term, and bring down the two next terms to the remainder for a dividend. 3. Divide the dividend by double the root, and set the result both in the quotient and divisor. 4. Multiply the divisor thus increased by the sum last put in the quotient, and subtract the product from the dividend, and so on as in common arithmetic. According to this rule, let us now extract the square root from the compound quantity : 4a+12 a3 x + 13 á2x2 + 6 a x3 + xa (2 a2+3 ax+x2 sq. root 4a+ Let us now work this sum according to the improved theorem, and mark the difference: Rule. 1. Arrange the compound quantity according to the dimensions of some letter, and set the root of the first term in the quotient underneath. 2. Multiply the root by 2, and divide the second term by the product, placing the quotient under the second term. 3. Multiply the last quotient by 2, and divide the third term by the product, placing the quotient under the third term. - : 4. Square the last quotient, and by the product divide the last term; if nothing remain, the square number is measured by the square root thus found. Like signs give +, unlike signs This rule answers for quadrinomials, pentanomials, and hexanomials; for by cutting off the mediate terms, as may appear from the following examples, pentanomials and hexanomials are reduced to quadrinomials. In trinomials the second or last term must be squared. According to this rule, extract the square root from a* z* ] + 0 square root. 4 a + 12 a3 x + 13 a2x2 + 6 a x + x+ 2 a2 + 3 ax Here the root of the first term 4 at is 2 a2, which we place in the quotient under the first term: secondly, we multiply this root by 2, giving for a product + 4a2, by which we divide the second term + 12 a3 x. The quotient + 3 a r is placed under the second term. We now pass over the mediate term + 13 aa xa placed within crotchets: and, thirdly, we multiply the last quotient + 3 ax by 2; the product is + 6 a x, by which we divide the third term + 6 a x3; the quotient + is placed under the third term fourthly, we square + x2 = + x2, by which we divide + x; nothing remains; therefore the quantity is truly measured by the square root 2 a2 + 3 a x + x2. Mathematicians must at once see the many advantages this simple theorem has over the complex one at present in use, especially where the higher geometry is connected with algebra. We remark that by the old method we are obliged, as in the foregoing example, to employ six lines, or 24 terms; in the other only two lines, or eight terms; the saving in time and calculation is, therefore, immense; for in fact we change compound into simple division. Let us now work a few sums according to this method, beginning with a trinomial. Extract the square root from Here the root of the first term at is a", which we place in the quotient: secondly, we multiply this root by 2, giving for a product + 2 a, by which we divide the second term - 4 a2x2; the quotient is 2 r, which we place under the second term: thirdly, as this sum is a trinomial, we square the last quotient -2x, giving + 4 x, which, subtracted from 4 x, leaves nothing; therefore the square root is a2 - 2x2 + 0. Extract the square root from the following quadrinomial: Here the root of the first term at is a2, which we place in the quotient: secondly, we multiply this root by 2, giving for a product 2 a2, by which we divide the second term 4 as b; the quotient is 2 a b, which we place under the second term: thirdly, 2 ab x 2 = 4 a band + 8 a b3 ÷ 4 a b gives 2 b for the third term. Lastly, 2 b2 × 2 b2 = + 4 b*, which subtracted from the last term leaves nothing. Therefore the root is a2 2 ab square 262 0. This last example is taken from the Philosophy of Arithmetic, an elementary work of considerable merit, by Mr. John Walker, late Fellow of Trinity College, Dublin. Extract the square root of the following pentanomial: a2 + 4 a3 x 1 + 6 a2 x2 + 4 ax2 + x2 a2 + 2 a x + x2+0 square root. Extract the square root from the following pentanomial, and prove it by involution: 24 4 23 2 x O square root Should the operator wish to save himself the tedious multiplications consequent to this proof by involution, he may either substitute a letter according to the binomial theory of Newton, or he may use the following more simple and concise method: Multiply the first by the last term of the square root, and double the product: secondly, square the mediate term of the root, and subtract from the mediate terms of the square quantity; if nothing remain, we may be certain the calculation is correct. Thus in this last example we multiply the first term of the root+x by the last term + 8 = + 8 x2, which doubled = + 16 x2, and the mediate term 2 x squared = + 4x2, which added = + 20 rs; subtracted, leave nothing. The advantage of this method may be well illustrated by extracting the square root according to the rule at present recommended by authors, and then involving the root back again to the square quantity, when it will be found that, according to the theorem herein recommended, the evolution and involution may be performed in two lines, instead of 12, required by the other. This, independent of its simplicity and accuracy, must, I presume, recommend its adoption. 4a + 8 b x + 4 br Here to measure the mediate terms + a x + 2 b = 2 a b and+2 a b doubled = 4 a b, and + 2x squared = 4 x", which subtracted from the mediate terms of the compound quantity leaves nothing. The square root of a pentanomial or a hexanomial never exceeds a trinomial. In a pentanomial the root is extracted from the first, second, and fourth terms: in a hexanomial, from the first, second, and fifth terms. Indeed a pentanomial is nothing more than an abbreviated hexanomial. Extract the square root of the following hexanomial: For the mediate terms ar doubled 2 and + 2 b squared ≈ 4 ba, which subtracted leaves nothing, For the mediate terms + a x + b2 = a be, which doubled = 2 a b2, and - 4 a b squared = 16 a b2, which subtracted leave nothing. For the mediate terms a x ̧x2 = + a2 x2, which doubled = 2 a2 x2, and a x squared = + a2 x2, which added = + 3 a2x2. Some mathematicians, sensible of the tediousness by evolution, have advised the extraction of the roots of the most simple terms, connecting them together by the signs+ or -, as may be judged most fit, then involving this compound root to the proper power; if it be the same with the giyen quantity, the square root is found as in the following example: a +2ab+ 2 ax + b2 + 2bc+c. According to this irregular method, like sailing without a compass, the first, fourth, and sixth terms, are supposed to give the square root; whereas in reality it is extracted from the first, second, and fifth terms, as may be seen according to the following working: as + 2 a b + 2 a c + b2 | + 2 b ¢ + ce a + For the mediate terms a x ca e, which doubled = + 2 ac, and b squared = + b2. Some may prefer the following method of working these sums. Extract the square root from laR For the mediate terms + ax + b2 = a2 ba, which doubled =2 a b3, and ab squared + a b3, which added = + 3 a b', exactly measuring the mediate quantity. Sir, I remain your obedient servant, SHOULD the following lemmas and propositions merit a place in your Annals of Philosophy, your inserting them therein will much oblige Your humble servant, JAMES ADAMS. Lemma I. The angles made by ordinates and tangents at different points of a curve are unequal, those being less that are nearest the vertex. Let A F B be a curve of continued curvature; A H, F G, ordinates to the abscissa BH; and let the tangents A D, FC, to any two points, A, F, on the same side of the vertex B, intersect each other in the point E, and the tangent BI in the points D and C. From this construction it will appear that the external angle, ADI, is greater than the internal angle, ICE; that is, the angle D A É greater than the angle CFG. Q. E. D. Lemma II. The arc of a circle is less than its corresponding tangent. A B, draw the tangent A D, and join C D. Per mensuration CA × A D = the area of the triangle CAD, and CA x A B area of the sector CAB; therefore, area of triangle C A D area of B A sector CAB AD: A B. But the triangle is greater than the sector; therefore A D is greater than A B; that is, the |