How do I show that there exists a real number that equals its cube plus its square plus 1? I was thinking $x = x^3+x^2+1$ then solve for $x$?
How do I show that there exists a real number that equals its cube plus its square plus 1?
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algebra-precalculus
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0That answer depends on what course it is. You are on the right track. – 2011-04-07
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2Note that you don't have to *solve* that equation, only to prove it has a real solution. – 2011-04-07
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0If the discriminant equals or is larger than zero can i claim a proof?But suppose i don't remember how to solve 3° equations what could i do? – 2011-04-07
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4I believe *continuity* is the word you're supposed to think about (I hope I'm not revealing too much :) – 2011-04-07
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0@Bill your edit left an extra +1 in the question. – 2011-04-07
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0@Zach: Thanks, I submitted a new edit proposal. I'm here slowly trying to learn the extra syntax rules the math editor adds. – 2011-04-07