# How to prove that the quartic equation $x^4-3x+1=0$ has exactly 2 real roots?

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.

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.

• Makes sense. Thanks a lot :) . Is this always true with quartics in general? – Future Math person Jul 26 '16 at 3:58
• you have 0, 2, or 4 real roots....all are possible – Alan Jul 26 '16 at 4:32

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 $2$.

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.

• 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 – Conrad Turner Jul 26 '16 at 4:15