Both values will answer the conditions of the question; for In all the above examples, after the question was put into equation, the first thing done, was to reduce all the terms containing x2 to one tena, and those containing into another, and to place them in one member of the equation, and to collect all the terms consisting entirely of known quantities into the other. This must always be done. Moreover a must have the sign + and its coefficient must be 1. The equation will then be in the following form. x2 + px = q. p and q being any known quantities and either positive or negative. Every equation, however complicated, consisting of terms which contain x2, and x, and known quantities may be reduced to this form. We consider 2 and p r as two terms of the second power of the binomial x + a in which Hence the binomial a + a is equal to x +2, and the third term of the second power is 1. In fact 4 x2+px+ ( x + ? ) ( x + 2) = x2 + p x + P2. 4 Therefore the first member of the above equation may be rendered a complete second power, of which a + is the root, by adding to it. The same quantity must be added to 4 the second member, to preserve the equality. The equation then becomes x*+px + 12=q+22. 4 From the above observations we derive the following general rule for the solution of equations which contain the first and second powers of the unknown quantity. ist. Prepare the equation, by collecting all the terms containing the first and second powers of the unknown quantity into the first member, and all the terms consisting entirely of known quantities into the other member. Unite all the terms containing the second power into one term, and all containing the first power into another. If the sign before the term containing the second power of the unknown quantity be not positive, make it so by changing all the signs of both members. If the coefficient this term is not 1, make it so by dividing all the terms by its coefficient. 2d. Make the first member a complete second power. This is done by adding to both members the second power of half the coefficient of x (or of the first power of the unknown quantity.) 3d. Take the root of both members. The root of the first member will be a binomial, the first term of which will be the unknown quantity, and the second will be half the coefficient of x as found above. The root of the second member must have the double sign ±. 4th. Transpose the term consisting of known quantities from the first to the second member, and the value of x will be found. 4. A and B sold 130 elis of silk (of which 40 ells were A's and 90 B's) for 42 crowns. Now A sold for a crown one third of an ell more than B did. How many ells did each sell for a crown? Let the number of ells B sold for a crown; then x + the number A sold for a crown; the price of 90 ells; To complete the second power of the first member, take one half of , and add its second power to both members. , which is 58 x 841 5 21 841 315 + 841 1156 21) 21)2 7 21)2 21)* 21) Taking the root of both members, Which give and 29 34 =± 21 21 The first value only will answer the conditions. Ans. B sold 3 ells for a crown, and A 33. The learner may observe, that in raising to the second power, I multiplied the numerator into itself, but expressed the power of the denominator by an exponent. This saved some work in this example. It may always be done when the number in the right hand member can be reduced to a fraction with the same denominator as the number added. In this case could be reduced to 21ths. The was reduced thus: When the second member is a whole number, it can be reduced to a fraction with any denominator; consequently this form may be used. 5. A man bought a certain number of sheep for 80 dollars: if he had bought 4 more for the same money, they would have come to him I dollar apiece cheaper. What was the number of sheep? 6. A merchant sold a quantity of brandy for £39 and gained as much per cent. as the brandy cost him. How much did it cost him? 7. Two persons, A and B, talking of their money, says A to B, if I had as many dollars as I have shillings, I should have as much money as you; but if I had as many shillings as their number multiplied by itself, I should have three times as much money as you, and 63 shillings over. How much money had each? 8. A colonel has a battalion of 1200 men, which he would draw up in a solid body of an oblong form, so that each rank may exceed each file by 59 men. What numbers must he place in rank and file? 9. A grazier bought as many sheep as cost him £60; out of which he reserved 15, and sold the remainder for £54, gaining 2 shillings a head by them. How many sheep did he buy, and what was the price of each? 10. A person bought two pieces of cloth of different sorts; of which the finer cost 4s. a yard more than the other. For the finer he paid £18; but for the coarser, which exceeded the finer in length by 2 yards, he paid only £16. How many |