## 4. From Bordeaux to Saint-Jean-de-Luz

Two cyclists set out simultaneously from Bordeaux to Saint-Jean-de-Luz (the distance between these cities is approximately 195 km). One of the cyclists, whose average speed is 4 km/h faster than the second cyclist, arrives at the destination 1 hour earlier. What is the speed of this cyclist?
4. Let $v$ (km/h) be the desired speed, and $t$ (h) be the time it takes for the cyclist moving at the higher speed $v$ to cover the entire distance.

The speed of the second cyclist, who is moving slower, is $v-4$. The time spent by him on the entire journey is $t+1$. Thus, we have

$$
195=v t=(v-4)(t+1),
$$

from which 1) $t=195 / v ;$ 2) $v-4 t-4=0$.

Expressing time in terms of speed, we get the equation

$$
v-4(195 / v)-4=0 \text {, or } v^{2}-4 v-4(195)=0 \text {. }
$$

Since the discriminant of the last quadratic equation is

$$
\Delta=4+195 \cdot 4=4 \cdot 196=(28)^{2} \text {, then } v=2+28=30 \text {. }
$$

(Clearly, the root of the equation $v=2-28=-26$ does not provide a meaningful solution.) Therefore, the desired speed is $30 \mathrm{km} /$ h.

The time it takes for the first cyclist to cover the entire distance is $195 / 30=6.5$ h. The time required by the second cyclist is $195 / 26=7.5$ h (i.e., exactly 1 h more than the first cyclist), which corresponds to the condition of the problem.

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