the given index, and the quotients will be the new indices for those quantities. Then, over the said quantities, with their new indices, place the given index, and they will be the equivalent quantities required. EXAMPLES. 1 1. Reduce 33 and 23 to quantities that shall have the index . Whence (31) and (22)ď, or 276, and 45, are the quantities required. 2. Reduce 5% and 63 to quantities that shall have the common index Ans. 125 and 36. 1 1 6° 3. Reduce 22 and 44 to quantities that shall have the com 1 mon index Ans. 166 and 165. 1 4. Reduce aș and a” to quantities that shall have the common index Ans. (29)+ and (22) 4: 5. Reduce a’ and 63 to quantities that shall have the common index Ans. (09)# and (14) 6. Note. Surds may also be brought to a common index, by reducing the indices of the quantities to a common denominator, and then involving each of them to the power denoted by its numerator. 1 1 EXAMPLES. 1 1 1. Reduce 32 and 43 to quantities having a common index. Here 37 = 38 = (39) = (27) Whence (27) and (16)& Ans. 2. Reduce 43 and 54 to quantities that shall have a com 1 mon index. Ans. 2567'2 and 1251'. 1 3. Reduce aż and a3 to quantities that shall have a common index. Ans. (a)and (a?). 4. Reduce a3 and 54 to quantities that shall have a common index. Ans. (au)73 and (19)7'. 5. Reduce añ and in to quantities that shall have a common index. Ans. (am) an and (bn) min. CASE III. To reduce surds to their most simple forms. RULE-Resolve the given number, or quantity, into two factors, one of which shall be the greatest power contained in it, and set the root of this power before the remaining part, with the proper radical sign between them.* EXAMPLES. 1. Let v 48 be reduced to its most simple form. Here 48 = V(16 x 3) = 43. Ans. 2. Let V 108 be reduced to its most simple form. Here 108 =3(27 X 4) = 33/4. Ans. Note 1. When any number, or quantity, is prefixed to the surd, that quantity must be multiplied by the root of the factor above mentioned, and the product be then joined to the other part, as before. EXAMPLES. 1. Let 2 v 32 be reduced to its most simple form. Here 2 v 32 = 2V(16 X 2) = 8 v 2. Ans. 2. Let 53/24 be reduced to its most simple form. Here 5/ 24 = 5; (8 X 3) = 10V 3. Ans. Note 2. A fractional surd may also be reduced to a more convenient form, by multiplying both the numerator and denominator by such a number, or quantity, as will make the denominator a complete power of the kind required; and then joining its root, with 1 put over it, as a numerator, to the other part of the surd.t * When the given surd contains no factor that is an exact power of the kind required, it is already in its most simple form. Thus, v 15 cannot be reduced lower, because neither of its factors, 5 nor 3, is a square. + The utility of reducing surds to their most simple forms, in order to have the answer in decimals, will be readily perceived from considering EXAMPLES. 1. Let vị be reduced to its most simple form. V 14. Ans. 2 14 1 Here ✓ =V x 49 2 2. Let 3 be reduced to its most simple form. 2 50 1 3 Here 3 3/ 3 v Х =-50. Ans. 125 125 3३/ EXAMPLES FOR PRACTICE. 3. Let ✓ 125 be reduced to its most simple form. Ans. 5 V5 4. Let 294 be reduced to its most simple form. Ans. 76. 5. Let 56 be reduced to its most simple form. Ans. 23/7. 6. Let ¥ 192 be reduced to its most simple form. Ans. 43/3. 7. Let 7 V 80 be reduced to its most simple form. Ans. 28 5. 8. Let 93 81 be reduced to its most simple form. Ans. 273/ 3. 3 5 9. Let ✓ be reduced to its most simple form. 121 6 Ans. z. v 30. 3 10. Let - 3 be reduced to its most simple form. 7 16 Ans. 1V12. 11. Let ✓ 98ačx be reduced to its most simple form. Ans. 7av 2x 12. Let ✓ ( 23 - aRa*) be reduced to its most simple form. Ans. a V (x — a®): the first question above given, where it is found that v*=V 14; in which case it is only necessary to extract the square root of the whole number 14, (or to find it in some of the tables that have been calculated for this purpose,) and then divide it by 7; whereas, otherwise, we must have first divided the numerator by the denominator, and then have found the root of the quotient, for the surd part; or else have determined the root both of the numerator and denominator, and then divided one by the other; which are each of them troublesome processes when performed by the common rules; and in the next example for the cube root, the labour would be much greater CASE IV. To add surd quantities together. RULE. When the surds are of the same kind, reduce them to their simplest forms as in the last case; then, if the surd part be the same in them all, annex it to the sum of the rational parts, and it will give the whole sum required. But if the quantities have different indices, or the surd part be not the same in each of them, they can only be added together by the signs + and EXAMPLES. 1. It is required to find the sum of v 27 and v 48. Here v 27 = V(9 X 3) = 3 / 3 Whence 7 v 3 the sum. Whence 8/4 the sum. Here 4 v 147 4v(49 X 3) : 28 V 3 and av Whence 43 v 3 the sum. 1 4. It is required to find the sum of 3 vand 2 5 10° 3 25 10 2 100 10 10 5V 10 5. It is required to find the sum of v 72 and V 128. Ans. 14 v (2). 6. It is required to find the sum of v 180 and V 405. Ans. 15 ✓ (5). y. It is required to find the surn of 3 V 40 and 135. Ans. 9V (5). 2 3 8. It is required to find the sum of 4 V 54 and 5 x 128. Ans. 32V (2) 9. It is required to find the sum of 9 / 243 and 10 v 363 Ans. 191 ✓ (3). 27 10. It is required to find the sum of 3 - and 7 V 50 Ans. 310 V (6). 1 1 4 32 Ans. 633 (2). 12. It is required to find the sum of V ab and 3 v 4bxo. 11. It is required to find the sum of 12 V and 3 V 32 1 Ans. 3 CASE V. To find the difference of surd quantities. tho quantities as in the last rule; then the difference of the rational parts annexed to the common surd, will give the whole difference required. But if the quantities have different indices, or the surd part be not the same in each of them, they can only be subtracted by means of the sign 1. It is required to find the difference of ✓ 448 and ✓ 112. Here ✓ 448 = V(64 X ?) ✓ (64 X 7) = 8V7 And ✓ 112= V(16 X 7)=477 Whence 477 the difference. 2. It is required to find the difference of V 192 and 3V24. Here V 192 = V(64 X 3)=4V3 Whence 23/3 the difference. 3. It is required to find the difference of 5 / 20 and 3 V 45. Here 5 v 20 =5N (4 X 5) =.10 V 5 Whence ✓ 5 the difference. 32 2 1 4 It is required to find the difference of 7 Vand ✓ 439 |