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

solve the inequality

x<=-1-√2

how?

what course are you taking!
now of some factoring thoerems

i meant do you know of some factoring theorems or not

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

|dw:1436640267462:dw|

The course I am currently taking is precalc @xapproachesinfinity

oh good then
what does it mean for ab<0

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

No idea??

When ab multiply they must be less than zero?

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

x<-1/2

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

just hold on before that

can we look at that parabola first

yes

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

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

first that parabola opens up yes

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

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

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

-1

not really
f(x)<0 when x1

there are irrational roots

and f(x)>0 for xx2

you got that part?

Yes

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

we solve for that case by itself!

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

or means union

you got it now?

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

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

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

why*

|dw:1436641991554:dw|

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

|dw:1436642079744:dw|

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

coming back i have a call

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

|dw:1436642261054:dw|

intersection here not union

root2 not root6

union is the last step

so how would the final answer look like?

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

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

x1= -3.44 x2= 1.44

do number line

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

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

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

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

that's what they share

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

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

you there?

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

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

you lost ? lol

Just on whether its a union or a intersection