Find the cube root of 26-32+5x3-3x-1. The root is placed above the given expression for convenience of arrangement. The first term of the root, x2, is obtained by taking the cube root of the first term of the given expression; and the first trial-divisor, 3a, is obtained by taking three times the square of this term of the root. The first complete divisor is found by annexing to the trial-divisor (3x-x)(x), which expression corresponds to (3a + b)b in the typical form. The part of the root already found (a) is now represented by x2-x; therefore 3a2 is represented by 3 (x2 — x)2 = 3 xa — 6x3 + 3x2, the second trial-divisor; and (3 a + b) b by (3x2 – 3 x − 1) (− 1); therefore, in the second complete divisor, 3a2 + (3 a + b) b is represented by (3 x1 − 6x3 + 3x2) + ( − 3 x2 − 3 x − 1) × (− 1) = 3 x1 − 6 x3 + 3x + 1. EXERCISE LXXX. Find the cube roots of: 1. x+6x2y+12xy2+8y3. 2. a3-9a2+27a-27. 3. x3 +12x2 + 48x + 64. 4. x6-3 ax5+5a3x3—3 a3x—ao. 5. x6+3x+6x+7x2+6x2+3x+1. 6. 19x39x2-99x+156x144x+64 26. 7. a6a9a1+4a3-9a6a-1. 8. 6426+ 1922+144x2 - 32x3-36x2+12x-1. 9. 1-3x+6x-10x+12x-12x2+10x-6x2+3x-29. 10. a9ab135 a3b3+729 ab5-7296o. 11. c-12bc5+60 b2ca — 160 b3ç3 + 240 b1c2 — 192b3c+64 lo. 12. 3a +48 a3b +60 a1b2 — 80 a3b3 — 90 a2b1 + 108 ab3 — 276o. CUBE ROOTS OF ARITHMETICAL NUMBERS. 216. In extracting the cube root of a number expressed by figures, the first step is to mark it off into periods. Since 1 = 13, 1000 = 103, 1,000,000 = 1003, and so on, it follows that the cube root of any number between 1 and 1000, that is, of any number which has one, two, or three figures, is a number of one figure; and that the cube root of any number between 1000 and 1,000,000, that is, of any number which has four, five, or six figures, is a number of two figures; and so on. Hence, if a dot be placed over every third figure of a cube number, beginning with the units' figure, the number of dots will be equal to the number of figures in its cube root. 217. If the cube root of a number contain any decimal figures, the number itself will contain three times as many. Thus, if 3 be the cube root of a number, the number is .3 .3 x .3 = .027. Hence, if the given cube number have decimal places, and a dot be placed over the units' figure and over every third figure on both sides of it, the number of dots to the left of the decimal point will show the number of integral figures in the root; and the number of dots to the right will show the number of decimal figures in the root. If the given number be not a perfect cube, ciphers may be annexed, and a value of the root may be found as near to the true value as we please. 218. It is to be observed that if a denote the first term of the root, and b the second term, the first complete divisor and the second trial-divisor is 3 (a+b), that is, 3a2+6ab362, which may be obtained from the preceding complete divisor by adding to it its second term and twice its third term, 3a2+3ab + b2 +3 ab+2b2 3a+6ab+362 a method which will very much shorten the work in long arithmetical examples. 219. Ex. Find the cube root of 14,102.327296. It will be observed that in this example a represents 20 and b represents 4; a', 10 (a+b), represents 240 and b' represents 1; a", 10 (a+b), represents 2410 and b" represents 6. It will be observed, also, that the trial-divisors are formed from the preceding complete divisors, according to the method explained in & 218. 21. Find to four figures the cube roots of 2.5; .2; .01; 4; .4. 220. Since the fourth power is the square of the square, and the sixth power the square of the cube; the fourth root is the square root of the square root, and the sixth root is the cube root of the square root. In like manner, the eighth, ninth, twelfth..... roots may be found. EXERCISE LXXXII. Find the fourth roots of: 1. 81a540 ab+1350ab2-1500 al3+62564. 2. 1-4x+102-162+19x16x+10x-4x+x8. Find the sixth roots of: 3. 64-192x + 240x2160x360x-12x+x6. 4. 72926-1458x+1215x-540x+135x2-18x+1. Find the eighth root of: 5. 1-8y+28y2 -- 56 y3 + 70 y1 — 56y+ 28 y — 8y' + y3. CHAPTER XIV. QUADRATIC EQUATIONS. 221. An equation which contains the square of the unknown quantity, but no higher power, is called a quadratic cquation. 222. If the equation contain the square only, it is called a pure quadratic; but if it contain the first power also, it is called an affected quadratic. PURE QUADRATIC EQUATIONS. Solve the equation 5a-482x2. 5a2-482x2 = 3x2=48 .. x +4 It will be observed that there are two roots of equal value but of opposite signs; and there are only two, for if the square root of the equation, 2216, were written ++ 4, there would be only two values of x; since the equation = 4. 4 gives x = gives x = - 4, and the equation Hence, to solve a pure quadratic, x= Collect the unknown quantities on one side, and the known quantities on the other; divide by the co-efficient of the unknown quantity; and extract the square root of each side of the resulting equation. Solve the equation 322-15=0. 3x2-15=0 3x2 15 It will be observed that the square root of 5 cannot be found exactly, but an approximate x2= 5 value of it to any assigned degree of accuracy ..x=±√5 may be found. |