The second power of a+b+c, or (a+b+c) (a+b+c) is a2 + 2 a b + b2 + 2 a c + 2 b c + c2 a2 + 2 a b + b2 + 2 (a + b) c + c2. To find the second power of 726 700, b=20, and c = 6. a2 = 700 X 700 Let a 490000 28000 400 8640 36 527076 726 726 4356 1452 5082 527076 The first three terms of the formula, viz. a2+2ab+b2, are the second power of a + b or of the hundreds and tens, viz. 720. The second power of 720 can have no significant figure below hundreds, and the significant figures of the second power of 720 and of 72 are the same; the former is 518400, the latter 5184. If from the whole number 527076 the two right hand figures be rejected, the number is 5270. This contains the second power of 72 and something more, viz. a part of the product 2 x (700+20) × 6 = 2 (a + b) c. The method of procedure then, is to find the largest root contained in 5270. The first three terms of the above formula, viz. a2 + 2 a b + b2, show, that this is to be found by the method given above for finding a root consisting of two figures. 52,70 (72 two figures rejected above, and it becomes 8676. This contains 2 (a + b) c + c2; that is, 2 × 720 × c+c2. If 8676 be divided by 2 x 720 1440, the quotient will be either c or a number larger by 1 or 2. The zero on the right of 1440, and the right hand figure in the dividend may be omitted without affecting the quotient. The quotient is 6. Put 6 into the root and raise the whole to the second power. 726 × 726 = 527076 2d dividend = 867,6 (144 =2d divisor. 726 × 726 = 527,076. There is, however, a method, which will save considerable labour in multiplying. In the last example, for instance, having found the second figure of the root 2, instead of raising the whole 72 to the second power, we may abridge it very much by observing, that the second power of the 70, answering to a2 in the formula, has already been found and subtracted; therefore it only remains to find 2 ab + b2, and subtract it also. But the 140 is 2 a, and the figure 2 found for the root answers to b; therefore if we add 2 to 140, it becomes 142 = 2a+b. If this be now multiplied by 2 or b, it becomes 2 x 142 = 284 = 2 a b + b2. This completes the second power of 72, which, subtracted from 370, leaves 86 as before. Prepare as before, and find the third figure of the root. Observe that the 2d power of 720 or a2 + 2 a b + b2 has already been found and subtracted; it only remains to find the other parts, viz. 2 (a + b) c + c2. The divisor 1440 answers to 2 (a+b). Add 6, the figure of the root just found, to this, and it becomes 1446, answering to 2 (a+b)+c. If this be multiplied by 6, it becomes 1446 x 6 86762 (a+b) c + c2. This completes the second power of 726, which, subtracted from 8676, the number remaining in the work, leaves nothing. The same principle will apply when the root consists of any number of figures whatever. What is the root of 533837732164? In the first place I observe that the second power of the tens can have no significant figure below hundreds, therefore the two right hand figures may be rejected for the present. Also the second power of the hundreds can have no significant figure below tens of thousands, therefore the next two may be rejected. For a similar reason the next two may be rejected. In this manner they may all be rejected two by two until only one or two remain. Begin by finding the root of these, and proceed as above. Operation. 53,38,37,73,21,64 (730642 49 43,8 (143 93,7 (1460 9377,3 (14606 8763 6 613 72,1 (146124 584 49 6 29 22 564 (1461282 29 22 564. After separating the figures two by two, as explained above, I find the greatest second power in the left hand division. It is 49, the root of which is 7. I subtract 49 from 53, and bring down the next two figures, which makes 438. Now considering the 7 as tens, I proceed as if I were finding the root of 5338; that is, I double the 7, which makes 14 for a divisor, and see how many times it is contained in 43, rejecting the 8 on the right. I find 3 times. I write 3 in the root at the right of 7, and also at the right of 14. I multiply 143 by 3, and subtract the product from 438. I then bring down the next two figures, which make 937. I double 73, or, which is the same thing, I double the 3 in 143; for the 7 was doubled to find 14. This gives 146 for a divisor. I seek how many times 146 is contained in 93, rejecting the 7 on the right, as before. I find it is not contained at all. I write zero in the root, and also at the right of 146. I then bring down the next two figures. I seek how many times 1460 is contained in 9377, rejecting the 3 on the right. I find 6 times. I write 6 in the root, and at the right of 1460, and multiply 14606 by 6, and subtract the product from 93773. I then bring down the next two figures, and double the right hand figure of the last multiplicand, and proceed as before; and so on, till all the figures are brought down. The doubling of the right hand figure of the last multiplicand, is always equivalent to doubling the root as far as it is found. From the above examples, we derive the following rule for extracting the second root. 1st. Beginning at the right, separate the number into parts of two figures each. The left hand part may consist of one or two figures. 2nd. Find the greatest second power in the left hand part, and write its root as a quotient in division. Subtract the second power from the left hand part. 3d. Bring down the two next figures at the right of the remainder. Double the root already found for a divisor. See how many times the divisor is contained in the dividend rejecting the right hand figure. Write the result in the root, at the right of the figure previously found, and also at the right of the divisor. 4th. Multiply the divisor, thus augmented, by the last figure of the root, and subtract the product from the whole dividend. |