Let x+A-1 +A,,x2+A,x=N, be a numerical equation which contains a real and incommensurable root, and let the first figure of this root (found by Sturm's Theorem, or any of the other known methods) be r, making x=x+r, and transforming the equation in æ into an equation in x', .... " or, transposing A and making N-A=N'," +A'x'2+A2x2+A=N; +A_ ̧‚x2+A'x'=N'. One root of this equation is composed of the remaining figures of that root whose first figure is r. Let be the first figure of the root x' of this equation; and let r′ be substituted for x' in the equation X'=N', and, conversely, if the polynomial in r' is made the divisor, the quotient is r'; r' being unknown, all the terms of this polynomial except the last, A', are also unknown. But since r=10r', and A'=mrTM-1+(m−1)A ̧TMTM−2+, &c., it is evident that the last term, A', must form a great part of the numerical value of the whole polynomial divisor in r'. The coefficient A of the first power of ' may, therefore, be employed as a trial divisor to find the value of r', the second figure of the root. By substituting this value for the symbol in the polynomial it can be ascertained whether this is or is not an accurate determination of the second figure of the root x. Making, in like manner, a'=x"+r', transforming the equation X'=N' into an equation in x", viz., +A" x2+A_x"+A,,, " transposing A,, denoting N'-A,, by N", and dividing N" by A, the coefficient of the first power of r", the first figure of the root r", and by consequence the third figure of the root x, is determined, &c. This method resembles the processes for the extraction of the square and cube root; and the coefficient of the first power of x', x", &c. is employed for the same purpose as 2a in the extraction of the square, and 3a2 in the extraction of the cube, root. 288. Let it be required to find, by this method, a positive root of the equation, x++3x+2x2+6x=148.6. The first figure of the root is found to be 2; making x=x+2, and effecting the transformation, x2++11x3+44x^2+82x'=88·6. The first figure of the root r' cannot exceed 9, otherwise the first figure of x must be greater than 2; therefore the quotient of 88-6+82-1 is too great. Substituting successively 9, 8, 7 for x' in the equation X'=88-6, it is found that 7 is the first figure of x', and consequently that the first and second figures of x are 2.7. Making x'=x"+7 and effecting the transformation, x+13·8x+70·04x2+161·142x"=5·6269. Dividing 5.6269 by 161 142 the quotient is 03, which is the correct figure. (The accuracy of this figure may be tested as that of the second figure of the root has been; it will, however, be found that the results given by the trial divisor after the second figure of the root are in general accurate.) Making "=""+03, and effecting the transformation, x+13.92x3+71·2874x2. +165.381768x=72923059. Dividing ⚫72923059 by 165-381768, the quotient is '004, which is the first figure of the root "" and the fourth of x. Making ""="""+004, and effecting the transformation, x'""'*+13·936x''''3+71·454536x''/2+165·952745616x'""'='066562038464. The first figure of x found from N""" and the coefficient of x'""' is 0004; and as the fifth and sixth decimal figures of the root are zeros, the value x 2.7344 is a very near approximation. The calculation of the transformations is annexed. In practice it will not be found necessary to preserve all the decimal figures in the coefficients of the transformed equations. It appears, in the annexed calculation, that after the third transformation the trial divisors or coefficients of the first powers of x'", x'"", &c. are nearly constant, and that consequently the last figures of the root may be found by division. In this example, if 72923059 is divided by 165.381768, the quotient is 004409, a result which differs by less than from that already 1 100000 1 +13.936 +71-454536 +165·952745616 =066562038464(0004 K K Required one value of x in each of the following equations? 1st. x3-12x2+45x-53=0?.............. 2d. x3-12x-28=0?............... 3d. x-12x2+12x=3 ?................ 4th. x+4x-4x2-11x+4=0?... .......Ans. x=5·879385+. .................Ans. x=4·30213+. 5th. x5+6x-10x3-112x2-207x-110=0?........Ans. x=44641016+. 6th. x6+2x+3x2+4x3+5x2+6x-654321=0?...Ans. x=8.9569795+. CARDAN'S RULE FOR THE RESOLUTION OF A CUBIC EQUATION. If in this reduced equation y+z is substituted for x, (y+2)+q(y+2)+r=0; or y3+3yz(y+2)+23+q(y+2)+r=0; or y+(3yz+q) (y + z)+23+r=0. In order that the second term of this equation may be reduced to zero, assume 3yz+q=0, Multiplying both terms of the second member of this equation by y and z differ only in respect of the sign prefixed to the radical. q3 + √ √ 2 1 + 27 4+27+ 3 + If x3-qx+r be the reduced equation, the value of x is This solution consequently appears under the form of the sum of two impossible quantities, expressible by Whence the sum of these quantities is possible. DES CARTES'S RULE FOR THE RESOLUTION OF A BIQUADRATIC EQUATION. 290. By rendering the coefficient of the first term 1, and the coefficient of the second term zero, the equation of the fourth degree is reduced to the form x++qx2+rx+8=0. This equation may be supposed the product of the two quadratic factors x2+ux+v, x2-ux+z, the coefficients of r being +u and -u, since the coefficient of r3 in the product of these factors is zero. Forming the product of these factors, and making it equal to the proposed equation, *+q+r+s=r^+(b+z−t) °+(uz−z) + Therefore (Art. 174) q=v+z-u2 and by subtraction »=3(9+u—") Multiplying equations 4 and 5, vz=4((q+u2)2 — *2), but, equation 3, vz=s; { ((9+u2)2 ——~~2), 4s=q2+2qu2+u*— u2 and 4us qu2+2qu*+u-r2, or u+2qu++(q2-4s)u2—r2=0. 5. This, which is a cubic equation in u2, can be resolved by Article 289. The value of u being known, the value of z is given by equation 4, and the value of v by equation 5. Lastly, u, z, v being known, two values of x are given by the equation x2+ux+v=0, and the two others by the equation x2-ux+z=0. No method of resolving literal equations of a higher degree than the fourth has been discovered. FINIS. LONDON: Printed by A. SPOTTISWOODE, |