Then, knowing the figure in the root to be in the place of tens, and therefore equal to 40, and that the second figure in the root must be such, that twice the product of the first and second terms, together with the square of the second, would complete the square, we took twice the root already found, viz. 40 x 280, for one of the factors, and using it for a divisor, found the second figure in the root by dividing the dividend by this factor.

Lastly, after finding that the second figure in the root was 5, we added it to the divisor, making 80+5=85, and multiplied the sum by 5, the last figure in the root, and thus obtained twice the product of the two terms, and the square of the last term; because, 80 is twice the first term of the root, and being multiplied by 5, which is the last term, the result is twice the product of the two terms; and 5 being multiplied by 5, the product is the square of the last term.

It will be observed, that 4, the first figure in the root, being in the place of tens, was called 40, and doubled for a divisor; but, if we had merely doubled the root without any regard to its place, making the divisor 8, and had cut off the right hand figure of the dividend and divided what was left, the result would have been the same; because, in this operation, both divisor and dividend would have been divided by 10. Thus 8 is contained 5 times in 42. The figure obtained for the root, in this abridged method, would be placed at the right hand of the divisor, instead of being added thereto; thus, 85, making the completed divisor the same as before. This course, being the most concise, will be adopted in the rule for extracting the square root, which we shall hereafter give.

Suppose 169 square rods of land are to be laid out in a square, and the length of one of its sides is required.

We know that the length of a side inust be such a number of rods, as, when multiplied by itself, will produce 169; therefore we must extract the square root of the given number of square rods, and that root will be the answer. √169=13 Ans.

We illustrate this last example by the square figure A B C D, each side of which is 13 rods long. This square is divided into 169 small squares, each of which is a square rod.

The whole figure is also subdivided into four figures, two of which, e f g D and hfi B, are squares; and the other two, A hfe and Cif g, are oblongs.

The square e f go D is 10 rods on a side, and, therefore, contains 100 square rods. The oblongs are each 10 rods long and 3 rods wide, and consequently each contains 30 square rods. The other square, B hf i, is 3 rods on a side, and contains 9 square rods.

To illustrate the process of extracting the square root by the above geometrical figure, we shall take the side A B, which is divided into two parts, the first of which, Ah, is 10 rods long; the other, h B, 3 rods long. A h being equal to e f, the square of A h, the first part, gives the area of the square D e f g; h B being equal to hf, the area of the oblong A h f e, is found by multiplying the two parts, A h and h B, together; the area of the other oblong i f g C, is the same; therefore, the area of the two oblongs is twice the product of the two parts, A hand h B. The square of the last part, h B, gives the area of the square h B i f.

We have therefore the square of the first part A h, 10X10=100 rods; twice the product of the two parts, A h and h B, 10X3X2=60 rods; and the square of the last part h B, 3X3-9 rods. These being added together make 169 rods, the square of the whole figure ABCD.

This illustration of a square corresponds exactly with that of the first example, and of course the extraction of the square root must proceed on the principles there exhibited.

From the illustrations of the two preceding examples, we give the following rule for the extraction of the square

root.

RULE. First--Point off the given number into periods of two figures each, by putting a dot over the place of units, and another over every second figure to the left; and also to the right, when there are decimals.

Secondly-Find the greatest square in the left hand period, and write its root in the quotient. Subtract the square of this root from the left hand period, and to the remainder bring down the next period for a dividend.

Thirdly-Double the root already found, for a divisor. Ascertain how many times the divisor is contained in the dividend, excepting the right hand figure, and place the result in the root, and also at the right hand of the divisor. Multiply the divisor, thus increased, by the last figure in the root, and subtract the product from the dividend, and to the remainder bring down the next period for a new dividend.

Fourthly-Double the root already found for a new divisor, and continue to operate as before, until all the periods are brought down.

It will sometimes happen, that, by dividing the dividend as directed in the rule, the figure, obtained for the root, will be too great. When this happens, take a less figure, and go through the operation again.

When the places in the decimal are not an even number, they must be made so, by continuing the decimal, if it can be continued; if it cannot, by annexing a cipher, that the periods may be full.

If there be a remainder after all the periods are used, a period of decimal ciphers may be added; or, if the given number end in a decimal, the two figures that would arise from a continuation of the decimal. The operation may be thus continued to any degree of exactness.

If any dividend shall be found too small to contain the d.visor, put a cipher in the root, and bring down the next period to the right hand of the dividend for a new divi dend, and proceed in the work.

When the square root of a mixed number is required, it will sometimes be necessary to reduce it to an improper fraction, or the vulgar fraction to a decimal, before extracting the root.

If either the numerator or denominator of a vulgar fraction be not a square number, the fraction must be reduced to a decimal, and the approximate root extracted.

1. Extract the square root of 4579600.

4579600 (2140 Ans.

2 What is the square root of 110 24?

i10.24(10.499+ Ans.

time.

3d. divisor, 208

Reducing to a decimal, we found it to be infinite, in the recurrence of 24 continually; therefore, in continuing the extraction of the root, instead of adding periods of decimal ciphers, we added the period 24 each The extraction of the root might have been continued indefinitely; but having obtained five places of figures in the root, we stopped, and marked off the three last places of the root for decimals; because we made use of three periods of decimals in the question,

3. What is the square root of 2704 ? 4. Extract the square root of 361. 5. What is the square root of 3025 ? 6. What is the square root of 121? 7. Extract the square root of 289. 8. Extract the square root of 400. 9. What is the square root of 4761? 10. What is the square root of 848241? 11. Extract the square root of 3356224. 12. What is the square root of 824464 ? 13. Find the square root of 49084036. 14. What is the square root of 688900? 15. Find the square root of 82864609. 16. Find the square root of 3684975616. 17. What is the square root of 44890000? 18. What is the square root of 165649? 19. Find the square root of 90484249636. 20. Find the square root of 26494625227849. 21. Find the square root of 262400.0625. 22. What is the square root of 841806.25 ? 23. What is the square root of 39.037504? 24. Find the square root of 213.715161. 25. Find the square root of .66650896. 26. What is the square root of 133407? 27. What is the square root of 1574 ? 28. Extract the square root of 318. 29. What is the square root of 51 30. Extract the square root of 27. 31. What is the square root of 55633? 32. Extract the square root of 10961. 33. Find the square root of 4120900. 34. Extract the square root of 5. 35. Extract the square root of 8. 36. Extract the square root of 84. 37. Extract the square root of 99. 38. Extract the square root of.101. 39. Extract the square root of 120. 40. Extract the square root of 124. 41. Extract the square root of 143. 42. Extract the square root of 1.5.

1 ?