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 below; and by omitting the terminal ciphers, the arrangement 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)'= a+4a3b+6a2 b2+4a b3+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+4a b3+b1. Then Vaa, figure in the tens' place, and the remainder, when a1 is removed, is 4a3b+6a2b2+4a b3+b1=(4a3+6a2b+4ab2+b3) b. The method of composing the divisor 4a3+6a2b+4a b2+b3, for the determination of b, the figure in the units' place, may be illustrated as follows: 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. Example. What is the fourth root of 1185921? 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 à rational quantity as a surd. 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 (a"); 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 102. When the given quantity is the product of a rational quantity and surd, we must represent the rational quantity in the form of the given surd and then express the product by means of the radical sign, or fractional index Thus we have ab = √a3× √b = √√a2b 3a√5b = √3a×3a × √√5b = √9a2×5b = √√45a2b EXAMPLES. (1.) Represent a2 in the form of a surd, whose index is . (6.) Represent as a surd the mixed quantity (x+4)√√4 (1.) a1o or (a10). (2.) √/7-4√3. (3.) /396. ANSWERS. (4.) √a3 ab or (a2—aab)1. (5.) √x2—y2 or (x2—y3)‡. (6.) √x+4 or (x+4)*. 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; and since the product of the sum and difference of two quantities is equal to the difference of their squares, we have, evidently, √ax √a = a; (√a−√b) (√a+√b) = a−b VIX √r2 = x; (x + √y) (x − y) = x2-y —y2. yxy=y; (√x − y) (√x + y) = x-y2. Hence it is obvious that, in these and similar equations, if one of the factors be given, the other factor or multiplier is readily known, and the proposed irrational quantity is thus rendered rational. In the same manner, since (x+y) (x2xy+y2)=x+y3 .. (Vx+Vy) (Vr°F√xy+Vy°)=x+y, and the expression +Vy may therefore be rationalized by multiplying it by Vr2 Vay + Vy, and √ √xy+Vy, multiplied by Vy, will * produce a rational result. .... .... Puta; then =√√a; r^~='√a"; ☛"——¦√aTM; &c. y^=b; then y=\/\/b; y2 =\/b2; ya='√/b3; &c.; hence, by substitution in the three preceding equations, we have Now the dividend being the product of the divisor and quotient, it is obvious that a binomial surd of the form Va-b will be rendered rational by multiplying it by n terms of the second side of equation (1), and a binomial sard of the form a+b will be rationalized by employing n terms of the second side of equation (2) or (3), according as ʼn is even or odd, the product in the former case being a-b, and in the latter a-b or a+b. Note. When n is an even number employ equation (2), and when it is an odd Lumber use equation (3), in order to rationalize Va+V√b. EXAMPLES. (1.) Find a multiplier to rationalize IT-7. Employing equation (1), we have a=11, b=7, and n=3; hence required maitipier = √/112+ VII.7+√7=V/121+√/77+/49. For 121 +/77 +1/19 VII-V 7 1331+/847+√/539 (2.) Rationalize the binomial surd /5+/4. Here we have a=5, b=4, n=3, an odd number; hence by equation (3) we have multiplier required=25-20+/16; for by multiplication (√5+V/4) (√25—20+/16)=5+4=9= a rational number. (3.) What multiplier will render the denominator of the fraction 104. To extract the square root of a binomial surd. Before commencing the investigation of the formula for the extraction of the square root of a binomial surd, it will be necessary to premise two or three lemmas. Lemma 1. The square root of a quantity cannot be partly rational and partly irrational. For if a=b+c, then by squaring we have -c a=b2+c+2b√/c: therefore c-a-b2- ; 26 that is, an irrational equal to a rational quantity, which is absurd. Lemma 2. If a±√b=x±√y be an equation consisting of rational and irrational quantities, then a=x, and √b=√y. For if a be not equal to æ, let a—x=d, then we have ±√yF√b=a-x; but a-x=d; therefore ±√y√b=d, which is impossible, .. a=x, and ✔✅b=√y. Lemma 3. If √a+√b=x+y; then √a—√b=x-y; where x and y are supposed to be one or both irrational quantities. For since a+/b=x2+y2+2xy; and since 2 and y2 are both rational, 2a y must be irrational, otherwise √/b=x2+y2+2xy--a, a rational quantity which is impossible by Lemma 1; hence by Lemma 2, we have a=x2+y2; √b=2x y |
