Well, when we have a transfer function that looks like:
$$\mathcal{T}:=\frac{\text{Y}\left(\text{s}\right)}{\text{X}\left(\text{s}\right)}=\frac{\text{G}\left(\text{s}\right)}{1+\text{G}\left(\text{s}\right)\text{H}\left(\text{s}\right)}\tag1$$
Where $\text{H}\left(\text{s}\right):=1$ because of the unity feedback and $\text{G}\left(\text{s}\right):=\text{K}\cdot\frac{1}{\text{s}-\alpha}$
We get a complete transfer function that looks like:
$$\mathcal{T}=\frac{\text{Y}\left(\text{s}\right)}{\text{X}\left(\text{s}\right)}=\frac{\text{K}\cdot\frac{1}{\text{s}-\alpha}}{1+\text{K}\cdot\frac{1}{\text{s}-\alpha}}\tag2$$
Now, for the poles and zeros:
- Poles:
$$1+\text{K}\cdot\frac{1}{\text{s}-\alpha}=0\space\Longleftrightarrow\space\text{s}=\alpha-\text{K}\tag3$$
- Zeros:
$$\text{K}\cdot\frac{1}{\text{s}-\alpha}=0\space\Longleftrightarrow\space\text{K}=0\tag4$$
For the settling-time we have that:
$$t\space_{2\text{%}}=-\frac{\ln\left(50\right)}{\lambda}\space\Longleftrightarrow\space\lambda=-\frac{\ln\left(50\right)}{t\space_{2\text{%}}}\tag5$$
We need to use:
$$\text{s}=\lambda+\omega\text{j}\tag6$$
Where $\lambda\space\wedge\space\omega\in\mathbb{R}$ and $\text{j}^2=-1$
Now, we need to solve for $\text{K}$ in the pole equation (assuming $\text{K}\ne0$):
$$1+\text{K}\cdot\frac{1}{\left(-\frac{\ln\left(50\right)}{t\space_{2\text{%}}}+\omega\text{j}\right)-\alpha}=0+0\text{j}\space\Longleftrightarrow\space\text{K}=\alpha+\frac{\ln\left(50\right)}{t\space_{2\text{%}}}\space\space\space\wedge\space\space\space\omega=0\tag7$$
Conclusion, we get a transfer function that looks like:
$$\color{red}{\mathcal{T}=\frac{\text{Y}\left(\text{s}\right)}{\text{X}\left(\text{s}\right)}=\frac{\alpha\cdot t\space_{2\text{%}}+\ln\left(50\right)}{\text{s}\cdot t\space_{2\text{%}}+\ln\left(50\right)}}\tag8$$
A general PID controller has the transfer function
$$K(s) = \underbrace{K_\text{P}}_{\text{proportional term}} + \underbrace{K_\text{I}\dfrac{1}{s}}_{\text{integral term}}+\underbrace{K_\text{D}s}_{\text{derivative term}}.$$
The proportional term is scaling the error $e$ between the desired output $y_\text{d}$ and the actual value of the output $y$ by the constant $K_\text{P}$ to generate an input to the plant. You can see this as the term that reacts to the present state of the error. As a rule of thumb, you can try to remember that higher $K_\text{P}$ values will make the controller react faster to deviations from the desired output, which might lean to unstable system behaviour. Imagine you are driving a car and as soon as you see that you are deviating from the middle of the road you react by steering very violently (high $K_\text{P}$) to the opposite direction. You can imagine that after a short time you will be deviating from the middle of the road but to the opposite direction, then you again react very violently into the other direction. I hope you can imagine that such an aggressive proportional reaction might lead to undesired dynamics of the car.
The integral term accumulates the error by integrating it over the past values of the error. It will generate an increasing input to the plant if the error is not vanishing. Hence, the integral term can be viewed as the term that takes into account the history/past of the error. An example for this is when you are standing under the shower. You start to open the valve for warm water if you see that the water is still not warm enough (error integrated over a time period) you open the valve for even more warm water. Two main characteristics of the integral term are that helps you get a better steady-state error but at the same time, it can lead to instability and increased oscillations. There is also the problem of integrator windup which is the reason why you should not use a standard PID controller (with the integral term) without an anti-windup.
The derivative term is reacting to changes of the error. The derivative term can be used to prevent oscillations because of the integral term. It can be viewed as a term that is reacting to the future of the signal. In practice, the PID controller is not implemented as I have written it because it is not a proper transfer function. The derivative term should not be used if it possible to achieve good performance without it as the derivative term can generate a lot of noise if your measurements are very noisy.
A PI-Controller is simply obtained if you do not use the derivative term $K_\text{D}=0$.
The reason why we are interested in the transfer functions that you have written is that they represent different input to output transfer functions.
See this following control circuit (adapted from]1)
The first transfer function is the reference $r$ to the error $e$ transfer function. It is also called the sensitivity function
$$S(s) = \dfrac{e(s)}{r(s)}=\dfrac{1}{1+K(s)G(s)}$$
the fourth function is called complementary sensitivity function (I think it is also called the transmissibility function)
$$T(s) = 1-S(s)=\dfrac{y(s)}{r(s)}=\dfrac{K(s)G(s)}{1+K(s)G(s)}$$
in which $y$ is the output.
The second transfer function is the transfer function from an additive disturbance between the controller and the plant $d$ and the output $y$. It is called disturbance transfer function
$$\dfrac{y(s)}{d(s)}=\dfrac{G(s)}{1+K(s)G(s)}.$$
The third transfer function is relating the reference $r$ to the controller output $u$. I am not sure how it is called in English, but it can be translated from German as actuation transfer function
$$\dfrac{u(s)}{r(s)}=\dfrac{K(s)}{1+K(s)G(s)}.$$
As you can see the stability of the closed-loop system is always dictated by $1+K(s)G(s)$. The study of these transfer function gives you additional information, for example, how does an additive disturbance between controller and plant manifest itself in the output ($y(s)/d(s)$). Or how does the actuation / control effort $u$ depend on the reference $r$ ($u(s)/r(s)$).
Best Answer
You made one incorrect assumption. Namely, that the coefficient of the quadratic term also has to be positive. However, if all coefficients are negative one can factor out the minus sign and the resulting characteristic equation can still have all roots in the left half plane.
So the relevant inequalities would become:
\begin{align} 1+50 K_d &< 0 \\ K_p - K_d &< 0 \\ -25 - 50 K_p &< 0 \end{align}
for which solutions can be found. But I will leave finding those solutions up to you.