Let 125000000 = supposed cube, whose root is, soo; 348003449 : 321006898 :: 500 : 500 [or root nearly 348003449)160503449000(461 = corrected root, 1392013796 2180206940 2088020694 421862460 348003449 73359011 2. Required the cube root of 210358. Here we soon find that the root lies between 20 and 30, and then between 27 and 28. Therefore 27 being taken, its cube is 19683 the assumed cube. Then 19683 210358 2 39366 420716 210358 19683 As 6040168 : 61754.6 : : 27 : 27.6047 27 4322822 1235092 16673742 60301.8) 604018)16673742(27.6047 the root nearly. •• 1208036 , 4593384 36525 36241 284 42 Again for a second operation, the cube of this root is 21035*318645155823, and the process by the latter method is thus : 21035'318645, &Co 2 42070-637290 210358 21035.8 21035'318645, 8c. As 6310643729 diff. .481355 :: 27.6047: the diff. '000210834 27.604910834 = the root required. 3. What is the cube root of 157464? 4. What is the cube root of ? Ans. •; 63, &c. 5. What is the cube root of 117 ? Ans. 489097 Ans. 54 TO EXTRACT THE ROOTS OF POWERS IN GENERAL RULE.* 1. Prepare the given number for extraction, by pointing off from the units place as the root required.directs. 2. Find This rule will be sufficiently obyious from the work in the following example: Extrac 2. Find the first figure of the root by trial, and subtract its power from the given number. 3. To the remainder bring down the first figure in the next period, and call it the dividend. 4. Involve Extract the cube root of a® +6a'--400' +962–64. @ +6a401 +962–64(a? +20-4 When the index of the power, whose root is to be extracted, is a composite number, the following rule will be serviceable : Take any two or more indices, whose product is the given in. dex, and extract out of the given number a root answering to one of these indices ; and then out of this root extract à root answera ing to another of the indices, and so on to the last. Thus, the fourth root = square root of the square root, The following theorems may sometimes be found useful in ex V ab 6 a a tracting the root of a vulgar fraction; V 4. Trivolve the root to the next inferior power to that, which is given, and multiply it by the number denoting the given" power for a divisor. 5. Find how many times the divisor may be had in the dividend, and the quotient will be another figure of the Toot. 6. Involve the whole root to the given power, and subtract it from the given number as before. 7. Bring down the first figure of the next period to the remainder for a new dividend, to which find a new divisor, and so on, till the whole be finished. EXAMPLES 1. What is the cube root of 53157376, 531573764376 3*X3=27)261 dividend. 506535378 53157376 2. What is the biquadrate root of 19987173376? Ans. 376 3. Extract the sursolid, or fifth root, of 307682821106 715625 . Ans. 3145 4. Extract 4. Extract the square cubed, or sixth root, of 43572838 1009267809889764416. Ans. 27534 :5. Find the seyenth root of 3448771746730751318249 2153794673 Ans. 32017. 6. Find the eighth root of. 11219163813204762362464 97942460481, Ans. 13527 TO EXTRACT ANY Root WHATEVER BY APPROXIMATION, RULE. 1. Assume the root nearly, and raise it to the same power with the given number, which call the assumed power. 2. Then, as the sum of the assumed power multiplied by the index more i and the given number multiplied by the index less 1, is to the sum of the given imber multiplied by the index more ! and the assumed power mul. tiplied by the index less i, so is the assumed root to the required root. Or, as half the first sum is to the difference between the given and assumed powers, so is the assumed root to the difference between the true and assumed 'roots ; which difference, added or subtracted, gives the true root nearly. And the operation may be repeated as often as we please, by using always the last found root for the assumed root, and its power as aforesaid for the assumed power. EXAMPLES. 1. Required the fifth root of 210358. Here |