CHAPTER XXIX.
Problems leading to Quadratic Equations.
306. We shall now discuss some problems which give rise to quadratic equations.
Ex. 1. A train travels 300 miles at a uniform rate ; if the rate had been 5 miles an hour more, the journey would have taken two hours less : find the rate of the train.
Suppose the train travels at the rate of x miles per hour, then the time occupied is {300}{x} hours. On the other supposition the time is {300}{x+5} hours ;
{300}{x+5} = {300}{x} - 2 (1). whence x^2 + 5 x - 750 = 0, or (x + 30) (x - 25) = 0, x = 25, or -30.
Hence the train travels 25 miles per hour, the negative value being inadmissible. It will frequently happen that the algebraic statement of the question leads to a result which does not apply to the actual problem we are discussing. But such results can sometimes be explained by a suitable modification of the conditions of the question. In the present case we may explain the negative solution as follows :
Since the values x = 25 and —30 satisfy the equation (1), if we write -x for x, the resulting equation, {300}{-x+5} = {300}{-x} - 2 will be satisfied by the values x = - 25 and 30. Now, by changing signs throughout, equation (2) becomes {300}{x-5} = {300}{x} + 2 and this is the algebraic statement of the following question :
A train travels 300 miles at a uniform rate ; if the rate had been 5 miles an hour less, the journey would have taken two hours more ; find the rate of the train. The rate is 30 miles an hour.
