What was the number of balls; and what the number of layers in each pile when complete? Let the number of balls in a side of the lowest layer; = Now since there were only 4 balls in one side of the second pile, there can only be four layers, which will contain 4x, 3. (x − 1), 2. (x − 2), and x-3 balls respectively; of the problem. Hence there were 6 layers in the first pile, and they contained 1, 3, 6, 10, 15, 21 balls, respectively; .. the whole number of balls in the first pile was 56, and in the second 50. ах (31.) In the preceding solutions it may be observed, that, in many instances, values of the unknown quantities are deduced, which do not agree with the conditions of the problems. This is always the case when the roots of the equations are negative; and the circumstance arises from that peculiar quality of an algebraic expression, by which it is denominated either positive or negative. The product of two or any even number of such quantities, whether all of them are positive or all negative, will only be affected with a positive sign: thus the quantity xy will represent the product of +x+y, or of xxy; and a2, of +ax +a, or of a; consequently, in the reduction of such quantities to their constituent factors by the rules of division or evolution, these factors may be considered either as all positive or all negative. But in common language, in which the conditions of a problem are expressed, quantity or number is from its very nature what in Algebra is meant by the term positive, i. e. it increases any homogeneous quantity to which it is added, and diminishes any one from which it is subtracted. Hence it may be understood, why, when quadratic equations are formed to express the conditions of a problem, the resulting roots may exceed in number what appear to be required as answers to the problem, and why such as are negative cannot be applied to its conditions. These roots or values, however, though inapplicable in their present shape, will, if assumed as positive, become correct answers to the problem under a different modification of the conditions. In the equations thence deduced, these former negative values will appear as positive roots, and the former positive values as negative roots. Thus, if Prob. 12, page 213, be transformed into the following, "A "detachment from an army was marching in regular column "with 5 fewer in depth than in front; but upon the enemy coming in sight the front was increased till it became 845 the original front; and by this movement the "detachment was drawn up in 5 lines. Required the num "ber of men;" from the solution of this problem the number is found to be 3900, answering to the number which would be found from using the negative value of x in the original problem; and the equation for determining this (x2 5 = 4225 5x) differs from the other only in the sign of x. In Prob. 19, page 217, x is found to be equal to ± 4, where the negative value shews that if the trading vessel had turned out of its first course in a direction contrary to CE, or on the opposite side of the line AC, it would have been taken after sailing 4 miles in that direction. In Prob. 1, page 227, the negative value of x is found to be 130. But if the problem be modified so as to be"A merchant sold a quantity of brandy, by which he "lost £.39 more than the prime cost, and found that his loss was as much cent. as the brandy cost him. What was come per "that price?" the equation for determining the price, is 2 x2 100 = S9+, which is deduced from the equation to the original problem by changing the sign of x, the positive value of which is in this case 130. Also, in Prob. 4, page 228, the negative value of x is 15 2 . Now if the problem were "Bought two sorts of linen, for the finer of which I gave 6 crowns more than for "the other. An ell of the finer cost as many shillings as "there were ells of the finer. Also 28 ells of the coarser 66 66 66 (which was the whole quantity) sold at such a price, that 8 ells cost as many shillings as one ell of the finer. How many ells were there of the finer; and what was the value of each piece?" an equation arises differing from the equation to the original problem only in the sign of x, and whose 15 positive root is -; whence there were 7 ells of the finer at 7s. 6d. per ell, the whole price of which was therefore £.2. 16s. 3d., £.1. 6s. 3d. and the whole price of the coarser was And in the very same manner, all the other problems may be transformed. (32.) The same reasoning will apply to the case in which all the roots of the resulting equation are negative. None of its values can in this case be applied to satisfy the conditions of the Problem; but if the conditions are properly modified, equations may be deduced, of which these values rendered positive will become roots, and will satisfy such conditions. The same observation holds, if the resulting values be the square roots of negative quantities, with this exception, that such roots can never be applied to satisfy the conditions of the problem under any modification whatever. |