Now, in the roots a+b and 12, a and 1 may be regarded as occupying the tens' position, b and 2 the units' position. 12 may be written 10 +2. Substituting these values for a and b in the cube a+ 3a2b+3ab2+b3, we have the result, 1000+ 600+120 +8 1728. = Subtracting a3 from the given cube, we have, 3 a2b+3 ab2+b3; and removing from this the second term of the root, we have, 3 a2 +3 ab + b2, which expression is the complete divisor. 209. Hence, the complete divisor equals 3 times the square of the first figure of the root (considered as tens), plus 3 times the first (as tens) by the second figure of the root, plus the square of the second figure of the root. 210. If the root consists of three or more figures, that part of the root already found should be treated as the first figure in applying the above rule. Again, decomposing 120, we have, That is, the cube of the first figure of the root is contained wholly in the first period of the cube; the cube of the first two figures in the first two periods, etc. (1) Extract the cube root of 256047875. Since there are three periods, there will be three figures in the root. (2) Extract the cube root of 7 to three places of decimals: In like manner the cube root of fractions may be found by taking the cube root of both numerator and denominator. 211. Since the fourth power is the square of the square, the sixth power the square of the cube, the eighth power the square of the fourth power, it follows that the fourth root may be found by extracting the square root twice, the sixth root by extracting the square root and the cube root, the eighth root by extracting the square root three times. In like manner any root may be extracted when the index of the power is a product of the factors 2 and 3. EXERCISE 73. Extract the fourth root of the following: 1. x+4x+6x2+4x+1. 2. 16x96x3y +216 x2y2 - 216 xy3 +81 y. Extract the sixth root of the following: 4. 646-192 25+ 240 x 160 x 60 x2 - 12x+1. Extract the eighth root of the following: 7. 216-8x11+28x12-56 x1o+70 x3-56 x6+28 x1 −8 x2+1. CHAPTER XVI. SIMPLE INDETERMINATE EQUATIONS. 212. If a single equation is given, involving two unknown quantities, with no condition imposed, the quantities may have an indefinite number of values. Such an equation is called an Indeterminate Equation. 213. If, however, on such an equation certain conditions be imposed, such that the value of the unknown quantities be positive, or be positive integers, the solutions may be limited. 214. Every equation of the first degree may be reduced to the form ax ± by = ± c. Here a, b, and c are integers and have no common factor; for if they have a common factor, the equation may be reduced still further by division. 215. If in the above equation a and b have a common factor which does not divide c, the solution for positive integers is impossible. For if we divide both members of the equations by the common factor of a and b, we shall have the left-hand member integral and the right-hand member fractional, which is impossible. EXAMPLE (1): Solve for positive integral values the equation Now, since x and y must be integral, x+8y-10 will be integral, and as a result 1 will be integral. y 8 Substituting this value of y in the original equation, wo have, = 8x+65-520 m x= 2+ 65 m. 81, Now, if m equal 0 or any positive integer, a will be a positive integer, but m cannot have a negative value if a remain positive. In the equation y = 1-8 m, if m equal 0 or any negative integer, y will be positive, but m cannot have a positive value if y remain positive. |