2.6180339887, and .3819660112; which two values of x are the reciprocals of each other. From the latter, viz. x2+1=o, we obtain x=±√−1, or + √−1, and −√-1, for the two remaining values of x. 2. Let x3-1=o be given, to find the values of x. Here it is plain that +1 is a root, or x-1=0, wherefore dividing the given equation by this, we have (2x)= = = = =) x2 +x+1=0, X 3. Given 23+1=0, to find the values of x. Ans. —1,1+√−3, 2 4. Let the equations x-1=0, x2+1==0, x3—1=0, and x+ 1=0, be proposed, to find the values of x in each. Literal equations, wherein the given quantity and the unknown one are alike affected, may be reduced to others of fewer dimensions, by the following rules. 52. When the given equation is of even dimensions. RULE I. Divide the equation by the equal powers of its two quantities in the middle term. II. Assume a new equation, by putting some letter equal to the sum of the quotients arising from the division of the given and unknown quantity, alternately, by each other. III. Substitute in the former equation the values of its terms found by the latter, and an equation will arise of half the dimensions of the given one, from the solution of which the roots of the given equation may be determined. EXAMPLES.-1. Required the roots of x-4 ax3+5 a2x2 — 4a3x+a+=o? First, dividing the whole equation by the equal powers in the x2 4x 4a a2 middle term, it becomes (- + +5- + =0; or, which is the 2 a2 a a2 ing, x 2 in the equation a +2=x2, and by substituting z and z for their values + 4. + +5=0, it becomes z'—4z+3 x2 a =0, whence z=3, or 1; but since ==+=== =z, if the former value a X -=3; whence x2—3 ax=— a2, which solved, ·3±√5=) 2.618034 a, or .381966 a. But if the 2. Given 7 x6-26 ax3—26 a3x+7a=o, to find the values These values substituted as before, we obtain 7 z3 — 26 z2 —21 z +52=0, one root of which (by Art. 47.) is 4, and by means of this, the equation may be depressed to the quadratic 7 z2+2z-13 =0, (Art. 32.) the two roots of which are +1.2273804, and +, or x2-azx=- - a2, by 1.5130947. Wherefore, since z= three values of z be successively substituted, the six roots of the given equation will be obtained. 3. To find the roots of a +6 ax3 - 20 a2x2+6 a3x+a+=o. 4. To find the roots of x4-20 ax3 + 12 a2x2 - 20 a3x+a1=o. 5. Required the roots of x-ax1 —aax+a¤—o. 53. When the given equation is of odd dimensions. RULE. Divide the equation by the sum of the known and unknown quantities, and proceed as before. EXAMPLES.-1. Given 5-3 ax2+6. a3 x3 + 6 a3 x2 −3 a1x+a3, to find the roots. First, dividing by x+a, the quotient is xa — 4 x3a+10 x3a2 4 xa3+a+=o; wherefore dividing this by x3a2, according to the x2 a2 x α preceding rule, the quotient is + 4. + +10=0; let z a x2 as વર્ x 2 α x + then z2===+=+2, and substituting these va e lues as before, z2-4 z+6=0; whence z=2±√2; but since and substituting for z its values found above, we obtain four of the roots, which together with a, (since x+a=o,) make up the five roots of the equation. 2. Given x3-ax3 — a3x+a3 —0, to find the roots. Ans. a, 3. Required the roots of x5 +4 ax1 — 12 a3 r3 — 12 a3 x2 + 4 a*x +a5=o? 4. To find the roots of x-ax1—a*x+a7=0. CARDAN'S RULE FOR CUBIC EQUATIONS. 54. Let x+ax=b be any cubic equation, wanting its second term; it is required to find one of its roots, according to Cardan's method P. P This rule bears Cardan's name from the circumstance of his having been the first who published it, namely at Milan in 1545, in a work entitled, Års Magna: but it was invented first, in or about the year 1505, by Scipio Ferreus, Professor of Mathematics at Bononia; and afterwards, viz. in 1535, by Nicholas Tartalea, a respectable mathematician of Brescia: from the latter Cardan contrived to extract the secret, which he afterwards published in violation of the most solemn protestations. The rules which Cardan thus obtained were for the three cases x3 + bx=c, x3=bx+ c, and x3 + c=bx ; and it must be acknowledged in justice to him, that he greatly improved them, extending them to all forms and varieties of cubic equations, in a manner highly creditable to his abilities as a mathematician. See Tartalea's Quæsiti et Inventioni diverse, ch. 9. Bossut's Hist. of the Math. p. 207. Montucla's Hist. des Muth. t. 1. p. 591. Dr. Hutton's Math. Dict. vol. 1. p. 68-77. The root obtained by this method is always real, although not always the greatest root of the equation: and it is remarkable, that this rule always exhibits the root under an imaginary form, when all the roots of the equation are real; and under a real form, when two of the roots are imaginary. See Dr. Hutton's Paper on Cubic Equations, in the Philosoph. Trans, for 1780. Assume y+zx, and 3 yz= -- a; substitute these values for r and a in the proposed equation, it becomes (y3 +3 y1z+3yz* +z3+a.y+z=y3 + z3 +3 yz.y+z+a.y+z=y3+z3—a.y+z+a •Y+z=) y3+z3=b; from the square of this take four times a 3 the cube of yz=- and the result is yo—2y3z3 +z°=b3 + the 4a3 4 as 27 square root of which is y3 —z3 = √62+· ·; but y3+z3=b; wherefore the sum and difference of these two equations being dan's theorem: but the rule may be exhibited in a form rather more convenient for practice; thus, because z= 55. RULE I. If the given equation have all its terms, let the second term be taken away by Art. 37. II. Instead of a and b in either of the above general theorems, substitute the coefficients of the corresponding terms, with their proper signs, in the transformed equation; then, proceeding according to the theorem, the root will be obtained. 1 If a be negative, and a greater than 6, the root 27 cannot be found by this rule ". 4 This is called the Irreducible Case; it exhibits the root, although real, under an impossible form: thus the root of the equation x3—15 x=4 is 4, but by Cardan's rule it is3 √2+ √−121 + 3 √2— √ 1 121, an impossible form. EXAMPLES.-1. Given x3 +6x=88, to find the value of x. Let the cube root of each of these imaginary expressions be extracted, they become 2+/-1+2−√ − 1, which being added together, the impossible parts destroy each other, and the sum is 4, agreeably to what has been observed. It is remarkable, that this case never occurs except when the equation has three real roots, as we have before observed. The irreducible case has exercised the abilities of the greatest algebraists for these three hundred years past, but its solution still remains among the desiderata in science. Dr. Wallis thought he had discovered a general rule, but it was afterwards found to apply only to particular cases. Baron Maseres gave a series, which he deduced by a laborious train of algebraic reasoning from Newton's Binomial Theorem, whereby this case is resolved without the intervention of either negative or impossible quantities. Dr. Hutton has likewise discovered several series applicable to the solution: (see Philos. Trans. vol. 68. and 70.) other series for this purpose may be seen in Clairault's Algebra, p. 5. Art. 19. Bossut's Algebra, Art. 178-9. Landen's Lucubrations, La Caille's Leçons de Math. Art. 399. &c. Lorgna's Memoirs of the Italian Academy, t. 1. p. 707. &c. The irreducible case may be easily solved by trigonometry; as early as 1579, Bombelli shewed that angles are trisected by the resolution of a cubic equation. Vieta, in 1615, shewed how to resolve cubics and higher equations by angular sections. In 1629, Albert Girard solved the irreducible case by a table of sines, giving a geometrical construction of the problem, and exhibiting the roots by means of the hyperbola and circle. Halley, De Moivre, Emerson, Simpson, Crakelt, Cagnoli, Wales, Maskelyne, Thacker, &c. have employed the same method of sines: and lastly, Mr. Bonnycastle, Professor of the Mathematics at the Royal Military Academy, Woolwich, has communicated additional observations on the irreducible case, and an improved solution by a table of natural sines. See Hutton's Math. Dict, vol. 2. p. 743-4. When one root is obtained by Cardan's rule, the two other roots may be derived not only by depressing the equation, as in ex. 1. but likewise as follows: let r= Cardan's root, and v and w=the two other roots, then will v+w= −r, |