Fourth degree equation with parameter

Question

For what values of the parameter a equation $$x^4-a*x^3+(3-2a)*x^2+a*x+1=0$$ has both positive and negative roots?

Answer 1

Since $$x=0$$ is not a solution, we can divide by $$x^2$$ and we get: $$x^2+\frac{1}{x^2}+a\left(-x+\frac{1}{x}\right)+3-2a=0$$. Now Substitute $$t=-x+\frac{1}{x}$$

Answer 2

The discriminant of your polynomial is $(8a^2 - 4a + 1)(a + 10)^2(a - 2)^2$.

Note that $0$ is never a root, so the numbers of positive and negative real roots can only change when $a$ is one of the roots of the discriminant, namely $a=2$ and $a=-10$.

For $ a < -10$ there are two positive and two negative real roots. At $a = -10$ the two positive roots and the two negative roots collide, making the polynomial $(x^2+5x-1)^2$. For $-10 < a < 2$, there are no real roots at all (e.g. for $a=0$ the polynomial is $x^4 + 3 x^2 + 1$). For $a=2$, the polynomial is $(x^2-x-1)^2$ which has two double roots, one positive and one negative. For $a > 2$, there are two positive and two negative real roots.

So the answer is: $a \le -10$ or $a \ge 2$.

Answer 3

Your equation can be written under the form :

$$\underbrace{\dfrac{x^4+3x^2+1}{x^3+2x^2-x}}_{f(x)}=a \tag{1}$$

which is the equation for the abscissas of intersection points of the curve with equation $y=f(x)$ and the horizontal line $y=a$.

Therefore our issue can be re-formulated equivalently into "not all the intersections points (if they are some...) have a same sign for their abscissas".

A quick look at the figure below shows that if

$$a \notin (-10,2)\tag{2}$$

the condition is fulfilled (same result as @Robert Israel).

Remark : in fact, line $y=a$ for $-10<a<2$ has even no intersection points with the curve.

The graphic below is convincing : we can even say that, under condition (2), there are always 2 positive roots and 2 negative roots.

BUT a rigorous proof would necessitate a study of the variations of function $f$ (through the study of the sign of $f'(x)$), knowing that

$$f(x)=x-2+\dfrac{8x^2-2x+1}{x^3+2x^2-x}=x-2-\dfrac{1}{x}+\dfrac{9x+2}{x^2+2x-1}\tag{3}$$