It may here, also, be farther remarked, that root of a negative quantity, is unassignable. any even Thus, a cannot be determined, as there is no quantity, either positive or negative, (+ or -), that, when multiplied by itself, will produce a2. EXAMPLES. 1. Find the square root of 9x2; and the cube root of Sx3. Here✓9x29× √x2=3×x=3x2 Ans. And /8x3/8X/x2=2Xx=2x. Ans. a2x2 2. It is required to find the square root of and 4c2 3. It is required to find the square root of 4a2x®. 4. It is required to find the cube root of- 125a3x®. 5. It is required to find the 4th root of 256a4x8. 6. It is required to find the square root of 7. It is required to find the cube root of 4a4 9x2 y2 8a3 125x 32a5x10 8. It is required to find the 5th root of 243 And for the cube root, fifth root, &c, of a negative quantity, it is plain, from the same rule, that (-a)X(-a)X(-a)——a3; and (-a3)x(+a2 )=-a5, And consequently a 3-a, and as CASE II. To extract the square root of a compound quantity. RULE. 1. Range the terms, of which the quantity is composed, according to the dimensions of some letter in them, beginning with the highest, and set the root of the first term in the quotient. 2. Subtract the square of the root, thus found from the first term, and bring down the two next terms to the remainder for a dividend. 3. Divide the dividend, thus found, by double that part of the root already determined, and set the result both in the quotient and divisor. 4. Multiply the divisor, so increased, by the term of the root last placed in the quotient, and subtract the product from the dividend; and so on, as in common arithmetic. EXAMPLES. 1. Extract the square root of x4 4x3+6x2-4x+1. x44x3+6x-4x+1(x2 -2x+1 Ans. x2-2x+1, the root required. 2 Extract the square root of 4a1+12a3x+13a2x2+ 6αx3+x1. 4a4+12a3x+13a2x2+6ax3+x+(2a2+3αx+x2 4a+ 4a2+3ax)12a3x+13a2x2 12a3x+9α2x2 4a2+6ax+x2)4a2x2+6αx3+x+ 4a2x2+6ax3+x4 * Note. When the quantity to be extracted has no exact root, the operation may be carried on as far as is thought necessary, or till the regularity of the terms shows the law by which the series would be continued. EXAMPLE. 1. It is required to extract the square root of 1 + x. x x2 x3 5x4 1+x(1+) ·+. &c. 128 Here, if the numerators and denominators of the two last terms be each multiplied by 3, which will not alter their values, the root will become 3x3 x2 3.5x4 3.5.7x5 1+- + + &c. 2 2.4 2.4.6 2.4.6.8 2.4.6.8.10 where the law of the series is manifest. EXAMPLES FOR PRACTICE. 2. It is required to find the square root of a4 +4a3x + 6a2x2+4ax3+x1 3. It is required to find the square root of x4 1 2x3+ 4. It is required to find the square root of 4x6 -4x1+ 13x2-6x+9 6 5. Required the square root of a +4x5+10x4+20x3+ 25x2+24x+16. 6. It is required to extract the square root of a2+b. 7. It is required to extract the square root of 2, or of 1+1. CASE III. To find any root of a compound quantity. RULE. Find the root of the first term, which place in the quotient; and having subtracted its corresponding power from that term, bring down the second term for a dividend. Divide this by twice the part of the root above determined, for the square root; by three times the square of it, for the cube root, and so on; and the quotient will be the next term of the root Involve the whole of the root, thus found, to its proper power, which subtract from the given quantity, and divide the first term of the remainder by the same divisor as before; and proceed in this manner till the whole is finished. (p) EXAMPLES. 1. Required the square root of aa—2a3x+3a2x2 — 2ax3 +x1. aa—2α3x+3a2x2 - 2ax3+x4 (a3—ax+x2 (†) As this rule, in high powers, is often found to be very laborious, it may be proper to observe, that the roots of various compound quantities may sometimes be easily discovered, as follows: Extract the roots of all the simple terms, and connect them together by the signs or, as may be judged most suitable for the purpose; then involve the compound root, thus found, to its proper power, and if it be the same with the given quantity, it is the root required. But if it be found to differ only in some of the signs, change them from to or from to till its power agrees with the given one throughout. Thus, in the third example next following, the root is 2a-3x, which is the difference of the roots of the first and last terms: and in the fourth example, the root is a+b+c, which is the sum of the roots of the first, fourth, and sixth terms. The same may Iso be observed of the sixth example, where the root is found from the first and last terms. F |