96. Second method of composing 3a2+3ab+b2, the true divisor. In this method we write a under a in the first column, and the sum 2a being multiplied by a, gives 2a to place under a2 in the second column, and the sum of 2a2 and a2 is 3a2 for the trial divisor. Again, under 2a in the first column write a, and the sum of 2a and a gives 3a. Now, having found b by the trial divisor, annex it to 3a in the first column, making 3a+b, which, multiplied by b, and the product placed in the second column, gives, by addition, the true divisor, 3a2+3ab+b2, as before. We shall exhibit the operation of extracting the cube root by both these methods. EXAMPLES. (1) What is the cube root of x—9x+39x1—99x3+156x2-144x+64? By the first method. (2) What is the cube root of x+6x5—40x3+96x-64 ? 94 (3) What is the cube root of a3+3a2b+3ab2+b3+3a2c+6abc+3b3c+3ac2 +3bc2+c3? Ans. a+b+c. (4) Extract the cube root of 26-6x3+15x1—20x3+15x2-6x+1. Ans. x-2x+1. 97. The same process is employed in the extraction of the cube root of numbers, as in the subsequent examples. 98. The local values of the figures in the root determine the arrangement of the figures in the several columns, as is exemplified by working the last example as on next page; by omitting the terminal ciphers, the arrangement is precisely the same as in the preceding example. 99. Extraction of the fourth root of whole numbers. The investigation of a method for extracting the fourth root of any number is similar to that employed for the cube root. Thus, since (a+b)*=a1+4a3b+6a2b3+4ab3+b1, we may conceive a to denote the number of tens, and b the number of units in the root of the number expressed by a1+4a3b+6a2b2+4ab3+b. Then Vaa, the figure in the tens' place, and the remainder, when a1 is removed, is 4a3b+6a2b2+4ab3+b1=(4a3+6a2b+4ab2+b3)b. The method of composing the divisor 4a3+6a2b+4ab2+b3, for the determination of b, the figure in the units' place, may be illustrated as follows: axa = a2 a'+4a3b+6a2b2+4ab3+b2 (a+b (4a+b)b-4ab+b2 (6a2+4ab+b2)b=6a2b+4ab2+b3 (4a3+6a2b+4ab2+b3)b=4a3b+6a2b2+4ab3+b*. 100. From this mode of composing the complete divisor we easily derive the following process for the extraction of the fourth root of any number. In the same manner, the student may readily investigate rules for the extraction of the higher roots of numbers, simply observing to use an additional column for each successive root. 101. To represent a rational quantity as a surd. 1 Let it be required to represent a in the form of a surd of the nth order; then, by (Art. 63), the form will be a", or ; for by raising a to the nth power, and then extracting the nth root of the nth power of a, we must evidently revert to the proposed quantity, a. Hence we have a= √ a2 = √ a3 = √√/ a1 = Wa15 = "/aTM= √√ a" 102. When the given quantity is the product of a rational quantity and a surd, we must represent the rational quantity in the form of the given surd, and then express the product with a single radical sign, or fractional index. Thus, we have a√b = √a°× √b= √ a2b 3a √5b= √3a × 3a × √5b = √9ax 5b = √45a2b a√xy=Vaxaxax √xу= √ a3× √xy=√a3ry αχα 12√7 =√144 × √7 = = √144×7 = = √1008 a(1—a—2x2)1— (a2)1 (1—a—2x2)1 = (aa—aox2)1 = √/a°—x2. EXAMPLES. (1) Represent a in the form of a surd, whose index is 5. (6) Represent as a surd the mixed quantity (x+4), (1) Valo or (@10)}. (2) √7-4√3. (3) √396. ANSWERS. x x+ (4) √a3—a2b or (a3—a2b)}. (6) √x+4 or (x+4)2. 103. To find multipliers which will render binomial surds rational. The product of two irrational quantities is, in many instances, a rational quantity, and, therefore, an irrational quantity may frequently be found, which, employed as a factor to multiply some other given irrational quantity, will produce a rational result; thus, Again, since the product of the sum and difference of two quantities is equal to the difference of their squares, we have, evidently, Hence it is obvious that, in these and similar equalities, if one of the factors be given, the other factor or multiplier is readily known, and the proposed irrational quantity is thus rendered rational. By a double operation of this kind, multiplying (√n+ √p+√9) by (√n+√p−√9), we have (√ñ +√p)2-q, or n+p-q+2√np; and multiplying this by n+p-q-2√np, the given expression, √n+ √p+ √q, is rationalized. In the same manner, since (x+y)(x2xy+y3)=x3±y3 .. (Vx±Vy)(Vx2=Vry+Vy3)=x+y, and the expression VrVy may, therefore, be rationalized by multiplying it by Vrry+Vy2; and Vray+Vy, multiplied by Vry, will produce a rational result. Again, by division [see Art. 23 (5), (6), (7)], Put .... 2"+y" x+y x=a; then x=Va; xa¬1=Va11; x"→2= √/ aa2, &c.; y"=b; then y=Vb; y2 = Vb2; y3=\/b3, &c.; hence, by substitution in the three preceding equalities, we have a-b Va−√b=Va¬1+ Vaa3b+ √ aa¬3b2+ √aa¬b3+ ... ... ... Now, the dividend being the product of the divisor and quotient, it is obvious that a binomial surd of the form a-b will be rendered rational by multiplying it by n terms of the second side of equation (1), and a binomial surd of the form Va+Vb will be rationalized by employing n terms of the second side of equality (2) or (3), according as n is even or odd, the product in the former case being a-b, and in the latter a―b or a+b. G |