Suppose two people start at points $A,B$ and move towards each other. The first time they meet is $60$ meters away from $A$ and the second time they meet is $75$ meters away from $B$. What's the distance between $A$ and $B$?
I think I'm missing something very simple here, but I don't know what...
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Here's a easier solution:
Let $u,v$ be the speed of the person starting from points $A,B$ respectively. Let the distance between $A$ and $B$ be $d$.
The idea is at the 1st encounter, the two people together traveled $d$ and it took them time $\frac{d}{u+v}$. We also know this is equal to the time for person 1 to travel 60, $\frac{60}{u}$.
At the 2nd encounter, the two people together traveled $3d$ at time $\frac{3d}{u+v}$. This also equals the time for person 1 to travel $d+75$, $\frac{d+75}{u}$.
We know the time for the 2nd encounter $\frac{3d}{u+v}$ is 3 times the time for the first encounter $\frac{d}{u+v}$. Hence,
$$3*\frac{60}{u} = \frac{d+75}{u}$$
Solving for $d$ gives us $d=105$. Note, no quadratic root finding involved.
Hint: Let $x_A(t)$ be the distance that the person starting at $A$ has walked after time $t$. Let $v_A$ be his speed. Similarly with $B$. They meet at $t=t_1$. Here we have $$x_A=v_At_1=60 \quad \text{and} \quad x_B(t_1)=v_Bt_1=|AB|-60.$$
Try writing down the equations for the second encounter. If it is still causing you trouble, let me know.