the reducing cubic of Descartes' solution of the biquadratic x+qx2+rx+8=0, then a2 82y2=r2, aßy=r, and −2q=a2+B2+g3.

x2 + ex+f= x2 + cx + 1} ( 1 + c2 -?'). Let ea one of the three values of e.

and x=(−a+B+7) and 1( − a − B-7), two of the required roots. In the same manner

.*.x=§(a+ß−7) and (a−8+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 + ß)x+aß=0, which is one of the quadratic equations of which the biquadratic is composed.

Let ช & 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 + d) aB − (a + B)yd= (a + B)(aß −– yô),

3. The equations are p2q - p(r− s) − q2=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, b, -b-1; and it may be shewn that a+a¬1=}(p+q), and b-b-1={(p − q).

5. First x-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+a* = 2x2 a2 - 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 x2 + 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=α, 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} b1{x2− (a+c)x+ac} c1{x2−(a+b)x+ab}

+
34

(b−c)(b-a)
C4

+

(a - b)(a−c) (b−c)(b − a) * (c − a) (c − b)

a (b+c)

b+ (a+c) c+ (a + b)
(b−c)(b− a) (c-a)(c-b)

+

(a−b)(a−c)

(c-a)(c-b)

a1 be

+

c'ab

+ + (-a)(e-)

(a - b)(a-c) (b−c)(b− a) (c− a)(c-b) ={(c+a)(c+b)+a2 + b2 }x2 − (a+b)(b+c) (c+a)x+abc(a+b+c),

... x1 - {(c+a)(c+b) + a2 + b2 } x2 + (a+b)(b+c) (e+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.

8

Assume x+, and by the necessary substitutions, 23 -pz2+(q − 3s)z +2ps − r=0

is the cubic equation.

8. The equation can be put into the form

(xo − 1) − px(x7 − 1)+qx2 (x3 − 1) − rx3 (x3 −1)+sxa (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)1⁄23 +(q-p+1)x2 − (p−1)x+1=0.

Dividing by x4 and arranging the terms,

(x* + _—12 ) − ( p − 1) (23 + 1 ) + (1−x+1)(x2 + 1) − (r−q+p−1) (x+1)

1

+(s-r+q-p+1)=0.

1

x2

х3

Making these substitutions and reducing the terms

≈4 − (p − 1): 3 + (y − 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) −9x2 (x3 + 1) + 3x 3 (x3 + 1) − 8×1 (x + 1) = 0. Obviously-1 is a root of this equation, and x+1=0. Depressing the equation it becomes 28 - 9x+12x5 - 20x1 +12x3 −9x2+1=0.

The nine roots are −2±√3, }(3±√/5), }(1±√−3), ±√−1, −1. 10. Since 12+ m2 +n2=1, the given equation may be put under the form (l2 + m2 + n2) v* — {(b2 + c2)l2 + (c2 + a2)m2 + (a2 +b2)n2 } v2

(v2 — b2)(v2 — c2)l2 + (v2 − a2)(v2 — c2)m2 + (v2 − a2)(y2 — b2)=0, and then dividing by (v2 – b2)(v2 — b2)(v2 — c2), the form required is obtained.

UNIV. OF MICHIGAN,

MAY 20 1912

SECTION XII. Cubic and Biquadratic Equations, pp. 66..6d.

..6d.

LONDON: LONGMANS AND CO.

Each Section of the Algebra may be purchased separately;
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ELEMENTARY ARITHMETIC,

WITH BRIEF NOTICES OF ITS HISTORY.
BY R. POTTS, M.A., TRINITY COLLEGE, CAMBRIDGE.
In Twelve Sections, demy 8vo.

EACH section of the Arithmetic may be purchased separately;
also the twelve sections together, done up in boards, with cloth
covers, at 48. 6d.

WORKS BY PERCIVAL FROST, M.A.,

FORMERLY FELLOW OF ST. JOHN'S COLLEGE, CAMBRIDGE;
MATHEMATICAL LECTURER OF KING'S COLLEGE.

NOTES AND ILLUSTRATIONS, AND A LARGE COLLECTION OF PROBLEMS, PRINCIPALLY INTENDED AS EXAMPLES OF NEWTON'S METHODS;

ALSO

HINTS FOR THE SOLUTION OF THE PROBLEMS.

THIRD EDITION.

BY

PERCIVAL FROST, M.A.,

FORMERLY FELLOW OF ST. JOHN'S COLLEGE, CAMBRIDGE;
MATHEMATICAL LECTURER OF KING'S COLLEGE.

The portion of Newton's Principia which is here presented to the Student contains the solution of the principal problem in celestial Mechanics, and must be interesting to all, even those who do not intend to follow out the more complicated problems of the Lunar and Planetary Theories.

The study of the geometrical methods employed by Newton cannot be too strongly recommended to a student who intends to pursue Mathematics, whether Pure or Applied, to the higher branches; for he will, under this training, be less likely to work in the dark when he uses more intricate machinery.

I have endeavoured in this work to explain how several of the results obtained in the Differential aud Integral Calculus can be represented in a geometrical form; and I have shown how, in a large class of problems, the geometrical methods are at least as good an 'open sesame' as the Differential Calculus.

In this, the third edition, I have given Solutions or Hints for the solutions of all the problems, in order that a student may, unaided by a tutor, have all the advantages which the book supplies.

MACMILLAN AND CO. London and Cambridge.