Find all possible pairs $(a,b)$ 
Suppose a quadratic equation $$f(x)=x^2+ax+b$$ $f(x)=0$ has roots as $\alpha$ and $\alpha^2-2$. Then find all possible pairs of $(a,b)$. Here $a,b,\alpha\in\mathbb{R}$

My work:
I found two pairs by equating $\alpha$ with $\alpha^2-2$. Also what I think that suppose there are two roots $c$ and $d$. Then if $c=\alpha$ then $\alpha^2-2$ is coming $d$ and if we put $d=\alpha$ then value of $c$ is coming $\alpha^2-2$. This is what should happen. Otherwise we can just put any real number in $\alpha$ and obtain $\alpha^2-2$ yielding infinite solutions.
Any help is greatly appreciated.
I have a solution but not sure whether it is right or wrong. Pls review.
https://www.sarthaks.com/1236891/the-numbers-of-pairs-a-b-of-real-numbers-such-that-whenever-is-a-root-of-the-equation-x-2-ax-b-0
 A: You're correct that any real $\alpha$ will result in real $a$ and $b$, but not necessarily all real values of $a$ and $b$ are possible. To check on this, note the roots being $\alpha$ and $\alpha^2 - 2$ means that
$$\begin{equation}\begin{aligned}
(x-\alpha)(x-(\alpha^2-2)) & = x^2 + ax + b \\
x^2 + (-\alpha - (\alpha^2-2))x + \alpha(\alpha^2-2) & = x^2 + ax + b
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Thus, by equating the coefficients of the powers of $x$, we get (note this could also have been obtained using the Vieta's formulas),
$$a = -\alpha^2 - \alpha + 2 \; \; \to \; \; \alpha^2 + \alpha + (-2 + a) = 0 \tag{2}\label{eq2A}$$
$$b = \alpha(\alpha^2-2) \tag{3}\label{eq3A}$$
Any real $\alpha$ gives a real value for $b$ in \eqref{eq3A}. However, \eqref{eq2A} is a quadratic in $\alpha$, which means $\alpha$ being a real value requires the discriminant to be non-negative, i.e.,
$$1^2 - 4(-2 + a) \ge 0 \; \; \to \; \;  9 - 4a \ge 0 \; \; \to \; \; a \le \frac{9}{4} \tag{4}\label{eq4A}$$
For each such real $a$ satisfying the above inequality, we then get using the quadratic formula that
$$\alpha = \frac{-1 \pm \sqrt{9 - 4a}}{2} \tag{5}\label{eq5A}$$
For $a = \frac{9}{4}$, this gives one value of $\alpha = -\frac{1}{2}$, while for smaller values of $a$ it gives $2$ values of $\alpha$. Using these value(s) of $\alpha$ in \eqref{eq3A} then gives the corresponding value(s) of $b$.
