Problem 1210

[Proof] Let $f(x)=x^{2}+p_{1} x+q_{1}$,
$$g(x)=x^{2}+p_{2} x+q_{2} \text {. }$$

If $\quad f(x)=g(x)$,
then
$$\begin{array}{l} x^{2}+p_{1} x+q_{1}=x^{2}+p_{2} x+q_{2}, \\ x=\frac{-\left(q_{1}-q_{2}\right)}{p_{1}-p_{2}} . \end{array}$$

At this point,
$$\begin{aligned} f(x) & =\left(\frac{q_{1}-q_{2}}{p_{1}-p_{2}}\right)^{2}-p_{1} \cdot \frac{q_{1}-q_{2}}{p_{1}-p_{2}}+q_{1} \\ & =\frac{\left(q_{1}-q_{2}\right)^{2}-p_{1} \cdot\left(q_{1}-q_{2}\right)\left(p_{1}-p_{2}\right)+q_{1}\left(p_{1}-p_{2}\right)^{2}}{\left(p_{1}-p_{2}\right)^{2}} \\ & =\frac{\left(q_{1}-q_{2}\right)^{2}+\left(p_{1}-p_{2}\right)\left(p_{1} q_{2}-p_{2} q_{1}\right)}{\left(p_{1}-p_{2}\right)^{2}} \\ & <0 \end{aligned}$$

This means that, in the Cartesian coordinate system, the intersection point of the two upward-opening parabolas (1) and (2) is below the x-axis. Therefore, each of these two parabolas intersects the x-axis at two points, and between the two roots of each polynomial, there must be a root of the other polynomial. Therefore, the quadratic trinomials $f(x)$ and $g(x)$ both have two real roots, and between the two roots of each polynomial, there is a root of the other polynomial.

[Proof] Let $f(x)=x^{2}+p_{1} x+q_{1}$,
$$g(x)=x^{2}+p_{2} x+q_{2}$$

If $\quad f(x)=g(x)$,
then

at this point,
$$\begin{array}{l} x^{2}+p_{1} x+q_{1}=x^{2}+p_{2} x+q_{2} \\ x=\frac{-\left(q_{1}-q_{2}\right)}{p_{1}-p_{2}} \end{array}$$
$$\begin{aligned} f(x) & =\left(\frac{q_{1}-q_{2}}{p_{1}-p_{2}}\right)^{2}-p_{1} \cdot \frac{q_{1}-q_{2}}{p_{1}-p_{2}}+q_{1} \\ & =\frac{\left(q_{1}-q_{2}\right)^{2}-p_{1} \cdot\left(q_{1}-q_{2}\right)\left(p_{1}-p_{2}\right)+q_{1}\left(p_{1}-p_{2}\right)^{2}}{\left(p_{1}-p_{2}\right)^{2}} \\ & =\frac{\left(q_{1}-q_{2}\right)^{2}+\left(p_{1}-p_{2}\right)\left(p_{1} q_{2}-p_{2} q_{1}\right)}{\left(p_{1}-p_{2}\right)^{2}} \\ & <0 \end{aligned}$$

This means that, in the Cartesian coordinate system, the intersection point of the two upward-opening parabolas (1) and (2) is below the x-axis. Therefore, each of these two parabolas intersects the x-axis at two points, and between the two intersection points of one parabola with the x-axis, there must be an intersection point of the other parabola with the x-axis. Therefore, the quadratic trinomials $f(x)$ and $g(x)$ both have two real roots, and between the two roots of each polynomial, there is a root of the other polynomial.