Ex. 2. Required the square root of a + 2ab +62+2ac+2bc +ca? a+2ab +62+2ac+2bc+co(a+b+c 2a +6 | 2ab +63 2ab+62 2a +26+c | 2ac+2bct 2ac+2bc+ca On comparing the root a+6+c, thus found with its power, the reason of the rule for deriving the root from the power is evident. And the method of operation is the same as in the last example. Thus, having tound the first two terms of the root as before, we bring down the remaining three terms 2a +2bc+c of the power, and dividing 2ac by 2a, it gives C, the third term of the root Next, let the last term (b) of the preceding divisor be doubled, and add c to the divisor thus increased, and it becomes 2a +2htc; multiply this new divisor by c, and it gives 2act abc+co, which being subtracted from the terms last brought down, leaves no remainder. In like manner the following Examples are solved. 89 Ex: 3. Required the square root of 4x4+63°+ +* + 15.0 +25 ? 89 3 4x4 +6x34 xP154 -5 4x4 Ex. 4. Required the square root of +4x+2x4+9x?–4x +4. Ans. 2+2r_:+2. Ex. 5. Required the square root of x'+4ax' +6a?r?+40°, tat. Aus. 2+ + 2ax ta. Ex. 6. Required the square root of a - 2a'+faé-fa+it. Ans. as-at Ex. 7. Required the square root of 4a4+12a’x+13a%22+ Bachc. Ans. 2u2 +3ax+x. Ex. 8. Required the square root of 9x4 +12x +34x2 +20x +-25. Ans. 3xo +2x+5. Ex. 9. Required the square root of a2 + ab +62+2act 2bc+c+ad+2bd+2cd+d. Ans. a+b+c+d. Ex. 10. Required the square root of a4+12a36+54a2b2 + 108ab3 +8164. Avs. a+6ab+962. Ex. 11. Required the square root of q@_6aRx+ 15a*r?— 20a’x3 + 15aRx4 bax+ 26. Ans. a’ – 3aʼr +3ax: — 23. Ex. 12. Required the square root of a'-_2aRx? +x4. Ans. a 22. CASE III. To extract the cube root of a compound Quantity. RULE. 300. Arrange the terms as in the last case ; and set the root of the first term in the quotient ; subtract the cube of the root, thus found, from the first term, and bring down three terms for a dividend. Next, divide the first term of the dividend by 3 times the square of that part of the root already determined, and set the result in the quotient; then, to 3 times the square of that part of the root, annex 3 times the product of the same part and the last result, and also the square of the last result, with their proper signs ; and it will give the divisor, multiply the divisor by the term of the root last placed in the quotient, and subtract the product from the dividend, bring down three terms or as many as may be necessary for a dividend, and proceed as before. Ex. 1. Required the cube root of a3 +3a25+3ab2 +63 ? a +3a+b+3ab +63 (a+b 3a2+3ab +62) 3026+3aba +63 3a+b + 3abo +63 The reason of the rule may be made evident from a comparison of the roots with its cube. Or, thus, if the quantity whose root is to be extracted, hás an exact root, the root of the leading term must be one term of its root; that is, the cube root of a3, which is a, is one term of the root, and the remaining terms being brought down, the root of the last term 63 is consequently another term of the root; but as the root may consist of more terms than two; the next term (b) of the root is always found by dividing 3a2b the first term of the dividend by three times the square of the divisor, and the two remaining terms of the dividend Saba +63=(3ab +62,6; hence 3ab+62 must be added to 3a2 for a divisor; and so on. Ex. 2. Required the cube root of xo + 6x5–40x? +963-64. 28 +6x5—40x3 +963-64 (x2+2x - 4 3x4 +6x3+4x2)6x5—40x3 6x”+ 12x4+823 3x4 +12x*— 24x+16)-12x*—48x3 +963-64 - 12x4_-48%%+96x—64 Ex. 3. Required the cube root of (a+b)3+3(a+b){c+3 (a+b)c+c?. Ans, atoto. Ex. 4. Required the cube root of x-6x +15x4_20x'+ 15x4-6x+1. Ans. x2x+1. Ex. 5. Required the cube root of x' +6x®y+15x*y*+20%* 73+15x+y4 +6xy + ye. Ans. 2a + 2xyty. Ex. 6. Required the cube root of 1-6x+12x2_8x*, Ans. 1-2x. CASE IV. . To find any root of a compound Quantity. RULE, 301. 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 ; by four times the cube of it, for the fourth root, &c. and the quotient will be the next term of the root. Involve the whole of the root, thus found, to its proper power, wbich subtract from the given quantity, and divide the first term of the remainder by the same divisor as before. Proceed in the same manner, for the next following term of the root ; and so on, till the whole is finished. 302. This rule may be demonstrated thus ; (a+b)n=ant nan-1b+ &c. (Art. 289). Here the nth root of an is a, and the next term nan='h contains b, (the other term of the root) nan-' times; hence, if we divide nan–16 by nan-s, we have b, or man-i=b; and so on, for any compound quantitity, the root of which consists of more than two terms. Now, if n=2; then, the divisor nan-1=2a, for the square nan-18 root ; root ; if x=4 ; then, . nan-=4a3, for the 4th . root; if n=5; then, nan-=5a", for the 5th root. And so on for any other root, that is, involve the first term of the root, to the next lowest power, and multiply it by the index of the given power for a divisor. Ex. 1. Required the square root of a4_2a': +3a2x2ax® +34 ? a-2a+3a2_20x+x*as-ax+ Es. 2. Required the 4th root of 16a“ – 96a36+216aRx?-216ax: +81.24. 16a4-96aRx+218a%x_216ax:+81x*(24—37 303. As this rule, in high powers, is often found to be very laborious, it may be proper to observe, that the roots of cer. tain compound quantities may sometimes be easily discover ed : thus, in the last example, the root is 2a-3x, which is the difference of the roots of the first and last terms; and so op, for other compound quantities. Hence, the following method in such cases ; extract the roots of all the simple terms, and connect them together by the signs + or -, as may be judged most suitable for the purposes, 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 t to -, or from - to t, till its power agrees with the given one throughout. However, such artifices are not to be used by learners, because the regular mode of proceeding is more advantageous to them; besides, a knowledge of those artifices which are used by experienced Algebraists, can only be acquired from frequent practice. Ex. 3. Required the square root of a + 2ab +62+2ac+2bc +0. Here, the square root of a:=a ; the square root of ba=b ; and the square root of cʻ=c. Hence, a+b+c, is the root required, because (a+b+c)2=a++ 2ab+ba+2ac+2bc=c2. Ex. 4. Required the fifth root of 32x'_80x4 + 80x3 - 40.co +10x-1. Aus. 2x-1. Ex. 5. Required the cube root of 2646x5 +15x4–20237 15x45x+1. Ang. 2_2x+1. Es. 6. Required the fourth root of a'-4aRx+6ar? - 4ax3 +x4. Ans. aEx. 7. Required the square root of r8+ 2x*y*+y8. Ans. x++y*. Ex. 8. Required the square root of 28 - 2x*y*+y®. Ans. a* - yo. Ex. 9. Required the cube root of a'-6a2x+12ax? -8x*. Ans. a - 22. Ex. 10. Required the sixth root of 26_6x +15x620x3+ 15x2-6 +1. Ans. 2l. Ex. 11. Required the fifth root of 20+15ay +902"y+ 270x4y +405x+y +243yło. Ăns 2:2 + 3y. Ex. 12. Required the square root of 2+2xy+y+6x2+ byzut 92. Ans. x+y+3z. § III. INVESTIGATION OF THE RULES FOR THE EXTRACTION OF THE SQUARE AND CUBE ROOTS OF NUMBERS. 304. It has been observed, (Art. 104), that, a denoting the tens of a number, and b the units, the formula a+2ab +62 would represent the square of any number consisting of two |