Log. r2. Log. tan. z. .... 9.9119523 Sum... Log. tan. u.... (tan. 2)=tan. u, Sum. Therefore u 2u-86° 7′ 50′′ .20.0000000 Log. x..... .3)29.9119523. Consequently = .3313139, the positive root of the equation, .9.9706507 as required. 43° 4′ 55′′, and 2. Given 3-3x=1, which is an equation falling under the ir reducible case, to find its three roots. Here, a being 3, and 61, we shall have, by taking the ra 9.5202890 Log. r... 10.0000000 1.5202890 36 3 dius r=1, cos. z= 2a a Therefore the three roots are 1.8793852, -1.5320888, and -.3472964.. And if the equation be x3-8x= -1, the three roots or values of x are -1.8793852, 1.5320888, and .3472964. Which are the negatives of the roots of the former equation. 1. Given 23-2x=-2, to find the root of the equation, or the value of x. See art. 94, p. 214. Ans. x=- -1.7693. 2. Given 3-300x=-1000, to find the root of the equation, or the value of x. See index. Ans. x 3.472964. 3. Given -9x=9, to find all the three roots of the equation, or values of x. Ans. 3.411474, -2.226682, and -1.184792. 4. Given x3-x2-2x+1=0, to find all the three roots of the equation, or the three values of x. Ans. 2 cos. n, —2 cos. In, and 2 cos. n, where n=180°, and rad. = 1. 98. This method of cubic equations by converging series, which in some cases will be found more convenient in practice than either of the former, consists in substituting the numeral parts of the given equations in the place of the literal one, in one of the following general formulæ to which it belongs; and then collecting as many terms of the series as are sufficient for determining the value of the unknown quantity to the degree of exactness required. 1. Given x+ax=b. 26 ✔{2(27b2+4a3) {1+6.9′2762+-4a 1., x 26 or z 2.5.8.11.14.17 2762 2.5 2762 2.5.8.11 2762 {1+6.9′276*+4a3)+6.9.12.15 6.9.12.15.18.21′2782+4a2)°+&c.} 276+4)+6.9.12.15.18.21 276+4)+&c.} 2.5 2762 8.11 2762 14.17 2762 √{2(2762+4a3)}' {1+6.92703 +420 + 4+12.15 20.23 2762 12.15 276+4)+18.21 270+4+24.27276+4a3) D In which case, as well as in the following ones, A, B, C, &c. denote the terms immediately preceding those in which they are first found. 2 Given x-ax±6, where 62 is supposed to be greater than a3 or 2762-4a3. 2 2762-4a3 2762 2.5.8 2762—4a3 2.5.8.11.14 2762 ·)· -)2 2762 -)B 9.12 2762 3.6.9.12.15.18 11.14,2762-4a3, 15.18 2762 2762 -)C⋅ )-&c.}, or 17.20.2762-4a3. 21.24 2762 )D-&c. 3.6 3.6 In which case the upper sign must be taken when b is positive, and the under sign when it is negative; and the same for the first root in the two following cases. 3. Given 23 ax= 6, where 62 is supposed to be less than a3, or 2762×4a3. Which series answers to the irreducible case, and must be used when 4a3 is less than 2762. root thus found be put = r, the other two roots may be expressed as follows: √(4a3—2762) 2.5,4a3-27b2 2.5.8.11 4a3-2762, 2.5.8.11.14.17 4a3-27b2 9/262 {1 6.9 2762 6.9.12.15 2762 8.11 4a3-2762 3.6.9.12.15.18.21 14.17 4a3-2762 2762 {1 -)B -) C+ 18.22 2762 √(4a3-2762) 2.5,4a3-27b62 9/263 6.9 2762 -)4+12.152762 Wherer, or+r, must be taken according as b is positive or negative; and the double signs must in the other, as usual. 4. Given x3-ax±b, where 162 is still supposed to be less than a3, or 276-4a3. 26 4. x2 or} {2(4a3—2762)} {1 {1 2.5 2762 6.94a3 27626.9.12.15 4a_2762 2.5 2762 2.5.8.11 2762 8.11 .2762 14.17 2762 2762 {2(4a3-2762)} )B ;)4+ 12.15 4a3-2762 18.21 4a3-2762 6.9 4a3-2762' C+ 20.23 2762 24.27′4a3_2762) D-&c., which series also answers to the irreducible case, and must be used when 4a2 is greater than 2762. Where the signs are to be taken as in the latter part of the preceding case. Of Biquadratic Equations. 99. A biquadratic equation, as before observed, is one that rises to the fourth power, or that is of the general form x+ax+bx2+ax+d=0, the root of which may be determined by means of the following formula, substituting the numbers of the given equation, with their proper signs, in the places of the literal coeffi cients a, b, c, d. RULE I. Find the value of z in the cubic equation z+(acb2 d) z = ròs b3 +(c2 + da2)-24b(ac+8d) by one of the former rules; and let the root, thus determined, be denoted by r Then find the two values of x in each of the following quadratic equations, x2 + ( {a + √ { { a2+2(r—}b)}) x=(r+fb)+~{(r+{b)2 —d} x2 + ( a −√ {{ a2+2(r—} b)}). ~~(r+zb)-—~ {(r+}b)2 —d}, and they will be the four roots of the biquadratic equation required. Or, if the equation be of the more commodious form, x2+bx2+cx+d=0, to which it can always be reduced, by taking away its second term, it may be resolved thus: RULE II. Find the value of z in the cubic equation z3( 1 1⁄2 b2 + d) z=1863+c2—}bd, and let the root thus determined be denoted by r. Then find the two values of x, in each of the two following quadratic equations, x2+√{2(r—}b)}x=—(r+žb)+✔{(r+¿b)2—d} 22 = √ {2(r—b) { x=(r+86)=√{(r+86)2_d and they will be the four roots of the biquadratic equation required. 1. Given the equation x-103+35x2-50x+24―0, to find its roots. Here a =—10, b=35, c=—50, and d—24; whence, by substituting these numbers in the cubic equation z3 + (fac—b2 —d)z=18gb3+{{(c2+da2)—24 (ac+8d), we shall have the following reduced equation, 2132 = 1888, 1z which being resolved, according to the rule before laid down for that purpose, gives z={(35+18—3)+(35—18/—3)}. But, by the rule for binomial surds, given in Art. I, Case 2, in the former part of the work, (35+18/—3)=3+}√—3, and (35-18-3)=√-3; wherefore z}+{√−3+}{√—3}=7. And if this number be substituted for r,-10 for a, 35 for b, and 24 for d, in the two quadratic equations, x2+ [a+ √ {{a2+2(r—}b)}]x=(r+{b)+~{(r+{b)2—-d}, x2+[}a — √ {{a2+2(r—}b)}]x——(r+¿b)—~ { (r+ {b)2 —d}, they will become, after reducing them to their most simple terms, x3-3x-2, and x2-7x-12; from the first of which —— +1+2 or 1, and from the second }±√}=}=} 4 or 3; whence the four roots of the given equation are 1, 2, 3, and 4. 2. Given +12x-17-0, to find the four roots of the equation. Here a=0, b0, c12, and d--17: Whence, by substituting these numbers in the equation z3( { b2 + d) x = 1 8 8 63+c2-bd, we shall have, after simplifying the result, z+17-18, where it is evident, by inspection, that z=1. And if this number be substituted for r, 0 for b, and -17 for a, in the two quadratic equations x2+√{2(r—{b)}x=—(r+{b)+~{(r+fb)2—d}, x2—√{2(r—{b)}x——(r+{b)—~ {(r+ {b)2 —d}, they will become, after reducing them, in the usual manner, to their most simple forms, 1+3/2, and x-2x-1-3/2, which being resolved, according to the general rule, we shall have, {√2+√(−1+√18)=—{√2+√(−1+3√2), √ √2√(+18)=√√√2—√(−1+3√2), x=+ {√2 + √(−{−√18)=+{√2+√(—1—3/2), +√√2 √(√√18)=√2—√(—1—3√√2), which are the four roots of the proposed equation. Where it is to be observed that the first two are real, and the two latter imaginary. RULE III. The roots of any biquadratic equation of the form x2+ax2+bx+c=0, may also be determined by the following, general formulæ first given by Euler, which are remarkable for their elegance and simplicity. Find the three roots of the cubic equation z+2az2+(a2—4c)z 62, by one of the former rules, before given for this purpose; and let them be denoted by r', r'', and r'''. = When bis positive, Then we shall have When b is negative, +wr+wr'+~~""" 101. If the three roots r', r', r'"', of the auxiliary cubic equation be all real and positive, the four roots of the proposed equation will, also, be real; and if one of these roots be positive, and the other two imaginary, or both of them negative, and equal to each other, two of the roots of the given equation will be real, and two imaginary; which are the only cases that produce real results. 3. Given 1-25x2+60x—36—0, to find the four roots of the equation. Here a=25, b-60, and c-36; whence, by substituting these values for their equals, in the cubic equation above given, we shall have z3-2×2522+(252+4×36)z=602, or z— 50z2+769z=3600; the three roots of which last equation, as found by trial, or by one of the former rules, are 9, 16, and 25, respectively; whence 2=1792/16/25)=(-345)6, |