How to prove that the roots of a quartic equation are not ALL real

Question

Given this equation: $$x^4 + x^3 - 3x^2 + 4x - 2 = 0$$ I wanna prove that not all roots are real. How can I go about achieving this?

Answer 1

Setting $$x^4+x^3-3x^2+4x-2=(x^2+ax+b)(x^2+cx+d)$$ where $a,b,c,d\in\mathbb Z$ such that $|b|\lt |d|$ gives you $$1=c+a$$ $$-3=d+ac+b$$ $$4=ad+bc$$ $$(b,d)=(1,-2),(-1,2)$$ Solving these gives you $(a,b,c,d)=(-1,1,2,-2)$, i.e. $$x^4+x^3-3x^2+4x-2=(x^2-x+1)(x^2+2x-2).$$ Here, note that $x^2-x+1=0\iff x=\frac{1\pm\sqrt{3}i}{2}$.

Answer 2

Just another way: If the quartic has all four real roots, then its derivative, must have all three roots real. So we must have three real roots for $$4x^3+3x^2-6x+4=0$$ But this cannot have a positive root as $3x^2+4 \ge 4\sqrt3x> 6x$, and Descarte's rule of signs does not allow more than one negative root.

Answer 3

We could also approach the question in this way. (I grant that this is not the tidiest method. I was looking for something that avoids computing discriminants -- which would answer the question directly -- or solving for zeroes of a cubic, but could exploit properties of the curve for the function.)

This quartic has a positive leading coefficient (thus, both "tails" go to "positive infinity"), but a negative $ \ y-$ intercept, so it must have at least two real zeroes. More than two real zeroes, or two real zeroes with multiplicities other than 1, would require that there be three "turning points" (horizontal tangents). [Plainly, there is not a single real root of multiplicity 4, since this polynomial is not of the form $ \ (x - a)^4 \ $ .]

The second derivative of the polynomial is the quadratic $ \ 12x^2 \ + \ 6x \ - \ 6 \ = \ 6 \ ( 2x^2 \ + \ x \ - \ 1 ) \ $ , which has zeroes at $ \ x \ = \ -1 \ $ and $ \ x \ = \ \frac{1}{2} \ $ . Investigating the sign of this function indicates that the original polynomial is "concave upward" over $ \ ( \ -\infty, \ -1 \ ) \ $ and $ \ ( \ \frac{1}{2}, \ \infty \ ) \ $ .

The first derivative $ \ 4x^3 \ + \ 3x^2 \ - \ 6x \ + \ 4 \ $ is a cubic polynomial and thus must have one real zero, so there is a "turning point" for the curve. We note that the sign of the first derivative changes (from negative to positive) in the interval $ \ ( \ -2, \ -1 \ ) \ $ , so it contains a turning point. We wish to show that there are no others.

On the "concave downward" interval $ \ ( \ -1, \ \frac{1}{2} \ ) \ $ , the first derivative is positive only, as there is no sign-change between the endpoints. Hence, there is no turning point in the concave-downward portion. (We can also eliminate the possibility of one real root for the function with multiplicity 3 , as this would require one of the zeroes to be a point of inflection [zero of the second derivative] as well.)

Having located only a single turning point, and no real root with multiplicity 3 , the curve of the function only has two $ \ x-$ intercepts. These cannot be two zeroes of multiplicity 2, since that would also require three turning points. Thus, the remaining roots must be a complex conjugate pair.

The analytic methods described by the other respondents are certainly briefer and, in a sense, encapsulate the argument given here. [I'll note that, upon consulting WolframAlpha, the discriminant of the quartic is $ \ -2916 \ < \ 0 \ $ , which also tells us that there are two distinct real roots and two complex conjugate roots.]