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SuicideFuel Problem (ITA 2021)

trying to ascend

trying to ascend

Oldcel KHHV
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Aug 30, 2020
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Let S ⊂ R be the solution set of the inequality (x² + x + 1) to the power of 2x²−x−1 ≤ 1. Therefore the set is?

[UWSL][/UWSL]
 
i aint no baby daddy to do yo homework boy
 
i aint no baby daddy to do yo homework boy
SocialzERo


IT'S NOT MY HOMEWORK, IT'S FROM AN ENTRANCE EXAM!!!!!!!!!!!!!!!!!!!!!!!!!
 
u applying to maths/physics or what?
NO, I'M TOO LOW IQ FOR THAT EXAM (AVERAGE IQ OF THE APPROVED ONES IS 125), HENCE WHY I'M HERE TRYING TO GET ANSWERS
 
Let S ⊂ R be the solution set of the inequality (x² + x + 1) to the power of 2x²−x−1 ≤ 1. Therefore the set is?
So from what I got the inequality is equivalent to one when x^2+x+1=1 or when 2x^2-x-1=0
Which give x=0 or x=-1 and x=-1/2 or x=1 (x^2+x+1 is always more then zero for any non imaginary x value)

Next take a=x^2+x+1
and b=2x^2-x-1
a^b<=1
Which means when a>1 and b<0
or a<1 and b>0
This gives us 0<x<1 and -1<x<-1/2
Which means the set that this is valid for is just
0<=x<=1 and -1<=x<=-1/2
 

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