymous

When ab multiply they must be less than zero?

- 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

Yes that is true so one of the variables has to be a negative

- 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

x<-1/2

- xapproachesinfinity

let's simplify first
we can write it as (2x+1)(x^2+2x-1)<=0

- xapproachesinfinity

just hold on before that

- xapproachesinfinity

can we look at that parabola first

- anonymous

yes

- xapproachesinfinity

x^2+2x-1 can this be <0?

- xapproachesinfinity

if so when does it be <0 if not it must be >0
just the kind of thinking you need here okay

- xapproachesinfinity

first that parabola opens up yes

- anonymous

Yes it does with minimum (-1,-2)

- xapproachesinfinity

here is a picture of that parabola
http://prntscr.com/7rixyx

- xapproachesinfinity

so it has part where it is <0 and as well as part where it is >0
good enough

- anonymous

-1

- xapproachesinfinity

not really
f(x)<0 when x1

- xapproachesinfinity

there are irrational roots

- xapproachesinfinity

and f(x)>0 for xx2

- xapproachesinfinity

you got that part?

- anonymous

Yes

- 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

we solve for that case by itself!

- xapproachesinfinity

second case (2x+1)<0 and (x^2+2x-1)>0

- 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

or means union

- xapproachesinfinity

you got it now?

- 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

so the final answer would be
|dw:1436641853588:dw|

- xapproachesinfinity

hm where did you [-1.0]
i said x1 and x2 are not -1 and 0

- xapproachesinfinity

and what -3 there? you have to explain your work

- xapproachesinfinity

why*

- anonymous

|dw:1436641991554:dw|

- xapproachesinfinity

roots are x1=-1-root2 , x2=root2-1

- anonymous

|dw:1436642079744:dw|

- xapproachesinfinity

so first case is x>=-1/2 and -1-root2<=x<=root2-1

- xapproachesinfinity

coming back i have a call

- anonymous

yes that is correct so i just need the roots of the second quadratic

- anonymous

|dw:1436642261054:dw|

- xapproachesinfinity

intersection here not union

- xapproachesinfinity

root2 not root6

- xapproachesinfinity

union is the last step

- anonymous

so how would the final answer look like?

- xapproachesinfinity

i don;t know yet, i'm doing with you lol

- xapproachesinfinity

so we have first case (-oo, -1/2] with [x1, x2]
can you find that interval what is ?

- anonymous

x1= -3.44 x2= 1.44

- xapproachesinfinity

do number line

- anonymous

(-oo, -3.44) U (-1/2, 1.44]

- xapproachesinfinity

oh wait a minute error you would have (-1/2, oo) not -oo, -1/2
first case! look above

- xapproachesinfinity

(-1/2,oo) intersection with [-3.4, 1.4]

- xapproachesinfinity

so that would be [-1/2, 1.44]

- xapproachesinfinity

that's what they share

- xapproachesinfinity

now second case x<-1/2 and x<-3.44 and x>1.44

- xapproachesinfinity

(-oo, -0.5) intersect (-oo,-3.44) intersect (1.44, oo)

- xapproachesinfinity

you there?

- xapproachesinfinity

i think we don't need [1.4, oo) there not intersection with that part

- xapproachesinfinity

just (-oo, -0.5] and (-oo, -3.4]
find the interval of intersection

- xapproachesinfinity

you lost ? lol

- anonymous

Just on whether its a union or a intersection

- anonymous

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