get 21 for the other factor of the root, so that the sought root equals (x 1) (≈ — 1) = xz x 2 + 1. - Hence, when the root r of " is composed of independent factors, we perceive how we must proceed in order to find them. Remark. It is clear that the preceding method of finding the roots of polynomials is essentially the same as that of finding the equal roots of equations. (7.) From what has been done, we deduce the following rule for finding any root of a monomial quantity. RULE. 1. Resolve or separate the quantity into as many equal factors as there are units in the index or number which denotes the root to be extracted, then one of the equal factors will be the required root. If the monomial has a coefficient, we may extract its root, as in Arithmetic; then the root thus found, being prefixed as a coefficient to the product of the letters in the quantity, after their indices have been severally divided by the index of the root, will be the root, as required, which is clearly the same as to take the product of the roots of all the factors of the quantity for its root. 2. If the monomial is a fraction, the root of the numerator, when divided by that of the denominator, will be the root, as required. 3. If the index of the root is an odd number, the sign of the root must be the same as that of the given monomial; but if the index of the root is an even number, the root must have the ambiguous sign± prefixed to it, to show that it may be either positive or negative. If the monomial is negative, and the index of the root an even number, then the root will clearly be an imaginary quantity. EXAMPLES. 1. To extract the square root of 9a3. Because 9a3a x 3a 3a x 3a, the root equals х 3a. 2. To find the cube root of 64ab, and the fifth root of 243a1015 Because 64a3b = 4ab2 × 4ab2 × 4ab2, and -243a5b10c1 = - 3ab2c3 × 3abic x 3ab2 × roots are 4ab2 and Zabc. 3abc x 6 3 3 3abic, the Otherwise, since 64a*b* = √43a*b* = 43a3¿§ = 4ab2, and we have 4ab2 and 3ab for the roots, as before. 4. To find the square root of 4ab5, and the cube root of 81a+b2. Here√4ab = ± √4a2b1 × ab = ± √4a2b1× √ab = ± 2ab2 Vab, and 81a*b* = √27a3l© × 3ab = √27a3l® × √3ab = 3ab2√3ab. Remarks. These roots are in part freed from the signs of the roots, and in part affected by them; consequently, because the roots can not be wholly freed from the signs of the roots (or from the radical signs), they are called surds or irrational quantities. Also, because 2ab2 √ab = 2ab2 × √ab, and 3ab3ab3ab2 × V3ab, 2ab2, and 3ab2, which are called the rational factors of the roots, are sometimes called the coefficients of Vab and V3ab, which are their irrational or surd factors. 5. To find the square root of of 64a12624. - 9ab, and the sixth root Here±√ — 9a2b* = ± √9a2b* × √ − 1 = ± 3ab2 √ — 1, = ± Remark. These roots are impossible or imaginary, because are called the coefficients of 1, in the roots. 1 is impossible in real numbers; and 3ab3, 2aab1 ́ 6. To find the square root of a2b3c2 = a2b2ca × bc. (8.) Before giving the usual rules for the extraction of the roots of compound or polynomial quantities, we will premise the following PROPOSITION. b a Any expression of the form 1 + in which a and b have like or unlike signs, may be resolved into factors of a similar form. For if n stands for any positive whole number, it is clear na + b na + 26. na + 3b = na + b na +26 = (1 + na + (n na × (1+ to n factors, (1), as Hence, having reduced any number or quantity to the form A" + B, such that A" is an exact n power of A, we If B is very small in comparison to A", then by omitting the terms which contain B in the denominators of the factors of the right member of (2), we shall approximately have A + B = A" × (1+ , or, extracting the nth root, we B (1+)", get VA" + B = A × (1+ = A + B nAn n It is clear from (2) that VA" + B, or the nth root of A"+ B, will be less than A + AB nA" + (n − 1)B ̊ B nAn-i and greater than A+ To perceive the use of what has been done, take the following EXAMPLES. Ex. 1.—To resolve 2 = 1+1=1+ into two factors, and thence find its square root. 1 1 Putting a 1, b = 1, and n = 2, (1) gives 2 = 1 + 1.5 and 1 + 1 = 1.333. are the required factors. 3 Because the square root of 2 is greater than 1.333. and less than 1.5, it is easy to perceive that the first part of it must be 1.41; indeed, if we represent the root by 1+x, we shall have (1+)2 = 2=1+ (1/3 + 1/1) 1 1 1 1 1 +0.5 +0.3333. + = 1.833. + or since (1)2 = 1 + 2x + ∞, we shall 6 have 1 + 2x + x2 = 1.833. + 0.166., or 2x + x2=0.833. + 0.166., which, by rejecting 2 and 0.166. is reduced to 2x = 0.833., which gives x 0.41 +, and of course we have 1+x=1.41 for the approximate value of the root, as above. we get, by resolving 1.0059856 into two factors as before, 1.0029928 and 1.0029838 for the factors; consequently (as before) we have 1.0029928 × 1.0029838 = 1.0029928 + 0.0029838 + 1.0059766 +, and 1 + .0059766 = : = 1.0029883 is the approximate square root of 1.0059856. Hence, we shall have 12 = √(1.41) × 1.0059856 1.41 x 1.0029883 1.4142135+ for the square root of 2; which is correct in all its figures. = Ex. 2.-To find the square root of 3. 2 1' Since 3 = = 1 + if we put a = 1, b = 2, and n = 2, (1) gives 3 (1 + - 1) (1 + 1), and in a similar way 1+1= (1 + 1)(1 + 1); consequently, 3 = = (1 + 1)2 × (1 + 1 }) . get 1 + Be 1 1 + X we 6 7' (1 + - ) 21.1547 for the approximate square shall get 131.5 x 1.1547 = 1.73205+ for the approximate square root of 3; which is correct in all its figures. Ex. 3.-To find the square root of 17.3056. Since 16 is the greatest integral square in 17, we shall put 16 × 17.3056 the number in the form 17.3056 = =4x 1.0816; 16 consequently, resolving 1.0816 into two factors by (1), we shall have 1.0816 = 1.0408 × 1.0392 = 1.0408 + 0.0392 + = 1.08; consequently, we shall have 1+ square root of 1.0816. Hence, we shall have 17.305642 × √1.0816 = 4 × 1.04 4.16 for the square root of the given number, the root being exact. Ex. 4. To resolve 11 into five factors, and thence obtain its fifth root. Because 11 =1+ 10 1 if we put a = 1, b = 10, and n= 5, |