1+x+4 ; 1+ x be of such a magnitude that each of these successive quantities under the root differs from 1+ less than the preceding one, the continued series + of terms 1+ -&c. will be an approximation to the true root of 2 8 16 1+x; that is, the more terms of the series are taken, the less will their sum differ from the square root of 1+x. And, since it is evident that no expression whatsoever, simple or compound, multiplied into itself, can produce 1+x, therefore an approximation to the root is all that can be required. But if a be not of such a magnitude as is here supposed, then it is clear that the operation performed upon 1+ leads to a result which is not even an approximation to its square root, because as more terms of that result are taken and squared we obtain a quantity which recedes farther and farther from 1+x. Its proper interpretation, in this stage of the subject, cannot well be given, and must be deferred. COR. If the approximate square root of any binomial, as a+b, be reand therefore √a+b=√ax√1+ (Art. 149); quired; since a+b=a(1+· b we have by the last Example, putting in the place of x, a b 152. It appears from Ex. 2, that a trinomial, a2−ax+ in which four times the product of the first and last terms is equal to the square of the middle term, is a complete square. The same relation is found to subsist between the parts of all complete squares of three terms arranged according to the powers of some one Thus 4x2+4cx+c3 is a complete square, viz. (2x+c)3, because 4×4x3×c3 =16c2x2=(4cx)2. Similarly x2-px+2* is a complete square, viz. 4xx2xl"=p'x'=(−px)'. because 153. The method of extracting the cube root is discovered in the same manner as that for the square root. The cube root of a3+3a2b+3ab2+ b3 is a + b (Arts. 140, 143); and to obtain a+b from this compound quantity, arrange the terms as before, and the cube root of the first term, a3, is a, the first term in the root; subtract its cube from the whole quantity, and divide the first term of the remainder by sa2, a3+3a2b+3ab2+b3 (a+b a3 3a2) 3a2b+3ab2+b3 the result is b, the second term in the root; then subtract 3a2b+3ab2+b3 from the remainder, and the whole cube of a + b has been subtracted. If any quantity be left, proceed with a + b as a new a, and divide the last remainder by 3(a+b) for a third term in the root; and thus any number of terms may be obtained*. Ex. To extract the cube root of 8x3+6xy3-12x3y—y3. 154. The rules above laid down for the extraction of the roots of compound quantities are but little used in algebraical operations; but it was necessary to give them at full length, for the purpose of investigating rules for the extraction of the square and cube roots in numbers. The square root of 100 is 10, of 10000 is 100, of 1000000 is 1000, &c. from which consideration it follows, that the square root of a number less than 100 must consist of only one figure, of a number between 100 and 10000 of two places of figures, of any * That the rule may be thus extended will be obvious from comparing the form of the cubes of a+b+c, a+b+c+d, &c., with that of a+b from which the Rule was deduced; For (a+b+c)3=(a+b)3+3(a+b)2c+3(a+b) c3 +c3, =a3+(3a2+3ab+b2) b + {3 (a+b)2 + 3(a+b)c+c2} c. Similarly (a+b+c+d)3=a3 +(3a2+3ab+b2)b+{3(a+b)2+3(a+b) c+c2} c and so on.-ED. +3(a+b+c)+3(a+b+c)d+d}d; number from 10000 to 1000000, of three places of figures, &c. If then a point be made over every second figure in any number, beginning with the units, the number of points will shew the number of figures, or places in the square root. Thus the square root of 4357 consists of two figures, the square root of 56478, of three figures, &c. NOTE 2. Ex. 1. Let the square root of 4856 be required. Having pointed it according to the direction, it appears that the root consists of two places of figures; let a + b be the root, where a is the value of the figure in the tens' place, and b of that 4356 (60+ 6 or 66 3600 120+6) 756 or 126] 756 in the units'; then is a the nearest square root of 4300, which does not exceed the true root; this appears to be 60; subtract the square of 60, (a), from the given number, and the remainder is 756; divide this remainder by 120, (2a), and the quotient is 6, (the value of b), and the subtrahend, or quantity to be taken from the last remainder 756, is 126×6, (2a+b)b, or 756. Hence 66 is the root required. It is said that a must be the greatest number whose square does not exceed 4300†: it evidently cannot be a greater number+ than this; and if possible let it be some quantity, a, less than this; then since a is in the tens' place and b in the units', a+b is less than a; therefore the square of a+b, whatever be the value of b, must be less than a2, and consequently a+b less than the true root. If the root consist of three places of figures, let a represent the hundreds, and b the tens; then having obtained a and b as before, let the new value of a be the hundreds and tens together, and find a new value of b for the units: and thus the process may be continued when there are more places of figures in the root. * It will be clearer to read "a the greatest multiple of 10 whose square does not exceed 4300." + Or, the greatest multiple of 10 whose square does not exceed 4300. ↑ Multiple. 155. The ciphers being omitted for the sake of expedition, the following rule is obtained from the foregoing process. 4356 66 36 126) 756 Point every second figure beginning with the units' place, dividing by this process the whole number into several periods; find the greatest number whose square is contained in the first period, this is the first figure in the root; subtract its square from the first period, and to the remainder bring down the next period; divide this quantity, omitting the last figure, by twice the part of the root already obtained, and annex the result to the root and also to the divisor; then multiply the divisor, as it now stands, by the part of the root last obtained, for the subtrahend. If there be more periods to be brought down, the operation must be repeated. Ex. 2. Let the square root of 611525 be required. 611525 (782 49 148)1215 1184 1562)3125 3124 1 remainder. The remainder in this example shews that we have not obtained the number which is the exact square root of the proposed quantity; but 782 is a near approximation to the square root, being in fact the square root of 611524; and 783 is too great, being the square root of 613089. 156. In extracting the square root of a decimal, the pointing must be made the contrary way, beginning with the second place of decimals, and the integral part must be pointed as before, beginning with the units' place: or, if the rule be applied as in whole numbers, care must be taken to have an even number of decimal places, by annexing ciphers to the right (Art. 41); because, if the root have 1, 2, 3, 4, &c. decimal places, the square must have 2, 4, 6, 8, &c. places (Art. 46). Ex. 3. To extract the square root of 64·853. 64 1605) 8530 8025 16103) 50500 2191 The remainder in this example appears to be great; but if the decimal point were retained throughout the operation, it would easily be seen that its real value is very small, and that it becomes smaller for every figure that is added to the root. For every pair of ciphers which we suppose annexed to the decimal another figure is obtained in the root. And in this and similar cases, when ciphers are added, the root can never terminate, because no figure multiplied by itself can produce a cipher in the units' place. of 157. The cube root of 1000 is 10, of 1000000 is 100, &c. therefore the cube root of a number less than 1000 consists of one figure, any number between 1000 and 1000000, of two places of figures, &c. If then a point be made over every third figure contained in any number, beginning with the units, the number of points will shew the number of places in its cube root. Ex. 1. NOTE 2. Let the cube root of 405224 be required. a3 = 343000 3a2 = 14700) 62224 the first remainder. 58800=3a2b 3360 = 3ab2 64 = b3 62224 subtrahend. By pointing the number according to the direction, it appears that the root consists of two places; let a be the value of the figure |