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5. One root between 2 and 3. 6. One root between 2 and 3.
7. One root between 0 and 1. 9. Two real roots, find the greater.
10. One root lies between 4 and 5.

XXV. 1. 3x2 - 2px+q=0 is the limiting equation of xs – px? + q&-p=0. The third root is

4pq-p3 – 9r

pr--39 2. Subtract the second from the first equation and there results (o-plæ? +(9-7X+(r – p')=0, a quadratic equation. See Art. 1, Section VIII.

1

m denote the three roots of the equation,

4. Let as

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1 2.-m.

a

a

a

1, and me

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C-a
Eliminate at

a

b-d 10. The equation whose roots are a, b, c

is x:-(a+b+c)? + (ab + ac + bc)x - abc=0). From the equation x2 – P,*+9,=0, P, =b+c, 9, =bc,

22 P,+92 = 0, P, =a +C, 9. =ac,

X2 - Pg2+13 +0, Pg=a+b, 93 -=ab. Hence 2(a+b+c)=,+Pa+Ps, .. a+b+c= }(P,+12+Ps),

ab+ac+bc=q, +92 +93,

a2b2c2 =9,9293, .. abc =(919293)* . . Making the substitutions

28 – 3(P,+p2+Ps)2? +(9, +92 +93)2 – (919293)1=0. 11. Eliminating x3 between the second and third equations, (q - r).x? + (1-9)x=0, from which q=r.

Eliminating x3 between the first and second equation,

(P-9)+ (9-r)x+ (r-P)=0, but g=r, .. (P-9)x? +(9-p)=0, and p=1, .. p, q, r are equal to one another, and p2 + y2 =2pq. If a be the given root, the equation x3 +px? + px +p=0 can be depressed, and the other two roots determined. 12. ax: +bx+c=1, b.r3 + cx+a=0, cx: + ax +b=1, adding the equations (a +b+c)x3 + (a +b+c)x+ (a +b+c)=0, -. (a +b+c)(x3 +x+1)=0,

and .. a+b+c=1, also a +ó=-C, .. --c3 =ía +b)3=a3 + b3 + 3abla +b)=as +63 3abc, .. q3 + b3 +69=3abc.

bc-a? From the first and second equation, x=

-62"

-62 from the second and third, x= and from the first and third, r=

al)-c? ab – co

be Also a, b, c are so related that a+b+c=0, whence it may be shown that the three values of x are equal to each other.

13. The limiting equation of x' +3px? + 3qx+r=0 is 3x? + 6px +39=0, or 22 + 2px +q=0, and <= =-p+(p? -9)", the values of x in the limiting equation, which will be impossible if q be greater than p?. Hence in this case the proposed equation has a pair of impossible roots.

14. Since a, a, c are roots of 28 +3 px? + qx+r=0, 2a+c=3p, a2 + 2ac=1, a'c=r, and a, c, c are roots of x2 + 3p,x2 +9,50 +r,=0, 2c+a=3p,, c+ 2ac=1;,

ac

(LC

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ac?=Tv
Then a cs=771,
and ac=(ry,), q=a(a + 2c)=3pa, qi=c(c+ 2a)=3pc,

991
.. 9pP,ac=991, and ac=;

991=(rr).

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9PP1

9pP

Subtracting the second equation from the first, 3(P-P2).2? +(9-9,)2+(r-9,)=0, a quadratic equation.

XXVI. 1. Each equation is reducible to x3 – 18.02 – 225x—656=0. 3. a=.

=. - 9, b=36, c= -60. 4. Since x= 3+/8, .. also x=3 - V8, and x2 - 6x +1=1, but us + px +qx+r=0, ..

** + px? +91 +1

-0,

X2 - 6x +1 .. X+ (P+6)= 0, (6p+36 +9 - 1)x+ (n - P-6)=0, from which x=-1. 5. The third root is if b3 + c3 - a'ctab? =0.

a'

XXVII. 3. Here pq=r, this equation becomes x3 – px? +qx Pq=0,

or x? (x − p) +9(x-p)=0, or (x-P) (22 +9). If a, b, c denote the roots of x3 - px+ qx -q=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+V-3). 6. Let = 3, then the equation becomes 23 +pz2 +2+r=0, if p, q, r be put for

y b cd

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3

9. Here (:2 + ax + m)(x2 +bx+ m)(x2 + cx+ m)
= 30 + (a +b+chu: +(3m + ab + ac + bc)x4 + {2(a+b+c)m + abc #3

+{ab +ac+bc +3m}m2? + (a +b+c)m2x+m".
a+b+c=1, ... a? + b2 +c? - -2(ab+ac+bc)=6m, .. ab +ac+bc +3m=0.
Hence x6 + abcx+ m%. To solve this equation, 26 – 20x3 + 343 =0.
Let x' =y, then the equation is reduced to a cubic y— 204 + 343=0.

10. By reducing the equation and arranging the terms according to the descending powers of a, a, +az +ag +a. is the coefficient of ra, and since a, tay+a, +,=0, a, +ag= -(a, +aa), and by means of this relation the coefficient of xp can be reduced to a,b, +a,b, +azb, +a.be =0, and the equation is thereby reduced to one of the first degree, which admits of only one value of ..

11. Since xy(x + y) +x=(x+z) + yz(y+z) = (+ y +z)(xy +xz+yz) – 3xyz, and xy(x2 + y2)+xz(+22)+yz(y2 +z2)=(x2 + y2 +za)(xy + x2 + yz) xyz(x + y +-). The second and third equations by substitution become

(x+y+z)(xy +xz+yz) 3xyz = 168,

and (.xo + y2 + 2)(x3 + x2 + yz) ryz(x+y+z) = 538. Then by substitution and reduction 3(x+y+z)? + f (x+y+z) = 52+, whence x+y+z is known, also xyz and xy + x2 + ym. Hence the solution is reduced to a well-known form by putting x+y+z=p, xy + x + yz=9, xyz=r, and then eliminating two of the unknown quantities y, z, the resulting, equation is a cubic, x3 - px2 + 2x - p=0.

12. I.et x, y, z denote the edges, then 2(xy + x2 + y2) = 22, (x2 + y2 +*2)=114, ryz = 6. Eliminating y and z, z3 – 6z2 +11x – 6 = 0, from which - will be found to be 1, 2, 3 lineal inches respectively. Also x and y can be found when the value of is known.

18. The criterion required for Cardan's rule to be applicable is, that be greater

than 23

; but

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(2abc) ; in this case, that

(a? +62 +c?)? (2alc)? be greater than

is 27 4

27 less than (a? +62 + c?)3

obviously, and consequently, the three roots of the equation

27 3 - (a’ +62 +ca)» – 2abc =() are possible, two of which are negative and one posi. tive. If a, b, c be supposed to be 1, 2, 3 respectively, the equation becomes 2-14x -- 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 A BCD is a quadrilateral inscribed in a circle, and AC, BD are the diagonals.

Then by Euc. VI. D. AC x BD = AB CD+AD BC. Let AB =a, BC=b, C'D=C, AD=X.

.: v(x? —co) XV (x? —a)=ac + bx,

and x*-a'r? -ca.c? +a?c? = a®c2 + 2abcx +bor,
also xo - (a? +62 +ca )c? 2abcx=0, : . *3- (a? + b2 +c)— 2abe =0.

a

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XXVIII. 2. Since the equations have two roots in common, each involves the same quadratic factor.

3. The limiting equation of 24 +rx+8=0 is 4x3 + r=0. The expressions ** +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)?(x—b)2=0, or æ* - 2(a+b)x3 + (a+ 4ab +62)x: - 2ab(a+b)x+aob2=0, is an equation identical with ** - px3 + qx r&+8=0. The values of a and b may be found in terms of any

b two of the four coefficients P, 9, 1, 8.

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 23 +- 1383 +33.2 +314+10 and 4x +39x? +66x +31 have a common quadratic factor x? + 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 suin 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, is one of the factors. Removing this factor, and let 204 – 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=1,

... x(x - 1)(x + 1)(x-2)(x+2)=25 – 533 + 4x ..2, 2-1, x+1, 2—2, 2+2 are the factors of x5-5x3 + fx. 3. If q=r+1, 9-1=r, and the equation can be put into the form

x* + 23% +2? +(9-1}x? +rx+s=0, or

(x2 + x)2 + r(x2 + x) +8=1, 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)3=4(x? + 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 93 4pq +8r=0, and when the product of two roots is equal to the product of the other two, pos=r?. In both cases the biquadratic can be solved by a quadratic.

2. The first equation can be reduced to (z? - ax)2 – 262(x2 – ax)=a2b?, and the second equation to 2x5 +8.c* – 14x3 – 44x2 + 6x=0.

7. When x' – 9x2 – rx--s is divided by x3 +ax+b, the quotient is w? ax +a? 6-9, and the remainder is – (as – 2ab qa +r)x bia? - 6-9) 8.

If ? +(2+6 be one quadratic factor of c* - 902 – rx – s, then must x? - ax +a? -- 6-9 be the other; and the two terms of which the remainder is composed must be respectively equal to zero.

And then as-(26+9)a+r=0, and ba? - 6-9) - s=0. When a is eliminated, the resulting equation becomes a cubic involving b 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 35 - prace +qx3 9.2 + px-1=0 be a recurring equation of five dimen. sions.

It may be put into the form (25 – 1) - pæ(73 - 1)+qx? (x - 1)=0, ..x—1=1, and 3=1, one of the roots, and the equation when depressed becomes **-(P-1).c3+(9-p+1)? – (p-1)x+1=0; divide each term by mc>,

1 1
..x?-(P - 1)x+(9-p+1)-(p-1). + 0,

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and

(**+2+ =

) -(p-1)(2+) – 2-q+p-l=p-9+1.

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Let +-= E", then z? -(P-1)==p-q+1, a quadratic by means of which the other roots can be determined.

(2) Let 36 - p.05 + qa* -r33 + qx? —pæ+1=0, a recurring equation of six dimen. sions. Divide each term by c', then 23 – px+qX+q.

1 1

0, x?

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+

or ( + )--(*+^)(+4) --=0.

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+

X

-r=.

23

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+

1
1

1 Let 2 +

=*– 2, x3 + =73 – 3:; make the substitutions,

x? ..:* -- p32+(9- 3)2 – (2p+r)=0, an equation of three dimensions. 5. Here w+ +252 +1=-(x3+x), and (** + x2 + 1)2 = (x3 + x)”, whence 95 +66 +** +x+1=0, let x?=y, then y' + y3 + y2 +y+1=1 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.

1 1 6. Form the equation whose roots shall be a, -,

b, and shew that it is

õ'
identical with the given equation when the latter is reduced to the form

24_Plq+1).23 - (q*+p2+1).x2 _Plq+1).x+1=0.
2
9

9 7. Let the roots be increased by e; assume y=x+€, x=y-, then yt – 4(0–1)43 +6c(e- 2)y? – 8(e2 — 1)y+ (e+ – 4e3 + 8e – 4)=0 is the transformacu equation. Anci in order that this equation may become a reciprocal equativn, 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(e? – 1), and e* - 4e3 + 8e-4=1.

From 4(e - 1)=8(eo – 1), 2c2-0-1=0, and e=l, e=-1, the former value satisfies the equation et – 4e3 +86-4=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 x* -- 4mx: +6m2q2 – 4m*x+m+=0.

9. The sum of the squares of the roots of the equation x4 +px: +qx? + pæ+1=0 is p? —29. Next divide the equation by x?,

1 1

1 then a + px +9+p. +

0, and xo +2+ 1

=%, then zo + pe=2-9 is the reducing quadratic. Let 7,, *, denote the roots of this equation.

+r(x+2)+9–2=0.

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If 3+

2

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Then z;2+z3= {- + VCkp? +2–0)}+{

+ V(łp?+2–9)}'+{-}-v(p2+2–21}" - PV(+p® +2 – 9) + mai +2–9+ +PV(} po +2–9)+*+2-1

2

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4

4

=p? – 29+4 .. (02 – 29+4) -(p? - 29)=4.

XXXII. 1. (1) When the roots of 23 – px? +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, ala-d)+(a-d)(a +d)+ala+d)=9, a(a +d)(- d)=T, or 3a=p, 3a? - d?=9, a(Q? - da)=r.

=P. p?

a=., -d=q, and do=

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- 9; also? {-d2} =r, and d’=

p? 3r

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2

For

I

1

v'

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33

3
Hence equating the values of d?,

22 .22 3r

-9

3 9 and 2p3 - 9pq +27r=0, the relation required. (2) When the roots of 23 - pre? +98—1=0 are in harmonical progression. * write then the given equation becomes y3 .

_qy? Py_1=0, which is an equation whose roots are in arithmetical progression, and the relation between the roots will be found to be 273 9pqr +27r2=0.

(3) When the roots of x3 – px? + qx – r=0 are in geometrical progressior.
Let a, ab, ab? denote the roots of the equation,

then a + ab + ab?=p, a·b + a?l2 + a 13 =9, a3b8=r,
c2b(1+b+ba) I
or ab=?, and a3b3--99,

but a3 1,3=r,
all +6+6) P

P q

=r, or 98 - p3r=0 is the required relation.

It is also obvious that rt=ab, the middle root. Also the relations which subsist between p, q, r, s, in xa-pw+qx? -- 92+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 who e roots shall be in arithmetical progression, or by assuming three quantities a,

2ac

a+c Larmonical progression to denote the roots.

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