Number of positive real roots in equation $x^6 - x - 1 =0$ [closed]

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How can I find the number of positive real roots in the equation $$x^6 - x - 1 =0$$ This question is given in my analysis book and I have no idea how to start

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3 Answers

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Or (more informally) re-write $x^6-x-1=0$ as $x^6=x+1$ and then graph the functions $y=x^6$ and $y=x+1$.

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Descartes' rule of signs is a not very well known, but sometimes surprisingly useful, theorem that helps with this problem. It says that if a polynomials' coefficients have $n$ sign changes when read from left to right, then the number of positive real roots is $n-2k$ for some integer $k\ge 0$. (That is, it is at most $n$, or if it is less than $n$ it is less by an even number.)

In this case, there is a single sign change, so there must be exactly one positive real root.

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Divide by $x$. The equation is equivalent to $$x^5 -( 1 + \frac{1}{x})=0$$ LHS strictly increasing on the interval $(0, \infty)$ from $-\infty$ to $\infty$ so the equation has a unique solution in $(0, \infty)$. Same for the interval $(-\infty, 0)$.

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