X^4-10x^2+24< x^3-4x
Write in interval notation, use bracket for single points

- anonymous

X^4-10x^2+24< x^3-4x
Write in interval notation, use bracket for single points

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- amistre64

x^2 = -6,4 for the LHS if I did it right in me head.
x = 0,2,-2 for the RHS.....
I wonder if that helps....

- amistre64

x^4 +x^3 -10x^2 +4x +24 < 0

- anonymous

thanks

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## More answers

- amistre64

that should be ...-x^3.... typo :)

- anonymous

So, what goes inside the brackets

- amistre64

dunno yet, still trying to factor it,
I got so far:
(x-2)(x+2)(x+2)(x-3)
We put these on a number line to see which of these makes everything less than 0....

- amistre64

if I did it right:
(-inf,-2) or (2,3)

- amistre64

the easiest way to determie the values are to draw a number line and put in the values that make the equation equal to zero.
Take each value and to the left of it, put a (-) and to the right of it put a (+); then multiply your signs together to get area that are (+) and (-).
then pick the areas that fit the equation :)

- anonymous

I got x3+3x^2-4x-12

- anonymous

Actually, once you have it factorised, it is probably easier to quickly sketch the graph of the quartic function.

- amistre64

or do some calculus :)

- amistre64

X^4-10x^2+24< x^3-4x subtract the RHS from the LHS.
x^4 -x^3 -10x^2 +4x +24 >0 right?

- amistre64

<0 .... typo again... its these fat fingers and this tiny little keyboard :)

- anonymous

yes

- amistre64

dont let the "hero" title fool ya, I am after all, just an idiot in disguise :)

- amistre64

the LHS factors to:
(x-2)(x+2)(x-3)(x+2) right? remultiply to check thatits right...

- anonymous

M Dot cancelled a factor of (x+2) at some point... best not to ever do that with a factor unless you can be sure it is >0 ∀ x ∈ R or it flips the inequality.

- amistre64

<.......-2.........2.........3.......>
- + + +
- - + +
- - - +
--------------------------
- + - +
if I did it right, these are my results for the number line....

- amistre64

but something seems off, the even degree should be positive at both ends...

- anonymous

The answer is only 2< x <3 , not x<-2

- amistre64

ahhh..... that seemes more plausible :)

- anonymous

x^4-10x^2+24\le x^3-4x.
The instructor wants x\[\epsilon\] and U

- amistre64

U is the universal set right?

- anonymous

right

- amistre64

can you do calculus on it? :)

- anonymous

This is an algebra class. I'm familiar with any calculus

- anonymous

I'm not familiar

- amistre64

bummer.... calculus would allow us to "see" what is happening at cetain points on the graph more easily

- amistre64

it appears that the graph touches then turns at x = -2 .... which is why it is positive on both sides of it....

- amistre64

so it "crosses" the x axis at 2 and 3 which make it go negative AND less than 0

- amistre64

so your notation would be [x|x is an element of (2,3)} or some such notation

- anonymous

x \[\in \cup\]

- amistre64

I am unclear about the "U" set, its been qwhile since I had to play in set theory :)

- anonymous

Ok, I'm going to go with my gut.

- amistre64

gut going is good :)

- anonymous

thank you for your help

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