volve the first power of the unknown quantity, 3o. terms consisting entirely of known quantities. An equation, which contains all three of these different kinds of terms is called a complete equation of the second degree. If the second of these different kinds of terms be wanting, the equation is then called an incomplete equation of the second degree. A complete equation of the second degree is sometimes called an affected equation, and an incomplete equation is sometimes called a pure equation of the second degree. 112. We are now prepared for the solution of incomplete equations of the second degree. Let there be proposed, for example, the equation Freeing from denominators, we have 12 x2 - 116 = x2 + 2040 Equations of the second degree, it should be observed, admit of two values for the unknown quantity, while those of the first degree admit of but one only. This arises from the circumstance, that the second power of a quantity will be positive, whether the quantity itself be positive or negative. Thus we have x in the preceding example equal + 14 or 14, or, uniting both values in one expression, we have x=14. Let us take, as a second example, the equation x2 whence x= 252 In this example 252 is not a perfect square; we can there fore obtain only an approximate value for x. Let us take, as a third example, the equation x2+25=9 Deducing the value of x from this equation, we have x = √16 To find the value of x, we are here required to extract the square root of 16. But this is impossible; for, as there is no quantity positive or negative, which multiplied by itself will produce a negative quantity, — 16, it is evident, cannot, have a square root either exact or approximate. -16 may indeed be considered as arising from the multiplication of +4 by -4; but 4 and 4 are different quantities; their product therefore is not a square. The result x16 shows then, that it is impossible to resolve the equation, from which it is derived. In general, an expression for the square root of a negative quantity is to be regarded as a symbol of impossibility. 113. Equations of the kind, which we are here considering, may always be reduced to an equation of the form a x2: b, a and b denoting any known quantities whatever, positive or negative. It is evident, that they may be reduced to this state, by collecting into one member the terms, which involve x2 and reducing them to one term, and collecting the known terms into the other member. Resolving the equation a x2b, we have This is a general solution for incomplete equations of the second degree. b If be a perfect square, the value of x may be obtained ex α actly, if not, it may be found with such degree of approxima b tion as we please. If be negative, we shall have V a a symbol of impossibility. From what has been done, we have the following rule for the solution of incomplete equations of the second degree, viz. Collect into one member all the terms, which involve the square of the unknown quantity and the known quantities into the other; free the square of the unknown quantity from the quantities, by which it is multiplied or divided; the value of the unknown quantity will then be obtained by extracting the square root of each member. QUESTIONS PRODUCING INCOMPLETE EQUATIONS OF THE SECOND DEGREE. 1. What two numbers are those, whose difference is to the greater as 2 to 9, and the difference of whose squares is 1258. Let 9 x the greater and 2 x the difference, then, &c. Ans. 18 and 14. 2. It is required to divide the number 18 into two such parts, that the squares of these parts may be in the proportion of 25 to 16. Ans. 10 and 8. 3. It is required to divide the number 14 into two such parts, that the quotient of the greater part divided by the less may be to the quotient of the less divided by the greater as 16 to 9. Ans. 8 and 6. 4. Two persons A and B lay out some money on speculation. A disposes of his bargain for £11 and gains as much per cent. as B lays out; B's gain is £36, and it appears that A gains 4 times as much per cent. as B. Required the capital of each Ans. A's £5, B's £129. 5. A charitable person distributed a certain sum among some 7ɔor men and women, the numbers of whom were in the proportion of 4 to 5. Each man received one third as many shillings as there were persons relieved; and cach woman received twice as many shillings as there were women more than men. Now the men received all together 18s. more than the women. How many were there of each? Ans. 12 men and 15 women. 6. In a court there are two square grass plots; a side of one of which is 10 yards longer than the side of the other; and their areas are as 25 to 9. What are the lengths of the sides? Ans. 25 and 15 yards. 7. A person bought two pieces of linen, which together measured 36 yards. Each of them cost as many shillings a yard as there were yards in the piece; and their whole prices were in the proportion of 4 to 1. What were the lengths of the pieces? Ans. 24 and 12 yards. whose length is to the 8. There is a rectangular field, breadth in the proportion of 6 to 5. A part of this equal to of the whole being planted, there remain for ploughing 625 square yards. What are the dimensions of the field? Ans. The sides are 30 and 25 yards. 9. Two workmen A and B were engaged to work for a certain number of days at different rates. At the end of the time, A, who had played 4 of the days, received 75 shillings, but B who had played 7 of the days, received only 48 shillings. Now had B played 4 days, and A played 7 days, they would have received exactly alike. For how many days were they engaged; how many did each work, and what had each per day? Ans. 19 days; A worked 15, and B 12 days, and A received 5s. and B 4s. a day. 10. Two travellers A and B set out to meet each other, A leaving the town C at the same time that B left D. They travelled the direct road C D, and on meeting, it appeared that A had travelled 18 miles more than B; and that A could have gone B's journey in 15 days, but B would have been 28 days in performing A's journey. What was the distance between C and D ? Ans. 126 miles. 11. A and B carried 100 eggs between them to market and each received the same sum. If A had carried as many as B he would have received 18 peace for them, and if B had taken only as many as A, he would have received only 8 pence. How many had each ? Ans. A 40, B 60. 12. What two numbers are those, whose sum is to the greater as 11 to 7, the difference of their squares being 132? Ans. 8 and 14. 13. A merchant sold for $969 a certain number of pieces of silk, for which he paid four-fifths as many dollars a piece as there were pieces. He gained $1000 by the sale, how many pieces did he sell? COMPLETE EQUATIONS OF THE SECOND DEGREE. 114. Let us take next the equation x2+8x= =209. This is a complete equation of the second degree. The solution of this equation, it is evident, would present no difficulty, if the left hand member were a perfect square. But this is not the case; for, the square of a quantity consisting of one term will consist of one term, and the square of a quantity consisting of two terms will contain three terms. Let us then see if x2+8x can be made a perfect square; for this purpose, it will be recollected, that the three parts which compose the square of a binomial are 1o. the square of the first term of the binomial, 2o. twice the first term multiplied by the second, 3°. the square of the second term. Thus (x + a)2 = x2 + 2 a x + a2 If then we compare x2+8 x with x2+2 a x + a2, it is evident, that x2+8x may be considered the first and second terms in the square of a binomial. The first term of this binomial will evidently be r; then as 8x must contain twice the first term by the second, the second will be found by dividing 8 x by 2 x, which gives 4 for the quotient. x2+8x is therefore the first two terms in the square of the binomial x + 4. If then we add 16, the square of 4, to x2+8x, the left hand member of the proposed, the result 2+8x+16 will be a perfect square. But if 16 be added to the left hand member, it must also be added to the right, in order to preserve the equality; the proposed will then become x2+8x+16=225 Extracting the root of each member of this last, we have whence x+4=15 Let us take as a second example, the equation x= 11, λ== 19. r2 — 3x=15} |