4. If a1=a, 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 +18a3b+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 − pœ2 +qx-r=0 are in harmonical progression. 1 1 1 a-b' a' a+b Let then denote the three roots, the sum of the reciprocals of the roots-a-b+a+a+b=3a, or q2x2 - (pq − 3»)qx + (q3 −3pqr+972)=0 is the equation which contains the roots 9. Since the roots of the given equation are in harmonical progression, cy by a y++ + + =0, whose roots are in arithmetical progression. d d d Let x+3y, x+y, x-y, x-3y denote the roots of this equation, c b d' α then 4x: 6x2-10y2: 4x3-20xy=0, x2 - 10x3y3 +9y4: d' 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 x3- qx + r x-α=0; .. 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 – 2ƒ)3 – 27q3ë3=0 is an equation of nine dimensions, and includes the three equations x3-3qx+2r=0, x3-3qwx+2r=0, and 3 - 3qw2x+2r=0, in which 1, w, w2 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 39 for q and 2r for r, when y = and z=2, the two equa tions will give values of y and z, from which 2 x=y+z={r+(r2 − q3)1 }3 + {r — (r2 − q3)+}‡. 1. See Art. 16, p. 20. XXXIV. У 2. Let a2,82,y2 denote the three values of e2 or y in y3 + 2qy2 + (22 − 48)y — p2=0, the reducing cubic of Descartes' solution of the biquadratic x*+qx2 +rx+8=0, then a2ß2y2=r2, aßy=r, and -2q=a2 + B2+y3. x2+ex+ƒ= x2 + ex+ +1 (4+e2). Let ea one of the three valucs of c. and x=(a+B+y) and ( − a − B-7), two of the required roots. ..x= ={(a+B−y) and =x2-ax +1(a2 – B2 — y2 + 2By)=0, 2 a ax + =1(B− y)2, .'. x 4 (a-B+y) are the other two roots of the biquadratic in terms of the roots of the reducing cubic. 3. See Art. 16, p. 20. 7. See Art. 17, p. 22. 8. See Art. 17, p. 22. 12. See Art. 16, p. 20, and Art. 18, p. 23. XXXV. 2. Since a, B are two roots of the equation, x2 − (a + B)x+aß=0, which is one of the quadratic equations of which the biquadratic is composed. Let Y, 8 denote the roots of the other quadratic. Since a+B+y+8=0, a+B= − (y+8), and x2+(y+8)x+yd=0, or x2 + (a + B)+yd=0 is the quadratic containing the other two roots, But q=+aßy+aßd — ayd — Byd= + (y +8)aB − (a + B)yô= (a + B)(aß – yd), 3. The equations are p2q - p(r− s) −22=0, and p2s+q(r−s)=0; and if r-s be eliminated, the resulting equation is p3s+p2q2 − q3=0. 4. First suppose the four roots real; since the last term of the equation is negative and equal to unity, the roots are of the form a, a−1, b, -b-1; and it may be shewn that a+a¬1=}(p+q), and b—b−1=}(p− q). 5. First x1=-4a3x+a*, add 2x2a2 to each side of this equation, and x2+2x2a2=2x2a2 - 4a3x+a1, complete the square of the left side, .*. x1+2x2a2 + a1 = 2x2a2 - 4a3x+2a1=2a2 (x2 − 2ax+a2). Extract the square root of each side, .'. x2+a2==±a(x− a)\/\/2, then x2+a2 = + a(x − a) 、\/2 and a2 + a2 = − a(x− a)√√/2 are the two quadratic equations from which two values of a will be found to be possible and two impossible. 6. It is obvious from inspection that a, b, c respectively, satisfies the equation; and, therefore, three of the roots are known, that is x=a, x=b, x=c, and x-a=0, x-b=0, x-c=0, .. (x− a)(x − b)(x − c)=0. Next the given equation can be put into the following form: _a^{x2 − (b + c)x+bc} = {(c+a)(c+b) + a2+b2 }x2 − (a+b)(b+c) (c+a).x+abc(a+b+c), = •°. x* − {(c+a)(c+b) +a2 +b2 }x2 + (a+b)(b+c)(c+a)x− abc(a+b+c)=0 (1), but (x-a)(x - b)(x − c), or x3 − (a+b+c)x2 +(ab+ac+bc)x - abc=0 (2). By dividing the equation (1) by the equation (2) there results the equation x+a+b+c=0, and consequently x=- (a+b+c) is the fourth root of the equation. 7. Divide each term of the equation by x3, and arrange the terms in pairs. Assume x+z, and by the necessary substitutions, z3 -pz2+(g - 3s): +2ps −r=0 is the cubic equation. 8. The equation can be put into the form 3. (xo − 1) − px(x7 − 1)+qx2 (x3 − 1) − гx3 (x3 −1)+sx1 (x-1)=0, which is obviously divisible by x-1, .. x=1, one of the roots; depressing the equation it becomes x3-(p-1)x2+(q−p+1)x® − (r−q+p−1)x3+(s−r+q−p+1)x+−(r−q+p−1)x3 +(q-p+1)x− (p −1)x+1=0. Dividing by x and arranging the terms, (x* + _—12 ) − ( p − 1) (x2 + 1 ) + (a−x+1)(x2 + 1) − (r−q+p−1) (x + 1) -p+ Making these substitutions and reducing the terms ≈4 − (p − 1)z3 + (q− p − 3) z2 − (r+q+2p−2)z+(s− r−q+p+1)=0 is an equation of four dimensions. 9. The equation is (x2+1)+x(x2+1) — 9×2 (x3 +1)+3×3 (x3 + 1) − 8x1 (x + 1) = 0. Obviously -1 is a root of this equation, and x+1=0. Depressing the equation it becomes 28 - 9x6 +12x3 - 20x +12x3-9x2+1=0. 1 49 1 4 4 and z2+z-12=0, ≈2+=+== 2+ == ;. . z=3 and -4; - 4, and x=-2±√3. = The nine roots are -2±√3, (3±√/5), }(1±√−3), ±√−1, −1. 10. Since l2+m2+n2=1, the given equation may be put under the form (l2 + m2 +n2)v1 — {(b2+c3)l2 +(c2+a2)m2 +(a2+b2)n2 }v2 and resolved into +b2c2l2+c2a2m2 + a2b2n2=0, (v2 —b2)(v2 - c2)l2 + (v2 − a2)(va − c2)m2 + (v2 − a2)(y2 —b2)=0, and then dividing by (v2 – b2)(v2 — b2)(v2 — c2), the form required is obtained. |