The speeds of trains A ($S_a$) and B ($S_b$), in km/h with $S_a S_b$, are roots of $x^2 - 130x + 4200 = 0$. Let $T_a$ and $T_b$ be their usual times to cover distance $D$. If the speeds are swapped, train A takes $(T_a + 2)$ hours and train B takes $(T_b - 2)$ hours. Find the sum of the

Question

The speeds of trains A ($S_a$) and B ($S_b$), in km/h with $S_a > S_b$, are roots of $x^2 - 130x + 4200 = 0$. Let $T_a$ and $T_b$ be their usual times to cover distance $D$. If the speeds are swapped, train A takes $(T_a + 2)$ hours and train B takes $(T_b - 2)$ hours. Find the sum of the original times $T_a + T_b$.

Accepted Answer

26. The roots of the quadratic are the two speeds, whose sum is 130 and product is 4200. Using the swapped-speed conditions, we get two equations in $D$, $T_a$, and $T_b$; solving them gives $T_a + T_b = 26$. The key is to use both the root properties and the time-distance relation together.

Solution

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