These products differ from the exercises in Section IV. in this circumstance—that the indices of the quantities here are fractional and not integral. The verification of these expressions involves the same operations as in p. 33, Section IV., the only difference being that in these examples the indices are fractional and not integral. These exercises are reduced in the same manner as those in pp. 13, 14 in Section V. 5. 7. 8. x3-2x2-6x-3 −2x(x+1)(x+1)3 x3-2x-1 x2-x-1 (x+1)(x2 - 3x-3)−2x(x+1)(x+1)1 ̄ ̄ ̄x3-3x−3− 2x(x+1)1 ́ = = x − 2x1yś+2x1y3—y ___ (x3 — y‡)(x2+xłyś — xłyś+y1) x-2xy+y XXVI. 9. a2"+a2”—1+1. The verification of these equivalents is effected by exactly the same processes as those in p. 24., Section V., the only difference being in the fractional indices. In Example 13, one side of the equivalent is omitted. It should be 2ab – { (1 − a1)(1 − b4) }1 1+b2+a3(1 − b3) XXVII. 1. See Sect. IV., Art. 3, and notes pp. 2, 3; and in this Section VI., Art. 3, and notes pp. 3, 4. 2. a2+b2=(a+b✓ − 1)(a − b✅✅/ − 1). 3. 2+3/-1_2+3√/-1 4-5-1 23 2-1. a+b√-1_a+b√-1c-dv-1_ac+bd, (bc-ad)-1 Hence the product involves (− 1)", an even root of -1, which is an impossible expression. Hence the product involving an odd root of -1 which is always possible and equal to -1. is +1; hence this expression exhibits the two 10. The third power of -1±√-3 symbolic cube roots of +1. And the third power of expression gives the two symbolic cube roots of -1. And -1+V-3_−1+√ −3 -1- ✔ - 3 2 +1±√-3 2 XXVIII. The first eight examples offer no difficulty beyond the exact performance of the operations indicated. 9. This may be shown in two ways: by extracting the square roots of each quantity and taking the sum of the results, and by squaring each side of the expression, and showing that the result when reduced is 32+68=100, an obvious identity. XXIX. The first five examples require no remark. 6. The expression may be put into the form X y-x√−1 ̧y - 2x√√ −1 2x − y√ −1 (x−y√ −1)(y + x√✓ −1 = -X y+2x-12 10. 7±4/3, 26±15/3, 97±56/3, 362±209/3. 11. 5±2/6, 11√✅2±9√3, 49±20/6, 109√2±89√3. 6, √36. 12. 77+30/6, 655/2+531/3, 11329+4620/6, 98225/2+80187/3. XXXIV. {√2}=(21)=2 or 1/2: {2}={2}-2 or 2/2. The square roots of the last four examples can be found as the example in the note, p. 15. The roots are respectively 3+1/3, √3+1, 3/3 −2√/2, 2+3√✓/−1. √3, XXXV. 1. Here 3/147=3√(49×3)=3x7√3=21√3 -3√75=-3√(25 × 3)=-3×5√3=-15√3 .. 3147-375-3√3 (21-15-1)/3-5/3. 2. 13. 3. 18+20/2-12/6. 5. 1+√3+√5. See XXXIX. 8, p. 36. 7. 24/3. 8. 75/2. 9. /3+2/2. 4. 8√2+7√/3+5√/5+2√30. 10. 3. 11. Let (√5+2)+(√√5-2)=x, and put (√/5+2)=a, (√√/5-2)=b, then x=a+b, and x=(a+b)3=a3+b3+3ab(a+b), .. x3=2\/5+3x, and x3-3x=2√/5. To find the value of x requires the solution of a cubic equation. XXXVI. Since (a+b)(√a−√√/b)=a−b, and (a√x+b\/y)(a√x−b\√y)=a2x—b2y; it is obvious that √a+√b may be made rational by multiplying it by √α-√b; and a/b by multiplying it by √/a+√√/b. Similarly for the factors ax+by and a√x-b√Y. since {(√a+√b)+√c}· {(√a+√b)−√c}=(√a+√√/b)3 —c=(a+b−c) +2\/ab, and {(a+b-c)+2a√√/b}. {(a+b−c) -2\√ab}=(a+b—c)2 — 4ab. And two factors will be required in general to render such a denominator a rational quantity. 1. 8-5√2_8-5√2_3+2√2_24+√√2−20−4+√√/2. 3-2√23-2√2^3+2√2 9-8 16. Hare (5.12)+= {1}= 8 102 103 + 1. √245+√/75√/245−√√/75_7√√/5+5√3_7√/5-5√/3 √5-No3 √5+√/3 √5-√/3 √5+√/3 -+ = 50+121550-12/15 2 = 2 100 =50. XXXVIII. 1. The expressions 2, 3, 4, 5, 6a, are respectively equivalent to (20), (320)'s, (415), (512), (610); and by expanding the powers 230, 320, 415, 512, 610, their numerical values will be known and the given expressions can be arranged in the order of magnitude. The extractions and divisions in each example are to be carried out to four places of decimals. |