fore the power, with the index of the root against it. Thus, the square root of 25 is expressed/25, and the cube root of 64 is expressed/364; and the 5th root of 16807,516807. The index to the square root is always omitted; the character only, being placed before it; thus, 16, the index 2, being omitted. When the power is expressed by several numbers, with the sign+, or, between, a line is drawn from the top of the sign over all the parts of it; thus the square root of 41-5, is/41-5, or thus,√ (41-5,) enclosing the numbers in a parenthesis. But all roots are now frequently distinguished by fractional indices; thus, the square root of 8, is 8, the cube root of 64 is 643, and the square root of 41-5, is 41—5' or (41—5). EXTRACTION OF THE SQUARE ROOT. Extracting the square root of any given number, is finding a number which multiplied by itself, would produce the given number; consequently multiplying the root into itself, is a proof of the work. RULE. Divide the given number into periods of two figures each, by setting a point over the place of units, another over the place of hundreds, and so on, over every second figure, both to the left hand in integers, and to the right hand in decimals. Find the greatest square in the first period, on the left hand, and set its root, on the right hand of the given number, (after the manner of a quotient in division,) for the first figure of the root, and the square number, under the period, and subtract it therefrom, and to the remainder bring down the two figures of the next following period, for a dividend. Place twice the root, already found, on the left hand of the dividend for a divisor. Seek how often the divisor is contained in the dividend, (exclusive of the right hand figure,) and place the figure in the root, for the second figure of it, and likewise, on the right hand of the divisor; multiply the divisor, with the last figure annexed, by the last placed in the root, and subtract the product from the dividend; to the remainder, join the next period for a new dividend. Double the figures already found in the root, for a new divisor, (or bring down your last divisor for a new one, doubling the right hand figure of it,) and from these find the next figure in the root, as last directed, and continue the operation in the same manner, till you have brought down all the periods. EXAMPLES. 1. What is the square root, or side of a square containing 36 square feet? Ans. 6 feet. FIG. I 6 feet. 6 feet. Operation. 36(6 Root. 36 6 36 Proof. DEM. The rule for the extraction of the sq. root, will appear obvious by attending to the proocess, by which any number is raised to the second power, or square.To find the 2d power of any number, we multiply the given number, or square root, by itself; therefore, to obtain the root from the power, we must inquire what number multiplied by itself, will produce the given number; we find on inquiring that 6 multiplied by itself will produce the given number; therefore we place 6 in the quotient, for the root, and multiply it by itself, placing the product directly below the given number; we find that it equals the given number, consequently, 6 feet is the square root or side of a square, containing 36 square feet; which must appear plain to the student on inspecting Fig. I., because he will readily discover, that the figure contains 36 square feet, which is the 2d power of 6 feet, the square root or length of one of its sides. 1 ft. NOTE.-There is a difference between square rods, square yards, and square feet, &c., and rods square, yards square, feet square, &c., thus: Three square feet may be represented by a Figure 3 feet long, and and one foot wide. And 3 feet square may be represented by a Figure 3 feet long, and 3 feet square are equal to 9 square feet. Such a Figure may be easily conceived of, by supposing two tiers of squares precisely like the one annexed, to be added to the bottom, which would give 9 equal squares. 3 feet. 2. A gentleman desirous of fencing into one square lot 625 square rods of land, wishes to know the length of one of the sides. 625(25 rods, the length of one side, or the square root of 625 square rods. 4 45)225 225 0 DEM.--First, as our rule directs, we place a period (.) over the unit figure of our given number, and then one over the second figure beyond it. This we do, because any one figure multiplied by itself, will never produce more than two figures in the product. These periods also show, that our root will consist of as many figures as we have periods over the given number. We next seek the largest square in the left hand period, which we find to be 4, placing it under the first period, and writing its root, 2, in the quotient, for the first figure of the root. We then subtract the 4, the square of the root already found, from 6, the first period; and to the right hand of the remainder, 2, we bring down 25, the next period, making 225 for our next dividend. Now it must appear obvious on inspecting the annexed Figure, that, at all stages of the work, the quotient expresses the side of a square formed from what has been subtracted from the dividend; therefore our quotient, 2, must express the side of a square inade up from what has been subtracted from our given number, 625; but the 2 in the quotient properly stands in the place of tens, because we must have another figure in the quotient, at the right hand of the 2, which gives it the local value of 20; consequently we may call the first figure of the. root 20, which expresses one side of the square rstu; therefore 20, the root, multiplied into itself, must give the number of square rods or area contained in the square r s t u already formed from what has been subtracted from the first period of our dividend. This will appear plain on inspecting the Figurer stu, 5=25 r. each side of which is 20 rods, as expressed by the root already. found; then the side of this square multiplied into itself, or the root already found, must give the number of square rods contain+ed in the square r s t u, which has been subtracted because 20×20=400, the number of square rods in the square r s t u, which is just equal to the number subtracted from the dividend; for the 4 under 6, the first period, stands in the place of hundreds, consequently it expresses 400. Now it is ev 20X5=100 20 rods C 5-25 r. ident, that 400 square rods of our dividend, are disposed of, in forming the square rstu, each side of which is 20 rods; and we have now left of our dividend 225 square rods, which we must so dispose of, as to keep our Figure in a square form after the additions are made, and also that our root may express one of the sides of the square after the Figure is completed. And in order to preserve our Figure r s t u, in a square form, the additions must be made on two sides; therefore, as our rule directs, we double the root, and place it at the left hand of 225, our dividend, for an imperfect divisor; imperfect, because another figure must be placed at the right of it, to complete the divisor.— We double the root for a divisor because the additions must be made on two sides of our Figure r s t u, to preserve it in a square form, each side being now 20 rods in length, the additions on the two sides must be 40 rods. And after doubling our root for a divisor, it expresses 40 in the divisor, because to complete our divisor another figure must be placed at the right of it; therefore, double the root is just equal to the additions which must be made on the two sides of our Figure r s t u. We next inquire how often our divisor 4, is contained in our dividend exclusive of the right hand figure; (we except the right hand figure of the dividend, because we have another figure to place in the divisor,) and we find that 4 is contained in 22, five times; we. then write five in the quotient for the second figure of the root, and also place the 5, at the right hand of the 4, in the divisor; we then have 45 for a divisor which is the whole length of the additions which must be made around the Figure r s t u, to preserve it in a perfect square form; because when we made the two first additions, that is, the long squares A and C, which equal in length the two sides of our first square r s t u, there was left a vacancy which is supplied by the little square, B; our divisor, 45, is then the whole length of A, B and C, and dividing by the length, our second quotient figure must express the width of the additions. We next multiply 45, our divisor, by 5, the last figure in the root, placing the product under 225, the remaining part of the dividend, which we find it equals, consequently we have no remainder. Our work is now finished, and it is plain, on inspecting the Figure, that the first figure of the root shows the length of one of the sides of the square r s t u, and the 2nd figure of the root, shows the width of the additions made to our first square, consequently 25, the root shows the length of one of the sides of the square after it is completed. The work may be proved by multiplying the root by itself, that is, one of the sides of the square by itself; or it may be proved by adding in one sum, the area of the different parts of the Figure, and if it be equal to the given sum, the work is right. rds. Proof by adding the parts. 20 20 Proof by raising the root to the second power or square. 400-Area of the square r s t u. 20×5=100=Area of the long square A. 20×5=100=Area of the long square C. 5X5 25 Area of the little square B. 25 rds. 25 125 50 625 Rods, area of the Fig. completed. 625 Rods, a number equal to the given sum. 3. A man has 26896 square rods of meadow ground; he wishes to know the side of a square equal in area to it. Proof, by adding the parts. 100X100=10000=Area of A. 100, 60, , 60, 160, 4, 60 3600 do. 640 do. 640 do. 160 4 , 2 B. C. n. D. E. NOTE.--We double the right hand figure of our last visor in order to get the divisor in forming a new dilength of the two sides of the little square placed in the corner, because the two additions to be made, must extend on two sides of the little square placed in the corner The attentive student must now discover on inspecting 26896 The area of this and the two first Figures whole Figure. under this rule, that for every figure in the root after the first, there must be an addition made around the square, and that the area of the Figure may always be calculated, by adding the area of the different parts of the Figure, giving to the figures representing the different parts their proper local value. 4. What is the square root of 81? Ans. 9 Ans. 36 Ans. 256. Ans. 654. Ans. 4802. |