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

Question

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.

Answer 1

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.

Answer 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.

Answer 3

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.

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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 $2$.

Answer 4

If we "drop" the constant term for the moment, the resulting quartic function $ \ g(x) \ = \ x^4 - 3x $ $ = \ x·(x^3 - 3) \ $ has (single) real zeroes at $ \ x \ = \ 0 \ $ and $ \ x \ = \ \sqrt[3]{3} \ \ . \ $ Since the curve for this even-degree polynomial "opens upward" (positive leading coefficient), there is a "turning point" (absolute minimum) between these $ \ x-$intercepts for which $ \ f(x_{min}) \ < \ 0 \ \ . \ $ We calculate that $ \ g(1) \ = \ -2 \ \ $ (should we wish more precision, from your first derivative, we obtain $$ \ x_{min} \ = \ \left(\frac34 \right)^{1/3} \ \ \Rightarrow \ \ g(x_{min}) \ \ = \ \ \left(\frac34 \right)^{1/3} · \left(\frac34 - 3 \right) \ \ = \ \ \left(\frac34 \right)^{1/3} · \left(-\frac94 \right) $$ $$ = \ \ \left(-\frac{2187}{256} \right)^{1/3} \ \ < \ \ \left(-\frac{2048}{256} \right)^{1/3} \ \ = \ \ -2 \ \ . \ ) $$

From this, we conclude that the minimum of $ \ f(x) \ = \ x^4 - 3x + 1 \ $ can be no larger than $ \ -1 \ \ . \ $ So the curve of this function also has just two $ \ x-$intercepts, hence $ \ f(x) \ $ possesses only two real zeroes.

Answer 5

If the discriminant of a quartic equation with real coefficients is negative then the quartic equation has two real roots and two complex conjugate roots.

The discriminant of the equation $ax^4+dx+e=0$ is $\Delta=256a^3e^3-27a^2d^4$. So, the discriminant of $x^4-3x+1=0$ is $\Delta=256+(-27)(-3)^4=-1931<0$.