2. Required the cube root of 11 + 5 √7 3 Required the cube root of 2 7 + 3 v3. 4. Required the fifth root of 29 √ 6 + 41 √ 3. 5. Required the cube root of 45 ± 29 √ 2. 9 V2. Ans. 32, and 3 6. Required the cube root of 9 ± 4v 5, or 9± Ans. +5, and 5. 3. 2 7. Required the cube root of 20±68 v V 80. 1 - -7. 6. 8. It is required to find the cube root of 35 ± 69 V- - 6. 9. It is required to find the cube root of 81 ± √ — 2700.* − 3 + 2 √ − 3, and Ans. CASE XIII. 3-2 √3. To find such a multiplier, or multipliers, as will make any binomial surd rational. RULE.t-1. When one or both of the terms are any even roots, multiply the given binomial or residual, by the same * Whenever it can be done, the operation, in cases of this kind, ought to be abridged, by dividing the given binomial by the greatest cube that it contains, and then finding the root of the quotient; which, being multiplied by the root of the cube by which the binomial was divided, will give the root required. Thus, in the example above given, 81+v-2700 ⇒27 × (3 + √ — 10), where the root of 3+1 being now more easily found to be-1+2v-, or -1+31-3, we shall have, by multiplying by 3, (which is the cube root of 27), −3+2√−3, as above. Also this is useful, in Cardan's rule for cubic equations; thus, (81+v (−2700))+3⁄4/(81—√(—2700))——3 × 2 ——6, or —=— -X 2=-3, or X2=9, the imaginary parts vanishing, by the contrariety of their sigus. See De Moivre's Appendix to Sanderson's Algebra, Universal Arithmetic, or Maclaurin's Algebra. + If a multiplier be required, that shall render any binomial surd, whether it consist of even or odd roots, rational, it may be found by substituting the given numbers, or letters, of which it is composed, in the places of their equals, in the following general formula: Binomial Va±√b. Multiplier an−1 = W/ an-2b+ Wan-3b2 = W/ an—4b3+, &c.; where the upper sign of the multiplier must be taken with the upper expression, with the sign of one of its terms changed; and repeat the operation in the same way, as long as there are surds, when the last result will be rational. 2. When the terms of the binomial surd are odd roots, the rule becomes more complicated; but for the sum or difference of two cube roots, which is one of the most useful cases, the multiplier will be a trinomial surd, consisting of the squares of the two given terms and their product, with its sign changed. sign of the binomial, and the lower with the lower; and the series continued to n terms. This multiplier may be derived from observing the quotient which arises from the actual division of the numerator by the denominator of the following fractions: thus, n ·=xn−1 + xn−2y+xn−3y2+, &c., +yn-1 to n terms, whether n be even or odd. II. xn-yn =xn−1. -xn−2y+xn−3y2—, &c., -yn-1 to 8 terms, where n is an even number. III. 2n + y2 = terms, when n is an odd number. Now, let xn = a, yn = b; then x = a, y="V√b, and these, fractions severally become a -b a-b and " n n-1 N--2 n n-2 = a+b &c., also y2 = b2, And, since −1 = y3b3, &c., therefore, a-b ton terms; where n may be any whole number whatever. And, a+b Na+ Vb · an-1 — njan-2b+nj an−3b2 —, &c., ton terms; where the terms b and an odd number; and the sign b-1, have the sign+, when n is when n is an even number. Now, since the divisor multiplied by the quotient gives the dividend, it appears from the foregoing operations, that, if a binomial surd of the form na - n nhbe multiplied by nan-ib nan-2b+, &c., ...+n/bn−1, (n being any whole number whatever), the product will be a— tional quantity; and if a binomial surd of the form tiplied by nan—1 — n/an-2b+ nan―2b2 ——, &c., a ra a+b be mul bn-1, the product will be ab, or a-b; according as the index n is an odd or an even number. See my Elementary Treatise on Algebra, Theoretical and Practical.-ED. EXAMPLES. 1. To find a multiplier that shall render 5 + √3 rational. Given surd 5+ √3 Multiplier 5√3 5 2. To find a multiplier that shall make √ 5 + √ 3 rational Given surd 5+ √3 3. To find multipliers that shall make 5+ 1✓ 3 rational. 5. To find a multiplier that shall make √5 — √x rational. Ans. 5+ √x. 6. To find a multiplier that shall make ✔a + √b rational. Ans. a b. 7. To find a multiplier that shall make a + √b rational. Ans. ab. 8. It is required to find a multiplier that shall make 1-32a rational. Ans. 1+2a +3⁄4/4a3. 9. It is required to find a multiplier that shall make 32 rational. Ans. 9+ 6+1 V4. 10. It is required to find a multiplier that shall make 3. 4 3 (a) + (b3), or a + +brational. Ans. (a®b3) + √ (a3b®) — √√ 6o. CASE XIV. To reduce a fraction, whose denominator is either a simple or a compound surd, to another that shall have a rational denominator. RULE.-1.-When any simple fraction is of the form b να' multiply each of its terms by va, and the resulting fraction 2. If it be a compound surd, find sucn a multiplier, by the last rule, as will make the denominator rational; and multiply both the numerator and denominator by it, and the result will be the fraction required. 7. Reduce 10 √(42) (18) Ans. 4 to a fraction that shall have a rational 3x x√x Ans. 9 -X to a fraction, the denominator of which a2 + b - 2a v b 3/7 - 3/5 Ans. a2 b to a fraction that shall have a ra tional denominator. Ans. 5 × [√/ (49) + 3/ (35) +3✓ (25)]. to a fraction that shall have a ra 8. Reduce 3/3 tional denominator. Ans. Ans. 4 { − √ 10 − 2 √2 + (2 + √ 5) × √ 5 } . OF ARITHMETICAL PROPORTION AND PROGRESSION. ARITHMETICAL PRORORTION, is the relation which two quantities of the same kind, have to two others, when the difference of the first pair is equal to that of the second. Hence, three quantities are said to be in arithmetical proportion, when the difference of the first and second is equal to the difference of the second and third. Thus, 2, 4, 6, and a, a + b, a + 2b, are quantities in arithmetical proportion. And four quantities are said to be in arithmetical proportion, when the difference of the first and second is equal to the difference of the third and fourth. |