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So I have this quartic equation here:

$x^4-3x+1=0$

I'm supposed to prove this equation has exactly 2 roots.

I defined $f(x)=x^4-3x+1=0$

Then I used the Intermediate value theorem at $f(0)$ and $f(1)$ to show it goes from $(0,1)$ to $(1,-1)$ so it must have crossed the x-axis since this polynomial is continuous. So there is at least 1 root.

The derivative is $f'(x)=4x^3-3$ and the second derivative is $f''(x)=12x^2$.

There is a critical point at $x=(3/4)^{\frac{1}{3}}$ where the function changes from increasing from $(-\infty,(3/4)^{\frac{1}{3}}]$ and decreasing from $[(3/4)^{\frac{1}{3}},\infty)$.

There is a point of inflection at $x=0$ where the concavity changes from concave up to concave down.

I'm not really sure how to use this information to determine how to show there is a second root though...

No I can't use Descartes's rule of signs. It would be easy otherwise.

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You have it's increasing then decreasing, so it can have at most two roots (Just by monotonicity). It can't have exactly 1 real root because complex roots come in pairs, so since you found 1, it must have a second.

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  • $\begingroup$ Makes sense. Thanks a lot :) . Is this always true with quartics in general? $\endgroup$ – Future Math person Jul 26 '16 at 3:58
  • $\begingroup$ you have 0, 2, or 4 real roots....all are possible $\endgroup$ – Alan Jul 26 '16 at 4:32
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The Sturm Chain for $x^4-3x+1$ is $$ \left\{x^4-3x+1,\,\,4x^3-3,\,\,\tfrac94x-1,\,\,\tfrac{1931}{729}\right\} $$ There are $2$ sign changes at $-\infty$: $\{+,-,-,+\}$.

There are $0$ sign changes at $+\infty$: $\{+,+,+,+\}$.

This means there are $2$ real roots.


In fact, there are $2$ sign changes at $0$: $\{+,-,-,+\}$.

This means that there are no negative roots and $2$ positive roots.

There is $1$ sign change at $1$: $\{-,+,+,+\}$.

There are $0$ sign changes at $2$: $\{+,+,+,+\}$.

Thus, one of the roots is between $0$ and $1$, and one is between $1$ and $2$.

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You could use the result learned in high school algebra. To find the number of positive real roots, check the number of sign changes in the coefficients of $$f(x)=x^4-3x+1.$$ We have positive, then negative, then positive. 2 sign changes, so 2 positive real roots.

To check the number of negative real roots, check the number of sign changes in the coefficients of $$f(-x)=x^4+3x+1.$$ There are no sign changes, so there are no negative real roots.

Therefore, there are two positive real roots, and two complex roots.

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  • $\begingroup$ Descarts rule of signs would give: two sign changes means there are at most 2 positive roots, but there may also be 0 positive roots. N sign changes means there are one of N, N-2, N-4, ... roots $\endgroup$ – Conrad Turner Jul 26 '16 at 4:15

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