Subtracting the second equation from the first, 3(p-p1)x2 + (q − 91)x + (r−r2)=0, a quadratic equation. XXVI. 1. Each equation is reducible to x3-18x2 - 225x-656=0. 3. α=- -9, b=36, c=-60. 4. Since x=3+8, .. also x=3-√√/8, and x2 − 6x+1=0, .'.x+(p+6)=0, (6p+36+q−1)x+(r−p−6)=0, from which ≈= −r. 5. The third root is if b3+c3 - a2c+ab2=0. a XXVII. 3. Here pq=r, this equation becomes x3-pa2+qx− pq=0, or x2(x − p)+q(x−p)=0, or (x−p)(x2+q). If a, b, c denote the roots of x3 − px2 + qx − r =0, it may be noted that since pq=r, (a+b+c)(ab+ac+b)=abc, from which may be deduced (a+b)(b+c)(c+a)=0. 4. One of the roots of x3-4x3 +6x-3=0 is +1. See Exercises IX. p. 29. The other two roots are (3±√3). 6. Let , then the equation becomes 3 +pz2 +qz+r=0, if p, q, r be put for b с d a a a y 9. Here (x2+ax+m) (x2 + bx+m)(x2 + cx+m) =x® + (a+b+c)x3 + (3m+ab + ac+bc)x4 + {2(a+b+c)m + abc ¦ x3 a+b+c=0, .* . a2 + b2 + c2 —— · 2 (ab+ac+bc)=6m, ... ab + ac+bc+3m=0. Hence x+abex3+m3. To solve this equation, x6 - 20x3 +343=0. Let x2=y, then the equation is reduced to a cubic y3-20y+343=0. 3 3 4 10. By reducing the equation and arranging the terms according to the descending powers of x, a,+a+a+a, is the coefficient of x3, and since а + a2+ a + a1 = 0, a2+a2 = − (a ̧ +α), and by means of this relation the coefficient of x2 can be reduced to a ̧b1+ɑb2+a ̧1 ̧+a ̧b=0, and the equation is thereby reduced to one of the first degree, which admits of only one value of x. 11. Since xy(x+y) +xz(x+z) +yz(y + z) = (x + y + z) (xy +xz+yz) − 3xyz, and xy(x2 + y2)+xz(x2+z2)+yz(y2 +z2) = (x2 + y2+z2)(xy+xz+yz) − xyz(x+y+z). The second and third equations by substitution become (x+y+z)(xy +xz+yz) −3xyz=168, and (x2 + y2+z2 )(xy +xz+yz) − xyz(x + y + z) = 538. Then by substitution and reduction (x+y+) 2 + } } (x + y + z)=52+, whence x+y+z is known, also xyz and xy+2z+yz. Hence the solution is reduced to a well-known form by putting x+y+=p, xy+xz+yz=q, xyz=r, and then eliminating two of the unknown quantities y,, the resulting equation is a cubic, x3 − px2 + qx − r = 0. 12. Let x, y, z denote the edges, then 2(xy+xz+yz) = 22, (x2 + y2+z2)i =√14, xyz-6. Eliminating y and z, z3 – 6z2 + 11x-6=0, from which z will be found to be 1, 2, 3 lineal inches respectively. Also x and y can be found when the value of z is known. 18. The criterion required for Cardan's rule to be applicable is, that be greater than 2; in this case, that (2abc) be greater than (a2+b2+c2)2; but (2abc)2 is 27 less than (a2+b2+c2)3 4 27 obviously, and consequently, the three roots of the equation x3 − (a2 + b2 + c2)x-2abc=0 are possible, two of which are negative and one positive. If a, b, c be supposed to be 1, 2, 3 respectively, the equation becomes x3-14x-12=0, which has all its roots real, and they lie between 0 and −1, -3 and 4, 4 and 5; and will be found to be 911, -3.201, 4.113 respectively. This problem occurs in Newton's Universal Arithmetic, Cap. 1, Sect. 4, of which he has given several solutions. The following one is taken as an illustration. Let AB, BC, CD denote three consecutive arcs of a semicircle, of which AD is the diameter. Join AC, BD. Then ABCD is a quadrilateral inscribed in a circle, and AC, BD are the diagonals. Then by Euc. VI. D. ACx BD=AB × CD + AD × BC. Let AB-a, BC=b, CD=c, AD=x. ··√(x2 —c2)× √(x2 — a2)=ac + bx, and x-a2x2-c2x2 + a2c2=a2c2 +2abcx+b2x2, also x1 − (a2 + b2 + c2 )x2 — 2abcx=0, .'. x3−(a2+b2+c2)x−2abc = 0. XXVIII. 2. Since the equations have two roots in common, each involves the same quadratic factor. 3. The limiting equation of x2+rx+8=0 is 4x3+r=0. The expressions x2+rx+s and 4x3 +r have a common factor, and the remainder is equal to zero. 4. Let a, b denote the equal roots, then (x-a)2 (x—b)2=0, or x1 − 2(a+b)x3 + (a2 + 4ab+b2)x2 — 2ab(a+b)x+a2b2-0, is an equation identical with x4-px3+qx−rx+8=0. The values of a and b may be found in terms of any two of the four coefficients p, q, r, s. 6. If there be equal roots in the equation, its limiting equation will contain one or more of them, and it will be found that x+13×3 +33x2 +31 +10 and 4x3 +39x2 +66x+31 have a common quadratic factor x2+2x+1, which indicates that the given equation contains three equal roots, each equal to -1, the fourth root being -10. It may be remarked that since the sum of the coefficients of the odd terms is equal to the sum of the coefficients of the even terms, the equation may be otherwise solved. XXIX. 2. Obviously, x is one of the factors. Removing this factor, and let x4-5x2+4=0. This is a quadratic equation of which the roots are ±1 and ±2, then x=0, x-1=0, x+1=0, x−2=0, x+2=0, .'. x(x − 1)(x+1)(x−2)(x+2)=x3 − 5x3 + 4x .. x, x-1, x+1, x−2, x+2 are the factors of x5-5x3 + x. 3. If q=r+1, g−1=r, and the equation can be put into the form x2+2x3 +x2 + (q−1)x2 + rx+8=0, or (x2+x)2+r(x2+x)+s=0, a quadratic equation. 4. Add 1 to each side, then (x+1)(x+2)(x+3)(x+4)+1 = (x + 1)2 + (x + 2)2 + (x + 3)2 + (x + 4)2+1, or (x2+5x+5)2 = 4(x2+5x+5)+11. 5. First find the roots common to the first and second equation, next those which are common to the first and third; and it will be found that 1, 2, 3, 4 are the roots of the biquadratic. XXX. 1. When the sum of two of the roots is equal to the sum of the other two, then p3 − 4pq+8r=0, and when the product of two roots is equal to the product of the other two, p2s=r2. In both cases the biquadratic can be solved by a quadratic. 2. The first equation can be reduced to (x2 - ax)2 — 2b2 (x2 — ax)=a2b2, and the second equation to 2x+8x-14x3- 44x2+6x=0. 7. When x-qx2 - rx-s is divided by x3 + ax + b, the quotient is x2 — ax+ a2 − b−q, and the remainder is — (a3 – 2ab - qa+r)x − b(a2 − b − q) – s. If x2+ax+b be one quadratic factor of x4 - qx2 - rx - s, then must x2 − ax+a2 −b−q be the other; and the two terms of which the remainder is composed must be respectively equal to zero. And then a3-(2b+ q)a+r=0, and b(a2 - b−q)—s=0. When a is eliminated, the resulting equation becomes a cubic involving 6 with q, r, s. XXXI. 1. When a reciprocal equation is of an odd number of dimensions, it has at least one root, +1 or -1. 2. (1) Let x3-pæ1 +qx3 − qx2 +px-1=0 be a recurring equation of five dimen. sions. It may be put into the form (x3 − 1) − рx(x3 − 1) + qx2 (x − 1)=0, .'. x—1=0, and x=1, one of the roots, and the equation when depressed becomes xa − ( p −1)x 3 + (q−p+1)x2 − (p−1)x+1=0; divide each term by x2, 1 Let x+ =, then 2-(p-1)=p-q+1, a quadratic by means of which the other roots can be determined. (2) Let x − px3 +qx1 —rx3 + qx2 −px+1=0, a recurring equation of six dimen sions. 1 Divide each term by xs, then x3 −px2+qx−r+q. —-p1 1 ..z3 — pz2 + (q − 3)≈ — (2p+r)=0, an equation of three dimensions. 5. Here x1+x2 + 1 = − (x3 + x), and (x1 +x2+1)2 = (x3+x)2, whence x2 +x€+x2+x2+1=0, let x2=y, then y1 + y3 + y2+y+1=0 is the equation whose roots are the squares of the roots of the given equation. It is obvious that the two equations are identical, and that the four roots of the transformed equation may be shewn to be equal to the squares of the four roots respectively of the given equation. 6. Form the equation whose roots shall be a, b, and shew that it is identical with the given equation when the latter is reduced to the form 7. Let the roots be increased by e; assume y=x+e, x=y—e, then y4-4(e-1)ys + 6e (e-2)y2 — 8(e2 — 1)y + (e1 - 4c3 +8e-4)=0 is the transformed equation. And in order that this equation may become a reciprocal equation, the coefficients of the first and last terms, as also the coefficients of the second and last term but one, must be equal; that is 4(e-1)=8(e2 - 1), and e* - 4e3 +8e-4=1. From 4(e-1)=8(e2-1), 2c2-e-1=0, and e=1, e=- - the former value satisfies the equation e* — 4e3 +8e-4=0. Lastly, substituting e=1_in_the_transformed equation, then y4 - 6y2+1=0 is the required reciprocal equation, whose roots can be found by a quadratic, also the roots of the given equation. 8. The equation can be reduced to the form 4 - 4mx3 + 6m2x2 - 4m3x+m1=0. 9. The sum of the squares of the roots of the equation x++pxs +qx2+px+1=0 is p2-2q. Next divide the equation by x2, then x2+p+q + p1 = + -2 1 1 1 If x+, then z2+pz=2-q is the reducing quadratic. 1. (1) When the roots of x3-px2+qx-r=0 are in arithmetical progression. Let the three roots be denoted by a-d, a, a+d. Then a-d+a+a+d=p, a(a−d) + (a − d) (a + d) +a(a+d)=q, a(a+d)(a− d)=r, or 3a=p, 3a2 - d2=q, a(a2 - d2)=r. p2 3r 3 9 Hence equating the values of d2, and 2p3 - 9pq+27r=0, the relation required. (2) When the roots of x3 − px2 + qx-r=0 are in harmonical progression. 3r =0, which is an equation whose roots are in arithmetical progression, and the relation between the roots will be found to be 27 - 9pqr+27r2=0. (3) When the roots of x3- px2+qx-r=0 are in geometrical progressior. Let a, ab, ab denote the roots of the equation, then a +ab+ab2=p, a3b+a2b2+a2b3=q, a3b3=r, It is also obvious that rab, the middle root. Also the relations which subsist between p, q, r, s, in x4-pæ3 +qx2 − rx+s=0, an equation of the fourth degree, may be found by the same process when the roots are in arithmetical, harmonical, and geometrical progression. The solution of cubic equations whose roots are in harmonical progression may be effected in two ways, (1) by transforming the given equation into one whose roots shall be in arithmetical progression, or by assuming three quantities a, harmonical progression to denote the roots. 2ac , in a+c 4. If a1 =α, a2=a+b, a ̧=a+2b, a=a+3b, the given equation assumes by substitution the form x2 + (4α+6b)x3 + (6a2 + 18ab+11b2)x2 + (4a3 +18a2b+22ab2 + 6b3)x +(a++6a3b+11a2b2+6ab3)=d. The left side of this equation will become a complete square by the addition of b. Let b4 be added to each side, and let the square root of each side be extracted. then x2 +(2a+3b)x+(a2 +3ab+b2)=±√(d+b1), from these quadratics the four roots can be found. This method may be illustrated by the solution of the equations (x+4)(x+6)(x+8)(x+10)=48, and (x+3)(x+7)(x+11)(x+15)=644. See example 2, under Exercise VII., p. 22 of Section VI. 5. Since the roots of x3-px2+qx-r=0 are in harmonical progression. Let a-b' a' then = -the sum of the reciprocals of the roots-a-b+a+a+b=3a, or g2x2 - (pg - 3r)qx + (q3 −3pqr+9r2)=0 is the equation which contains the roots 9. Since the roots of the given equation are in harmonical progression, from which by elimination may be found c2=4bd and c++ 400ad=0. XXXIII. 7. Let a denote the possible root of the equation x3-qx+r=0, then x=a, and x-a=0; x3 − qx+r=0, or x2+ax+a2 −q=0, the quadratic equation which ... x-a contains the two impossible roots, and taking a=y+z, the required expression will be found. 8. The equation (x3 – 27)3 – 27q31⁄23=0 is an equation of nine dimensions, and includes the three equations x3-3qx+2r=0, x3-3qwx+2r=0, and x3-3qw2x+2r=0, in which 1, w, wa denote the three cube roots of unity. The first of these equations, x3-3qx+2r=0, is only another form of x3 − qx+r=0, obtained by writing 3q for q and 2 for r, when y = and z=2, the two equations will give values of y and from which - 2 x=y+z={r+(y2 − q3)+ }1 + { s` − (»2 − q3)i}š, 1. See Art. 16, p. 20. XXXIV. 2. Let a2,82,72 denote the three values of e2 or y in y3 +2qy2 + (2a − 46)y −7·2=0, |