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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....
x^4 +x^3 -10x^2 +4x +24 < 0
that should be ...-x^3.... typo :)
So, what goes inside the brackets
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....
if I did it right: (-inf,-2) or (2,3)
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 :)
I got x3+3x^2-4x-12
Actually, once you have it factorised, it is probably easier to quickly sketch the graph of the quartic function.
or do some calculus :)
X^4-10x^2+24< x^3-4x subtract the RHS from the LHS. x^4 -x^3 -10x^2 +4x +24 >0 right?
<0 .... typo again... its these fat fingers and this tiny little keyboard :)
dont let the "hero" title fool ya, I am after all, just an idiot in disguise :)
the LHS factors to: (x-2)(x+2)(x-3)(x+2) right? remultiply to check thatits right...
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.
<.......-2.........2.........3.......> - + + + - - + + - - - + -------------------------- - + - + if I did it right, these are my results for the number line....
but something seems off, the even degree should be positive at both ends...
The answer is only 2< x <3 , not x<-2
ahhh..... that seemes more plausible :)
x^4-10x^2+24\le x^3-4x. The instructor wants x\[\epsilon\] and U
U is the universal set right?
can you do calculus on it? :)
This is an algebra class. I'm familiar with any calculus
I'm not familiar
bummer.... calculus would allow us to "see" what is happening at cetain points on the graph more easily
it appears that the graph touches then turns at x = -2 .... which is why it is positive on both sides of it....
so it "crosses" the x axis at 2 and 3 which make it go negative AND less than 0
so your notation would be [x|x is an element of (2,3)} or some such notation
x \[\in \cup\]
I am unclear about the "U" set, its been qwhile since I had to play in set theory :)
Ok, I'm going to go with my gut.
gut going is good :)
thank you for your help