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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 $x$?

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That answer depends on what course it is. You are on the right track. –  Aryabhata Apr 7 '11 at 14:33
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Note that you don't have to solve that equation, only to prove it has a real solution. –  lhf Apr 7 '11 at 14:38
    
If 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? –  user7143 Apr 7 '11 at 14:45
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I believe continuity is the word you're supposed to think about (I hope I'm not revealing too much :) –  user8268 Apr 7 '11 at 14:46
    
@Bill your edit left an extra +1 in the question. –  Zach Langley Apr 7 '11 at 16:34
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4 Answers

To make things as simple as possible: you want $x$ such that $x = x^3 + x^2 + 1$, that is $x^3 + x^2 - x + 1 = 0$. And every cubic with real coefficients has a real root (because it has different signs at $x$ and $-x$ for large enough $x$).

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Let $f(x) := x^3 + x^2 -x + 1$. Consider the following,

$$\mathrm{sgn}\left(\lim_{x \to \infty} \quad f(x) \right),$$

and,

$$\mathrm{sgn}\left(\lim_{x \to -\infty} \quad f(x) \right).$$

What can you conclude?

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That it at some point it crosses the x axis so we can conclude that –  user7143 Apr 7 '11 at 17:21
    
there exists a real root –  user7143 Apr 7 '11 at 17:22
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Another way is to show that it is continuous, and then find a value of the curve $y=x^3+x^2+1$ below the $y=x$ line and another point above the $y=x$ line.

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As every polynomial is continuous on $R$ and use the Intermediate value theorem? –  user7143 Apr 7 '11 at 14:51
    
Yup you got it. –  picakhu Apr 7 '11 at 14:51
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and use \mathbb{R} $\mathbb{R}$ –  picakhu Apr 7 '11 at 14:52
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If a polynomial has real coefficients then its solutions will be either real or come in complex conjugate pairs. That implies that every polynomial of odd degree with real coefficients has at least one real root.

The reason that the non-real solutions must come in conjugate pairs is because they must have the same symmetry as the polynomial: The polynomial does not change when you switch i with -i so the solution set cannot change when you do that either.


A polynomial (with real coefficients) is a statement in the language of rings that defines a set of points $X$ in $\mathbb C$. These sets are called $\mathbb R$ definable and the symmetries of $\mathbb C$ that fix $\mathbb R$ are relevant, let $\sigma$ generate them then $x \in X \iff x \in \sigma(X)$.

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This is certainly true, but its justification seems to require some analytic reasoning with (e.g.) the Intermediate Value Theorem. Thus the latter answer seems more fundamental. On the other hand, your answer is accessible to a precalculus student and the OP has not specified her background, so there is a good chance this is exactly what her instructor is expecting to hear. –  Pete L. Clark Apr 7 '11 at 15:30
    
@Pete, I have tried to express it clearer I don't think it needs any analysis. –  quanta Apr 7 '11 at 15:55
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Hmm. First, a polynomial with real coefficients is not a statement in the language of rings for two reasons: (i) the variable $x$ is unbound by quantitifers, so it is a rather a formula, (ii) but it is not a formula either, because the names of the individual real numbers are not part of this language. Second and more to the point: you seem to be assuming that $\mathbb{R}[\sqrt{-1}]$ is algebraically closed. This is an analytic fact; one famous proof starts by showing that every odd degree real polynomial has a root and then applying Sylow theory. –  Pete L. Clark Apr 7 '11 at 17:01
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(And by the way, why are you bringing model-theoretic language into the question? This question lies somewhere in the closed interval from precalculus to calculus.) –  Pete L. Clark Apr 7 '11 at 17:05
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@Pete: What precisely do you mean when you say "$\mathbb R[\sqrt{-1}]$ is algebraically closed" is an "analytic fact"? –  Bill Dubuque Apr 7 '11 at 19:13
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