Divide the resolvend by four times the cube of the assumed root: to the quotient add three fourths of the assumed root, and the sum will be the true root, or an approximation to it. Let the biquadrate of 20736 be required, as before? 10x10x10x4=4000)20736(5.184 Add 2 of 10=7.5 Approximated root 12-684, to be made the assumed root for the next operation. EXTRACTION OF THE SURSOLID ROOT BY APPROXIMATION. A PARTICULAR RULE.* 1. Divide the resolvend by five times the assumed root, and to the quotient add one twentieth part of the fourth power of the same root. 2. From the square root of this sum subtract one fourth part of the square of the assumed root. 3. To the square root of the remainder add one half of the assumed root, and the sum is the root required, or an approximation to it. Note. This rule will give the root true to five places, at the least, (and generally to eight or nine places) at the first process. Required the sursolid root of 281950621875? 200 assumed root. 5 1000 281950621-875 quotient. Add 200×200×200×200-÷20/=80000000 √361950621 875=19025nearly, Subtract 200X 200-4=10000 A GENERAL RULE FOR EXTRACTING THE ROOTS OF ALL POWERS. 1. Prepare the given number, for extraction, by pointing off from the unit's place, as the required root directs. 2. Find the first figure of the root by trial, or by inspection into the table of powers, and subtract its power from the left hand pe riod. 3. To the remainder bring down the first figure in the next period, and call it the dividend. 4. Involve the root to the next inferiour 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 root. 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 remain der for a new dividend, to which find a new divisor, as before, and, in like manner, proceed till the whole be finished. EXAMPLES. 1st. What is the cube root of 20346417? 03 20346417(273 2x2x2 8 root of the 1st. period, or 1st. Subtrahend. 8 1st. Subtrah.2×2=4(=next inferiour power,) and, 4x3 (the index of the given pow.)= 12 1st. Divisor. 22x3=12)123=-Dividend 27x27x27=19683-28. Subtrahend. 273 = 27X27=729 (next inferior power) and, 19683=2d. Subt, 729×3 (=index of the given pow.)= 2187 2d. Divisor. 272x3=2107)6634=2d.Di:273×275×273=27346417=3d. Subtra. The extracting of roots of very high rowers will, by this rule, be a tedious operation: The following method, when practicable, will be much more convenient. When the index of the power, whose root is to be extracted. is a composite number, take any two or more indices, whose product is equal to the given index, and extract out of the given number a root answering to another of the indices. and so on to the last. Thus, the fourth root=square root of the square root; the sixth root=square root of the cube root; the eighth root square root of the fourth root; the ninth root the cube root of the cube root; the tenth root square root of the fifth root; the twelfth root the cube root of the fourth, &c. The general rule for extracting the roots of all powers, may be illustrated in the same way, as those for the square and cube roots. Any student may at once, sce the truth of the rule, in exhausting the several products of the case illustrating the rule for the cube root. And the same will be evident by raising the number to any higher power, 2d. What is the biquadrate root of 34827998976? Ans. 431·9+. 3d. Extract the sursolid, or fifth root of 281950621875? Ans. 195. 4th. Extract the square cubed, or sixth root of 1178420166015625? Ans. 325. A GENERAL RULE FOR EXTRACTING ROOTS BY APPROXIMATION. 1. Subtract one from the exponent of the root required, and multiply half of the remainder by that exponent, and this product by that power of the assumed root, whose exponent is two less than that of the root required. * The general theorem for the extraction of all roots, by approximation, from whence the rule was taken, and the Theorems deducible from it, as high as the twelfth power. Let G resolvend whose root is to be extracted. tre root required; r being assumed as near the true root, and m-exponent of the power-then the equation will stand thus. By this Theorem the fraction is obtained in numbers to the lowest terms in all the odd powers; and in the even powers only by having the numerator and denominator found by this equation. 2. Divide the given number by the last product; and from the quotient subtract a fraction, whose numerator is obtained by subtracting two from the exponent, and multiplying the remainder by the square of the assumed root; and whose denominator is found by subtracting one from the exponent and multiplying the square of the remainder by the exponent. 3. After this subtraction is made, extract the square root of the remainder. 4. From the exponent subtract two, and place the remainder as a numerator; then subtract one from the exponent, and place the remainder under the numerator for a denominator. 5. Multiply this fraction by the assumed root; add the product to the square root, before found, and the sum will be the root required, or an approximation to it. EXAMPLE. What is the square cubed root of 1178420166015625? 6-1 Exponent of the required root is 6. Then,X6=15. 3004 8100000000 and this multiplied by 15=121500000000. 1178420166015625-121500000000—9698.9314, from this 325.43 And the sum is the approximated root= For the 2d operation, let 325-43 = assumed root. ANOTHER METHOD BY APPROXIMATION.* RULE. 1. Having assumed the root in the usual way, involve it to that power denoted by the exponent less 1. A rational formula for extracting the root of any pure power by approximation. Let the resolvend be called G, and let r+e be the required root, r being assumed in the usual way. 2. Multiply this power by the exponent. 3. Divide the resolvend by this product, and reserve the quo tient. 4. Divide the exponent of the given power, less 1, by the exponent, and multiply the assumed root by the quotient. 5. Add this product to the reserved quotient, and the sum will be the true root, or an approximation. 6. For every succeeding operation, let the root last found, be the assumed root. EXAMPLE. What is the square cubed root of 1178420166015625 ? 1-4580000000000) 1178420166015625(80-824. Add x 300=250 330-824 approximated root. For the next operation, let 330 824 be the assumed root. SURDS. I. SURDS are quantities, whose roots cannot be obtained exactly, but may be approximated to any definite extent by continuing the extraction of the roots. Surds are expressed by fractional indices or exponents, or by the radical sign. Thus, 33, or √3. |