CHAPTER VIII. SURDS AND IMAGINARY QUANTITIES. α 85. In the Elementary Algebra, Art. 272, it is proved that the denominator of any expression of the form can be rationalised by multiplying the numerator and the denominator by bc, the surd conjugate to the denominator. a Similarly, in the case of a fraction of the form √b + √c + √d' where the denominator involves three quadratic surds, we may by two operations render that denominator rational. For, first multiply both numerator and denominator by √b+ √c-√d; the denominator becomes (√b+ √c)3 − (√d)3 or b+c-d+2bc. Then multiply both numerator and denominator by (b+c-d)-2bc; the denominator becomes (b+c-d)2 - 4bc, which is a rational quantity. 86. To find the factor which will rationalise any given bino mial surd. CASE I. Suppose the given surd is a – Jb. Let /ax, by, and let n be the L.C.M. of Ρ and a" and y" are both rational. 9; then Now "y" is divisible by x-y for all values of n, and x" n-2 - y" = (x − y) (x′′− 1 +x”¬3y + x" ¬3y3 + +y"−1). Thus the rationalising factor is ...... CASE II. Suppose the given surd is a + /6. Let x, y, n have the same meanings as before; then (1) If n is even, x" - y" is divisible by x + y, and x” x” ̄2y Thus the rationalising factor is n-2 ...... and the rational product is " — y". + xy"−2 − y"−1 ; (2) If n is odd, x" + y" is divisible by x+y, and Example 1. Find the factor which will rationalise √3+ 3/5. Let x=32, y=53; then x and y are both rational, and x6 - y = (x + y) (x5 − x1y + x3y2 − x2y3 + xya — y3) ; thus, substituting for x and y, the required factor is 1 Example 2. Express (531+9) ÷ (59) as an equivalent fraction with a rational denominator. 1 1 1 To rationalise the denominator, which is equal to 52-34, put 52=x, x-y=(xy) (x3+x2y + xy2+y") 1 34-y; then since and the rational denominator is 52 - 34-52-3=22. We 87. We have shewn in the Elementary Algebra, Art. 277, how to find the square root of a binomial quadratic surd. may sometimes extract the square root of an expression containing more than two quadratic surds, such as a + √b + √c + √d. Assume √a+ √b+ √c + √d = √x + √y+ √z; .: a + √b + √c + √d = x + y + z + 2 √xy + 2 √xz + 2 √yz. If then 2 √xy = √b, 2 √xz= √c, 2 √yz = √d, and if, at the same time, the values of x, y, z thus found satisfy x + y + z = α, we shall have obtained the required root. Example. Find the square root of 21-4/5+8√3 – 4/15. Assume Put .. 21-4/5+8/3-4√15=x+y+z+2√xy-2√xz-2√yz. by multiplication, 2xy=8/3, 2xz=4/15, 2√yz=4√5; xyz=240; that is √xyz=4√15; whence it follows that √x=2√3, √y=2, √z=√5. And since these values satisfy the equation x+y+z=21, the required root is 2/3+2-√5. 88. If Ja + √b= x + √y, then will Ja−√b=x-√y. For, by cubing, we obtain а a + √b=x3 + 3x2 √y + 3xy + y √y. Equating rational and irrational parts, we have Similarly, by the help of the Binomial Theorem, Chap. XIII., it may be proved that if "/a+ √b=x+ √y, then "Ja-√b=x−√y, where n is any positive integer. 89. By the following method the cube root of an expression of the form ab may sometimes be found. The values of x and y have to be determined from (1) and (2). In (1) suppose that /a-b=c; then by substituting for y in (2) we obtain If from this equation the value of x can be determined by trial, the value of y is obtained from y = x2 C. NOTE. We do not here assume r+y for the cube root, as in the extraction of the square root; for with this assumption, on cubing we should have a+b=x/x+3xy+3y√x+y√y, and since every term on the right hand side is irrational we cannot equate rational and irrational parts. |