19. a2 + 2 ab + 2 ac + b2 + 2bc + c2. 20. 1+ 2x + 3 x2 + 2 x3 + x1. 21. 14x + 10 x2 + 20 x3 + 25x + 24x + 16x°. 23. et + 2e312 + e22 + 2 e2 + 2 + e-22. 24. 1+ 2 gr + 3 q2μa2 + 4 q3μ»3 + 3 q*p* + 2 q3μ5 + q¤μ·o ̧ 25. 1+ 2+ 3 e2а + 4 e3a + 5 e1d + 4 e5 +3 ebа + 2 e1а + e8d 30 ах · 3 ax + 25a2 + 5 a3 + a1 26. 9x2 38. p3z* — 2 p3z2 + p1z® + 2y ( 1 − 1 ) z 1 — 2 y( p − y)z3 p How can one of these roots be derived from the other? 42. Calculate to seven decimals the values of 10 by means of one of the series in 41. Here 10 = 31/1 107 1 in 41. 0 1 1, and put x = 43. Calculate in a similar manner to five decimal places 1/2. 45. Calculate likewise 1 3, 5, and 6 to four decimal places. 46. Calculate to five decimal places 1/7 and 1/13. THE SQUARE ROOT OF ARITHMETICAL NUMBERS 288. The rule for finding the square root of an algebraic expression makes it possible to derive a rule for finding the-square root of numbers in Arithmetic. The square root of 100 is 10, of 10000 is 100, of 1000000 is 1000, and so on; therefore the square root of a number less than 100 consists of one figure, of a number between 100 and 10000 consists of two figures, of a number between 10000 and 1000000 consists of three figures, and so on. If, therefore, a dot be placed over the figure in units' place of a number equal to or greater than 1, and over every alternate figure, the number of dots will be equal to the number of figures in the root of the number. Thus, the square root of 4096 consists of two figures, the square root of 611524 of three figures, and so on. 289. Find the square root of 4489. Point the number according to the rule; hence the root will consist of two figures. 4489 60+ 7 120 +7 889 889 Let a + b denote the root; then a may be taken as the value of the figure in tens' place, and the figure in units' place. Then a is the greatest multiple of 10 whose square is less than 4400; this is found to be 60. Subtract a2 or 3600 from the given number and the remainder is 889. Divide the remainder by 2 a, that is, by 120, and the quotient is 7, which is the value of b. Hence (2 a+b)b, which is (120+7)7, or 889, is the number to be subtracted. Therefore, since there will be no remainder, the conclusion is that the required root is 67. The ciphers may be omitted for the sake of brevity and the following rule be derived from the process: 4489 67 36 127 889 889 Point off the number into periods of two figures each, beginning with units' place. Find the greatest number whose square is contained in the first period; this will be the first figure of the root; subtract its square from the first period and to the remainder bring down the next period. Drop the right hand figure of the remainder and divide the number so found by twice that part of the root already found. Annex this quotient to the part of the root already found and also to the trial divisor. Then multiply the divisor as it now stands by the figure of the root last found and subtract the product from the last remainder. If there are more periods to be brought down the operation must be repeated. 290. Extract the square roots of 481636 and 11566801. NOTE. The student should note the occurrence of the cipher in the root. 291. If the square root of a number has decimal places, the number itself will have twice as many. Thus, if .23 is the square root of some number, this number will be (.23)* = .0529; and if .113 is the square root of some number, the number will be (.113)2 = .012769. Therefore, there is an even number of places in a decimal which is a perfect square, and the number of decimal places in the root will be half as many as in the given number itself. Hence this rule for extracting the square root of a decimal may be deduced: alternate Place a dot over the figure in units' place, and over every figure, continuing to the left and to the right of it; now proceed as in the extraction of the square root of whole numbers, and mark off as many decimal places in the result as there are periods in the decimal part of the given number. EXAMPLE. Find the square roots of 556.0164 and 0.667489. It follows from the dotting that the root of the first example will have two integral and two decimal places, and that the root of the second example will have no integral but three decimal places. 292. The student will readily see that many integers have, strictly speaking, no square root. Take, for example, the integer 7. It is clear that 7 can not have a square root; for the square of 2 is 4 and of 3 is 9, therefore the square root of 7 lies between 2 and 3, and consequently the square root of 7 can not be an integer. The square root of 7 can not be a fraction, for if any fraction, which is strictly a fraction, be multiplied by itself, its square will be a fraction. 293. If the square root of a number consists of 2n+1 figures, when the first n + 1 of these figures have been found by the usual method, the remaining n may be obtained by division. Let N represent the given number; a the part of the square root already found, that is, the part in the first n+1 figures found by the rule, with n zeros annexed; and x the part of the root which remains to be found. Then Now N-a is the remainder after n+1 figures of the root, represented by a, have been found; and 2 a is the corresponding trial divisor. Equation (1) shows that N-a divided by 2 a gives x, the rest of the square root desired, increased by 2 a |