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\[2x^3+ 5x^2 -1 \le 0\]
solve the inequality
what course are you taking! now of some factoring thoerems
i meant do you know of some factoring theorems or not
\(2x^3+5x^2-1=0\) has one solution we start by factoring this equation first so we can draw conclusion about the inequality
i found a factor but don't know how to continue.
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
When ab multiply they must be less than zero?
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
Yes that is true so one of the variables has to be a negative
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
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
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
not really f(x)<0 when x1
there are irrational roots
and f(x)>0 for x
you got that part?
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
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
or means union
you got it now?
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
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
roots are x1=-1-root2 , x2=root2-1
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
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)
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