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I'm starting to learn proof and logic, all my knowledge in mathematics is: basic discrete boolean logic, introductory logic, basic Calculus, Introductory Linear Algebra.

I've been trying to use what I know to prove this statement, but I simply can't.

If $x$ is real number such that $x^4 + 2x^2 -2x \lt 0$, then $0\lt x\lt 1$.

I dont know enough Calculus to find its roots, and if use Newton's method, I am able to find $x=0$ and $x\approx 0.77777$. But I wont know if there are two more real roots.

Any tips?

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The other two roots are complex. – AnonymousCoward Jan 7 '11 at 17:53

The highlighted statement can be proved without calculus or root approximations.

$\rm\ x^4\: +\: 2\:x^2\: -\: 2\:x\ \ge\ 0\ \ $ for $\rm\ \ x \le 0\ $ since all three summands are $\ge 0\ $ there.

$\rm\ x^4 + 2\:x\: (x-1)\ \ge\ 0\ \ $ for $\rm\ \ x \ge 1\ $ since the $\:$ two summands are $\ge 0\ $ there.

Therefore, if $\:$ it$\:$ is $\: <\: 0\ $ then $\rm\ 0 < x < 1\:.\quad$ QED

The algebra has a vivid geometric interpretation: shifting the quadratic $\rm\: f = 2\ x\ (x-1)$ upward by adding some function $\rm\:g\ge 0\:$ (here $\rm\: g = x^4\:$) clearly preserves the property that $\rm\ f \ge 0\ $ outside $[0,1]$.

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+1 for a super elementary answer. – AnonymousCoward Jan 7 '11 at 18:29

The second derivative is $12x^2+4$ which is positive for all $x$ so the function is convex. Since you showed the two roots are in $[0,1]$ it follows that $f(x) <0 \implies x\in [0,1]$, since by convexity the function is only negative between its two roots.

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Or more simply, after you divide out the $x$, the first derivative is positive, the function is negative at 0, positive at 1. – Aryabhata Jan 7 '11 at 17:57

The equation can be written as $x(x^3 + 2x -2) = 0$. For a general cubic of the form $x^3 + px + q=0$, you can calculate the discriminant $\Delta$ as

\begin{equation} \Delta = -4p^3 -27q^2 \end{equation}

Here, $p=2$ and $q=-2$. Since the discriminant is negative, the cubic equation has only one real root. The other two roots are complex conjugates (

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