Choosing parameter such that the roots of a cubic polynomial have negative real parts

Question

I need to figure out the values of $a\in \mathbb{R}$, which gives roots with negative real parts.

$$P(\lambda) = −\lambda^3 − 3 \lambda^2 + (a − 2) \lambda − 1$$

What technique to apply?

According to the exercise, the correct answer is $a < 5/3$

Answer 1

You can use the Hurwitz criterion after multiplying by (-1) you get

$$G(\lambda)=\lambda^3+3\lambda^2+(2-a)\lambda+1$$

The Hurwitz criterion for a cubic polynomial $$a_3\lambda^3+a_2\lambda^2+a_1\lambda+a_0$$

says. The real part of the roots is negative if $a_3>0$, $a_2>0$, $a_1>0$, $a_0>0$ and $a_1a_2>a_0a_3$.

From this, we obtain $2-a>0$ (third inequality).

Additionally, we need to have $3(2-a)>1$ (last condition), which implies $2-a>1/3$. Can you do the rest?

Answer 2

If we restrict our attention to real roots, we can use the intermediate value theorem.

The coefficient of $\lambda^3$ is negative, $\lim_{n\to-\infty}P(\lambda)=\infty$. By the intermediate value theorem, if there exists a $k<0$ such that $P(k)>0$ then there exists a negative real root. This is not just a sufficient condition, but also a necessary one. If a polynomial has a negative real root, then there exists a negative number $k$ such that $P(k)>0$.

Answer 3

I don't know if the following is correct. And if it is, it is only partial solution.

You are interested for which $a$ the function $f(x)= {x^3+3x^2+2x+1\over x}$ cuts the line $y=a$ only in $x$ with negative real part. Thus all real solutions must be also negative. Calculating the $f'$ we see that it has only local exstremum (minimum) at $x={1\over 2}$ and $y= {23\over 4}$ so for $a\geq {23\over 4}$ the $p$ has solution with positive real part. So $a$ must be smaller then $23/4$