Can anyone help me plot the graph of the set $$B=\{(a_1+a_2, a_1a_2):a_1,a_2\in\mathbb{R},|a_1|<1,|a_2|<1\}.$$
Ok so I can shade the region $|x|<2$ and $|y|<1$. But do not know how to proceed after that. And above sets is real roots of the equation $x^2-\alpha x+\beta$ whose modulus is less than 1. Will it be the intersection of the shaded area above and the parabola $x^2-4y>0$?
On
As noted in the comments, the set in question is the image of the open unit square $S_{0} := (-1, -1) \times (-1, 1)$ under the polynomial mapping $$ f(a_{1}, a_{2}) = (a_{1} + a_{2}, a_{1}a_{2}). $$ The open square is contained in the square $S := [-1, 1] \times [-1, 1]$, which is compact, and therefore has compact image under the continuous mapping $f$.
Lemma: If $y = f(x)$ is a point of the boundary of the image $f(S)$, then $x$ is either a critical point of $f$ or lies on the boundary of $S$.
Proof: Contrapositively, if $x$ is an interior point of $S$ and is not a critical point of $f$, the inverse function theorem guarantees some disk about $x$ maps diffeomorphically to a neighborhood of $f(x)$. Particularly, $f(x)$ is an interior point of the image. (Note the similarity with the situation for local extrema of functions of one variable.)
The critical points of $f$ are the line $a_{1} = a_{2}$ (easily checked), the images of the boundary of $S$ and the critical set are easily found (see also Ovi's answer), and these curves border a unique bounded plane region. Finally, in this example, we have $$ f(S_{0}) = f(S \setminus \partial S) = f(S) \setminus f(\partial S). $$ To prove this, it may help to note that $f(a_{2}, a_{1}) = f(a_{1}, a_{2})$ for all $(a_{1}, a_{2})$ in $S$, so the image of $S$ is the image of the triangle $T \subset S$ where $-1 \leq a_{2} \leq a_{1} \leq 1$; and to show $f$ is bijective on $T$.
I don't have time now to write a very formal answer, but hopefully this should help:
Let's examine what happens at the extreme cases, when one of $a_1, a_2$ is $\pm 1$.
$$\begin{align*} a_1 &= +1: &\{(a_2 + 1, a_2): |a_2| \le 1 \}\\ a_1 &= -1: &\{(a_2 - 1, -a_2): |a_2| \le 1 \}\\ a_2 &= +1: &\{(a_1 + 1, a_1): |a_1| \le 1 \}\\ a_2 &= -1: &\{(a_1 - 1, -a_1): |a_1| \le 1 \}\\ \end{align*}$$ These four sets form the red part:
Now instead of fixing $a_2$ as $\pm 1$, fix it as some $b$ with $|b| < 1$ and let $a_1$ range over $[-1, 1]$. This gives the blue line. As $b$ changes, the ends of the blue line remain on the red part. I encourage you to play with the $b$ slider on Desmos.
So it seems like the region will be bounded by the red part below, and by some sort of blue parabola-like curve above.
To find the blue parabola above, let's fix the $x$ coordinate as $c$ (with $|c| < 2$) and find the corresponding point in the region with the largest $y$ coordinate. This is the optimization problem
$$\begin{align*}\max a_1a_2\\ \text{s.t. } \ a_1 + a_2 = c\\ -1 \le a_1 \le 1\\ -1 \le a_2 \le 1 \end{align*}$$
We solve this problem by eliminating $a_1$. We have $a_1 = c - a_2$, and the requirement $-1 \le a_1 \le 1$ turns into $c - 1 \le a_2 \le c+1.$ So the optimization problem turns into
$$\begin{align*}\max (c - a_2)a_2\\ \text{s.t. } -1 \le a_2 \le 1\\ c - 1 \le a_2 \le c+1 \end{align*}$$
Assume $c \ge 0$. The case for $c < 0$ is similar. The problem turns into $$\begin{align*}\max (c - a_2)a_2\\ \text{s.t. } c-1 \le a_2 \le 1\\ \end{align*}$$ which is simply a quadratic over the interval $[c-1, 1]$. For $0 \le c \le 2$, the maximum occurs at $a_2 = \frac c2$ and it is $\frac {c^2}{4}.$ Therefore the upper boundary for the region should be $f(c) = c^2/4$. So the region should look like this: