Let's represent on the $(p ; q)$ plane the points for which the coordinates $p, q$ satisfy that the equation $x^{2}-2 p x+q=0$ has:

a) two roots;

b) two as a root;

c) two as the only root.
Solution. a) A quadratic equation has two distinct roots if and only if its discriminant, $D=4 p^{2}-4 q$, is positive. In the $(p ; q)$ plane, this is true for the points outside the parabola $q=p^{2}$. This set is shown in Figure 1.

![](https://cdn.mathpix.com/cropped/2024_05_02_79e848200058d3eb85a9g-1.jpg?height=391&width=352&top_left_y=268&top_left_x=881)

Figure 1

b) The equation has "2 as a root" if substituting 2 for $x$ results in an equality: $2^{2}-4 p+q=4-4 p+q=0$. In the $(p ; q)$ plane, this is the equation of a line, as shown in Figure 2.

![](https://cdn.mathpix.com/cropped/2024_05_02_79e848200058d3eb85a9g-1.jpg?height=347&width=369&top_left_y=892&top_left_x=867)

Figure 2

c) If 2 is the only root of the equation, then in addition to the condition found in part b) $(4-4 p+q=0)$, the discriminant must be 0, i.e., $q=p^{2}$. This gives us the equation $p^{2}-4 p+4=(p-2)^{2}=0$, from which $p=2$ and thus $q=4$. Therefore, the condition in c) is satisfied by only one pair of numbers, which is $(2 ; 4)$ (Figure 3).

![](https://cdn.mathpix.com/cropped/2024_05_02_79e848200058d3eb85a9g-1.jpg?height=398&width=290&top_left_y=1525&top_left_x=907)

Figure 3

Remark. If we represent the three sets of points in a single coordinate system (Figure 4), it is visible that the line from part b) is tangent to the parabola bounding the set of points from part a) at the point from part c).

![](https://cdn.mathpix.com/cropped/2024_05_02_79e848200058d3eb85a9g-1.jpg?height=352&width=290&top_left_y=2120&top_left_x=907)

Figure 4