Let $P(x) = x^2 + Bx + C$, for $B,C \in \mathbb{R}$, have real roots. Represent the set of all other $p(x)= x^2 + bx + c$ that share a root with $P(x)$ by a plot in the $bc$-plane. It forms a nice picture: for example, here it is for $B=3, C=-1$, when the roots are $-\frac{1}{2} \left(3 \pm \sqrt{13}\right)$:
Lines intersect at $b=3,c=-1$.
All those $(b,c)$ on the two lines share a root with $x^2 + 3x -1$. The lines are tangent to the paraboloid $b^2 = 4c$, where the discriminant is zero. Two questions:
Q1. Can you explain the characteristics of this picture geometrically, without relying heavily on explicit algebraic calculations?
By "the characteristics" I mean: Two straight lines meeting at the $B,C$ point, tangent to the discriminant parabola.
Q2. What is the analogous picture for $P(x) = x^3 + B x^2 + C x + D$, the set of cubic polynomials that share at least one root with $P(x)$, plotted in $bcd$-space?
I would welcome further generalizations: To coefficients and roots in $\mathbb{C}$, and/or to higher degree polynomials.
Fix a root $\alpha$ of $P(X)$. Then every polynomial of that form with root $\alpha$ must be of the form $(x-\alpha)(x-\beta)$ and $b=-\alpha-\beta, c=\alpha\beta$. From this formula we get that the dependance between $c$ and $b$ is linear. More specifically $c=-\alpha b-\alpha^2$. This way you obtain a line through $(B,C)$ and if you chose the other root you obtain the other line.
Now, about the parabola, it is clear that these lines must intersect it in the point $(-2\alpha_1,\alpha_1^2 )$ and $(-2\alpha_2,\alpha_2^2)$ and they certainly can't go in the $b^2<4c$ area because a point in this zone corresponds to a polynomial with no real root. It follows that the lines must be tangent to the parabola.