3. To extract the cube root of 99 + 70 V/2. 2 Here A 99, B = 70 12, and VA-B2 = 1; and since n = 3, (a') becomes a3 + 3x(x2 - 1) = 4x3 — 3x = 99, which gives 3; consequently, from (b) ya C=9-1= x = 8, or y = √8=212, and of course 99 + 701/2 = x + y = 3+212, as required. 8 4. To extract the cube root of 10-91-3. 2 Here A = 10, B 9-3, and VA — B2 = = - = = = 100-(9-3)2= √100 + 243 = √343 7 C; consequently, since n = 3, (a) becomes a3 + 3x(x2 - 7) = 4x23 — 21x10, which gives x 2, and y = x2-C=4−7= - 3 gives y=-3. Hence, -10 -9 √3 =x—y= 2-3 is the required answer. 5. Extract the fourth root of 1561 ± 696 √5. 6. Extract the fourth root of — 7 ± 24 √ — 1. Ans. 3±215. Ans. 2-1. Ans. 5±√3. 7. Extract the cube root of 80 ± 72 √ — 3. - CASE IX. If we have an expression consisting of any number of terms, then if any number of its terms are surds, it is proposed to find the form of another expression (called the multiplier), such that the product of the given expression and the multiplier shall be rational, or free from surds. RULE. 1. When the given expression is a monomial surd, which has 1 or any rational number for a coefficient. Raise the surd to such a power as is denoted by the index (of the surd), and the surd sign will be removed; then divide the power by the surd, and the quotient will clearly be the multiplier. Thus, if m va represents the given surd having m for its = Van-1 = the multiplier, and we shall have a x Va-1= Vana, which is rational. 2. When the given expression consists of more than one term, being a binomial or a polynomial. If the terms under any of the surds are numbers, represent them by letters, using different letters for different numbers, and suppose the expression is represented by P. Putting P equal to 0, we shall get the equation P = 0, (1); then, by the common method of freeing the terms of an equation from any surds which may affect them, free the terms of (1) from surds, and then bring all the terms of the resulting equation into the first member of the equation, and let the equation thus obtained be denoted by Q = 0, (2). Since (1) and (2) exist together, it is clear that Q must be exactly divisible by P; consequently, if we divide Q by P, it is clear that the quotient will be the required multiplier, such that P, being multiplied by it, the product will be Q, a rational product. When letters are used for numbers in P, restore the values of the letters in P, Q, and the multiplier; then they will be expressed as they ought to be in the final result. EXAMPLES. 1. To find multipliers which shall free 3 Va and 4 va from surds. Ans. Va and Va. 2. To find a multiplier which shall free 3 Va - Vb from surds. Assuming the equation 3 va√b=0, or 3√ a = then, by squaring, we have 9ab, or 9a-b=0; conse quently, we have 9a-b and we shall have (3 Va√) × (3 √a + √b) = 9a - b, which is rational. 3. To find multipliers which will make 5 – 2√3, 4+5√3, 217-3 15, and 4 √2 + √3, rational. Ans. 5+ 2√3, 4 − 5 √3, 2 √7 + 3 √5, and 4 √2 — √3. 4. To find a multiplier which will free Wa+ √o + √c from surds. = Assuming the equation va + √ʊ + √c = 0, or √ a + võ Vc, we get a + b + 2 √ ab = c, or 2 √ ab = c a — b, which gives 4ab = (a + b − c)3, or (a + b — c)2 — 4ab = 0; consequently, we have (a+b-c)2-4ab = = (a+b―c+2 √ ab) x (a+b-c-2 Vab × (√a + √b − √e) × (a + b − c − 2 √ ab) = the required multiplier. 5. To find a multiplier which will free 15+ √3 — √2 from surds. Ans. (√5 + √3 + √2) × (5 + 3 − 2 − 2 √15) = (√5 + √3 + √2) × (6 −2/15.) 6. To find a multiplier that shall free Va - from surds. 7. To find the multiplier which will free 4-3 from surds. Ans. V16 + V12 + √9 = 2√2 + V12 + √9. 8. To find a multiplier that shall free 32-23= √54 24 from surds. Ans. 1542 + V54 × 24 + √242. 9. To free a + b from surds. Here we have a + b ab or a+b=0; consequently, √ a + √ b = √ a2 - √ ab + = the multiplier, which will free the given expression from surds. 10. Find the multiplier which will free 13+ V2 from surds. Ans. V9 — V6+ √4. 11. Find the multiplier which will free 23+ 3√2 = 124 + 54 from surds. 12. To find a multiplier that will make a + rational. Here we have a + b = 0, or Va= == α -b Vb, which gives a=b, or a − b = 0; consequently, a + b + Vab2 - b3 the multiplier. = Remark.—It may be shown in like manner that Va+ Va2b + Vab2 + VÃ3 is the multiplier that will free Vaib from surds. 13. To find the multiplier which will free 13+2 and 4-3 from surds. × x Ans. 133-132 x 2 + √3 × 22-23 and 143 + 42 × 3 + √4 × 32 + √3. CASE X. To reduce a fraction which contains surds to a fraction whose numerator or denominator shall be free from surds. RULE. Multiply the numerator and denominator by a multiplier which will free either the numerator or denominator (as required) from surds. IMPOSSIBLE OR IMAGINARY ROOTS. 1st. Because all even powers of positive or negative numbers or quantities are positive, it follows that a negative number or quantity can not be an even power of a real number or quantity; consequently, when we consider and treat a negative number or quantity as being an even power, its root must be impossible in real numbers or quantities, or the root must be imaginary. Thus, √—1, 4√— 2, 6√ — 3, V—5, √—a, √—b, — 3 √ — 2, — 5 √ — 6, mi — 6 are all |