Solutions for Hall & Knight

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SOLUTIONS OF THE EXAMPLES IN

HIOHEE ALGEBEA.

SOLUTIONS OF THE EXAMPLES

I2T

HIGHEE ALGEBEA

H.

S.

HALL,

M.A.,

FOBMEELT SCHOLAB OF OHKIST's COLLEGE, CAMBRIDGE, UASTEB OF THE MILITARY AND ENGINEEEINO

S.

R

KNIGHT,

SIDE, CLIFTON

COLLEGE

B.A.,

FOBMEELT SCHOLAR OP TRINITY COLLEGE, CAMBRIDGE, LATE ASSISTANT-MASTER AT MARLBOEOUGH COLLEGE.

Uonlfon

MACMILLAN AND

CO.

AND NEW YOEK. 1889 [The Right of Translation

is

reserved.]

;

PKINTED BY

C.

J.

CLAY, M.A.

AND SOXS,

AT THE DNIVEP-ailY PBES8.

m^'"'

^^r

PREFACE.

This work forms a Key or Companion

to the

and contains

all

many

cases

full solutions

more than one

of nearly solution

out the book frequent reference illustrative

Examples

is

is

Higher Algebra,

the Examples.

In

given, while through-

made

to the text

and

The work has been

in the Algebra.

undertaken at the request of many teachers who have introduced the Algebra into their classes, and for such readers

it is

mainly intended

diciously used, the solutions

;

but

may

by that large and increasing

it is

also

class

hoped

that, if ju-

be found serviceable

of students

who read

Mathematics without the assistance of a teacher.

H. S.

June, 1889.

S.

R.

HALL, KNIGHT.

HIGHEE ALGEBEA. EXAMPLES. Letr = ^=^ = ^;

8.

stituting for a,

h,

c in

Let

9.

A:

then

terms of

=

= dr,

c

Pages 10—12.

I.

b = cr=dr',

we have

d,

— q+r-p

a=br=dr^; and by

^^.TT ,. ¥c ^ + d* + b^cd^

= r°=

sub-

-

d2"

'

p+q-r

r+p-q

{q-r)x+(r~p)y + {p-q)z=h{[q-r){q+r-p) + ... + ...}=0.

then

_y_^y + x^x

10.

x—z

y

z

—

sum of numerators _, -i ^''-' = 2(x+u) Each ratio= smn 01s-^,denommators x+y ..

,

-

^

Thus each

ratio is equal to 2 unless

x+y

=-=2: y

z

In the second

case,

X

y=

whence x

,

..

T, Eaohratio=

:

'

-x, and a:

,, 11.

i + ?/ = 0.

—2

—

=-

,

y "

In the :

:

2

:

case

3.

whence a;:v:z=l: -1:0.

2/

——

sum of numerators = sum ot denominators :

z=i

first

iix+y + z) ,

^

J ^\ + c)

(p + q)(a +

'

1). '

^

li

Multiply the numerator and denominator of each ratio by and add, then each of the given ratios

a, 6, c respec-

tively

_ [b + e)x+{c + a)y+{a + i)z {p+q){bc + ca+ab) From

equations

12.

(1)

and

See Example

H. A. K.

(2)

*"''

the result follows.

2, Art. 12.

»,

1

[CHAP.

RATIO.

2

iy + 2z-x _ 'iz+2x-y _ 2x + '2y-z Multiply the numerator and c o 6 I, I, eacn denominator of each of the given ratios by 1, 2, 2 and add ; then ratio

_ -(2y + 2z-a;)+2(2z + 2a!-y) + 2(2g + 2y-z) _ -a + 2b + 2e

5x 2b + 2c-a'

Similarly each of the given ratios is equal to 2c + 2a-b

14,

and to

;

2a+2b-e

Multiplying out and transposing, b'h?

that is

+ chj^ - 2bcyz + cV + a'n^ - 2eazx + aV 5V - 2a5a:i/ = (bz -cyY + (ex - azf + (ay - bxf = 0. bz-ey = 0, cx-az=0, ay-bx = 0.

;

.•.

15,

Dividing throughout by Imn,

my + nz-lx mn

nz + lx-my

_ lx+my-nz Im

nl

{lx+my-nz} _ 2lx _ x '~ nl + lm m + n' nl+lm

_(nz + lx-'my) + ~

Thus we have x _

y

_

m+n~ n+l~ Hence the 16,

y+z-x

" l

+ m"

(n +

l)

+ (l + m)-(m+n) ~

y + z-x 21

result.

From ax+cy + hz=0, cx+by + az=0, we have by

cross multiplica-

tion, "^

—

—

y

g

_7.

Ti^^^ ~ U^-' ~ ab^^~

^^'^• '

Substituting in the third equation bx + ay + cz=(i,

we have l(ac-W) + a(bc-d?) + c(db-c^) = Q.

17,

From

the

first

two equations we have by cross multiplication, as

_

hf-bg

_

3/

z

gh-af~ ab-h?'

Substituting in the third equation,

we

get

g(M-''>9)+f{ah-af) + c(ab-W)=Q. 18.

From

From

the

first

the

first

and second equations, X _ y __£_ ac + b bc + a 1-c'

and third equations, X _ y ab + c~

_

.(1).

z

l-b^~ bc + a

.(2).

I-]

[chap.

PROPORTION.

4 From

27.

From

the

first

two equations, X ^ y db + a ab + b

z (1).

1-ab

the second and third equations,

_

X 1

From

_

and

(1)'

(2)'

*

a

^

-

,f' (6

,, + 1)

^

y

x .j-^

1-bc

= .p^ 1-ab

_

a^

.

z

(2).

bc + c

bc + b

6c

x

-yy^, 0(6 + 1)

z^

a(l-bc)~ c(l-ab)'

From

28.

From

the

and second equations, ^ = y - ^ hf- bg gh -af ab- W

first

the second and third equations,

^

^

the

first

_^_= (2)

and

y ca-g^

^-J—

(3).

gh-af

^^ -^ =^5^^ ^^ x

x

(3)

_

x^

From

(2). ^'

hf-bg

and third equations,

fg-ch

From

^

^

y

fg-oh

ftc-/"

From

(1).

;

y^

bc-f'-'~ ca-g^' and (3) z X z X _ y y be -/^ ca-g^' ab-b? fg- eh gh-af hf-bg'

(1), (2),

'

'

hy equating the denominators the second

EXAMPLES, n. Examples

Example

6,

4, 5, 6,

7

may

and put ^ =

-5

a-c _ bTc-dlc _ b-d b-d Examples Put -

=-

8, 9,

=

10

may

=

all 7c,

result follows.

Pages

be solved in a similar manner; thus take

so that

a=bk, c=dk; then

Tcjj^ + d^ ^ Jk'b-^ Jb'^ + all

+ Tt'd' _ Ja' + ^/P+T^' e''

sjb^ + d^

d''

be solved in a similar manner.

- = fc, so that c=dk,

2a + 3d_2dP + Sd 3a - 4d ~ 3dk^ - 4d

19, 20.

b = dW, a = dTt^, then in

2k^ + 3

Example

_ 2 63^ + 36' _ 2a^ + 3h^ _ ~ SA* - 4 ~ Sft'/jS - 46^ ~ 3a^ - 46^

9,

PROPORTION.

II.]

j = - = k;ttiena = bk,c^r, k

Put

11.

be

" ~" "/ _ j^+^__H!Lllil. -= ^ """-'" •^'-

and

.

54

.Kp-'-") Componendo and

13.

2a:-3 x+1

Componendo and

15. tions

we obtain a simple

whence i— =s5s -13 i-r x+1 _ 3x-l

dividendo,

.

5a;

-13'

71X

"1"

—5.

a;

= 0, '

or

:;^_ww

diyidendo,

fix

"T"

j^^

clearing of frac-

C

equation.

We have

16.

a(a~b-c + d) = a^-ab-ac + ad=a'-ah-ac + hc = {a-b){a-c);. (a-b){a-c) a

a- b-c+d=-

J^

The work done by

18. .^/

1^ 1, (a;-l)(x+l); hence /

a;

-

(a;-l)(x + l)

1

men

in x

+1

days

is

Denote the proportionals by x, y, 19-2/, 21 -x. x(21-x)=2/(19-2/) x« + 3/2 + (19-^)2 + (21-x)2=442 and x2 - 2/2 - 21x + 191/ = 0. From (1) 19.

From Add

proportional to

9 = ^.

Then (1),

(2).

a:°+2/2-21x-19i/ + 180 = 0.

(2)

x2-21x + 90=0,

x=6

or 15.

y^-19y + Q0 = 0,

Subtract

y = 9 or 10.

Let the quantities taken from

20. tively.

21. are

A and B

be x and y gallons respec-

Then

x-9

Suppose the cask contains x gallons; after the first drawing there gallons of wine and 9 gallons of water. At the second drawing

X—9 — — X 9 gallons of wine are taken, and therefore the quantity X (x

- 91

— 9fx ^

9)

=

(x

—

of

wine

left is

9)2 .

Hence the quantity of water

in the cask is

[CHAP.

PEOPOETION.

6

X

X

(x-^Y :18x-81 = 16 (a; -9)2 =16 (2a; -9).

:.

or

Denote the quantities by

22.

Difference between

first

a, ar,

and

:9,

wfi.

a'fi,

last

DifEerence between tbo other two

_ ar'~a ~ or* ~ r

—

„ = r^+r + 1 = 3r+(l-r)2 — o+

r

r

and

this is greater

23.

Let

than

T amd C

II.

,

—

(1-^)' r

. 1

3.

denote the town and country populations 18

the increase in the town

population

is

4

country

24.

On

m^^^^'

18r+4c=15-9(T+C).

Let 5x and x denote the amounts of tea and coffee respectively. the

first

J of coffee is

-ryrj;

supposition, the increase of tea is

XX

;

and the

the second supposition,

r^ x 5a;

(1)

and

(2),

the increase

r^ x 6a;.

5o + 5=42c

(1).

we have

5S + a = 18c

From

;

7c

.

total increase is

.-.

On

„

Tnn^'^ 100

total

.:

^j^ ^>

5a + 6

7

+a

3

56

(2).

25. Suppose that in 100 parts of bronze there are x parts of copper a; of zinc ; also suppose that in the fused mass there are 100a parts of brass, and 1006 parts of bronze. 100a parts of brass contain ax parts of copper and a(lOO-x) parts of zinc. Also 1006 parts of bronze contain 806 parts of copper, 45 parts of zinc, and 165 parts of tin. Hence in the fused mass there are aa; + 805 parts of copper; a (100 -a;) + 45 parts of zino, and 166 parts of tin.

and 100 -

a.^

+ 806 _ g (100 - a + 46 _ 165 " 10 ~ 16 74 )

VARIATION.

111.]

.-.

10 (oa; + 805) = 74x166; that

10 {o (100 -a;)

Also

+ 44}=16xl66;

10aj;= 3846. 10a (100 -x) = 216 J.

ia

is

3846

lOflx

• lOa(lOO-x) 26.

that

,

2166'

or:

X

16

lOO-i"

9

Let X be the rate of rowing in stiU water, y the rate of the stream,

and a the length of the course.

Then the times taken

to

and with the stream are

row the course against the stream, in

x-y

.

-

X

,

x+y

still

•'

-^=84 x-y

Thus

(1), " ^

^--^ =9 X x+

(2). ' ^

y

From

water,

minutes respectively. "^

a=8i{x-y),

(1),

S4(a;-y)

.

X .:

'

2Sy{x-y)=3x[x+y), 3i' - 25w/

or

U(x-y) ^^ x+y

+ 28i/» = 0,

x = 7y, or Sx=iy. If

x=7y, then a=84x6«, and time down stream=2-=63 minutes. Sy

Similarly in the other case.

EXAMPLES, m. 7.

P=-^

where

.

thus

f»

9.

and

-=-

o

10.

.

hence j =

26, 27.

mx

= l, and Q=PJ?=,y48 x ^75

Here y = mx + -

= 3m + 5

m is constant;

Pages

,

From

where

to

= x

=60.

and n are constants

these equations

we

find

-=-

m = 2,

whence

;

6

= 4m + -

.

n = - 8.

o

Here

m=5, n = 36.

y=mx+ -^,

so that

19=2m + j, and

19

= 3m + -^

;

whence

[CHAP.

VARIATION.

8 11.

^=^^, BO that 3=^^^^, and therefore m = I

12.

Here

From

x + y=m[z+-\

,

x-y=n

and

I

—

;

hence

|.

the numerical data,

4=mx24, and 2=raxlJ; x + y = -^(z + -

thus

By

1

and x-2/ = -(z

2x=j^z +

14.

from

j.

4

44

addition,

—

yf'

•

y=m + nx+px-.

Here

the numerical data,

0=m + n+p; l=m+27i + 4^; 4=m + 3?i + 92); whence m=l,

m=-2, p = l; and y=l-2x + x^={x~iyK

15. Let s denote the distance in feet, so that s=mt^. Now 402J=m x 5', hence

In 10 seconds, In 9 seconds,

The

s s

t

the time in seconds : then s

oc «',

m=16-l.

= 16-l x 102=1610. = 16-1 X 92=1304-1.

difference gives the distance fallen through in the ID"" second.

16.

Focr^

Let r denote the radius in so that

the volume in cubic feet; then

179| = mx(|j

Hence 7

when

V

feet,

V=mr^.

r= J

,

1 r=m x n\^ =g jj

x

;

/TX**

x

"»

^

(^^ j

1

= 8 x 179J=22H.

17. Let w denote the weight of the disc, r the radius and t the thickness then w varies jointly as r' and t; hence w=mtr^. If w', r', t' denote corresponding quantities for a second disc, w'=mt'r'^.

„ Hence t

If

->

r

w — = tr^ w tr^ ,

-r-r-,

9^2 w = 39 and — =2, we have 2= 8/2 — 8 w

y^

,

that

is

3r=4r'.

18. Suppose that the regatta lasted a days and that the days in question were the (^-l)'^ x"", and [x + )"*. Then the number of races on the a;"' day varies as the product x(a- x-1). Similarly the numbers of races on the (s-l)"' and (a; + l)"' days are proportional to (a; - 1) (a - a; - 2). and (x + \){a~x).

VARIATION.

in.J

Hence

9

{x-l){a- x-2)=6k

(1),

x{a- x-l)=5k

(2),

{x

+ l){a-x) = 3k

(3).

2x-2-a = Subtract (3) 2a;-a = 2A: (2), Subtract (4) 2 = k. (5), Hence from (5), 2x-a = 4; that is o = 2a;-4 Also from (2), a;(a-x + l) = 10, - 3) = 10. and substituting from (6) Thus a;=5anda = 6. Subtract

(2)

from from from

(1),

lc

(5).

(6).

a; (a:

,

19.

(4).

Let £p be the cost of workmanship w carats the weight of the ring;

£x £y

the cost of a

the value of

diamond of one carat; a carat of gold.

a=p + {w-3)y + 9x, b=p + {w~i)y + l&x, c—p + {w-5]y + 2Sx,

Thus

.'.

a + c-2b = 2x, whence

x=—-

20. Let £P denote the value of the pension, then by the question P oo ^/Y, that is

Also

and

21.

Let

Y

the

number

of years;

P=m^Y

(1).

P + 50=mjY+y

(2),

JY

'

From this last P + 50 = 5m.

J.

^"

8

'

r=16, and

equation

F denote the

therefore from (1)

force of attraction,

T

and

(2),

P=im,

the time of revolution; then

i^'xg.andT^xJ. Tl-iusDocJ'TS; that

is

M T^^ D x -^

oi

MT^xD'; fmf

Thus 7n^t^=hd^, and m^t^^kd^^; that Using the numerical data,

d ^^

35

= oT

^=343, and

•

f2= 27-32 days.

m, SiSti^ •'•

{27-32)2

is

35x35x35

~ 31 X 31 X 31

:.

2

^=

MT^=kDK (7 3

—

-j^-

TflnZn

tin

[CHAP.

ARITHMETICAL PEOGRESSION.

10

27-32 X 5

Let

22.

then

/

13-66

5

,

__

,^^

'-^r- V 3i=3r72^=''^' ^^^'-

•••

'^

re be tlie rate of the train ia miles per hour, in tons; g the quantity of fuel used per hour, estimated q = kx^;

2=ix(16)i;

but

2

•••

.•.

2

= 256^

the cost of the fuel per hour

2 =

£

is

„ x ofifl^'"

\

.-.

cost of fuel per mile is

* 256

X

x^

£- x 256~^256

'

Also cost for journey of one mUe. due to " other expenses,"

*

cost of journey per

.-.

and

mUe

16a;' ^^

£

(

ggg

+ jg^ )

>

has to be as small as possible.

this

Now

this

expression=/'^- T-T-j + KTy and

'Jl--1^=Q; lO

k"

20

is

thatis,

therefore is least

when

x = 12.

i^JX

Hence the miles

is

least cost of the journey per mile is f/^, 7s. 6d.

and the cost for 100

£^^=£9.

EXAMPLES. 18.

«

= 5 (a +

J)

;

thus 155

Again Z=o + (m-l)(J, that

IV.

= 5 (2 + 29),

is,

Pages

a.

and

31, 32.

n= 10.

29=2 + 9(?.

20.

Here 18 = a + 2(i, 30=o + 6i, so that a=12, d=3.

21.

Denote the numbers by ^ -

then

Hence 22.

i-d, 4, Thus

li, a, a+d, 3a=27, thatis, o = 9.

(9-d)x9x The middle number 4+d.

is

(9

+ d) = 504.

clearly 4, so that the three

(4-d)3 + (4)3 + (4 + d)3=408.

numbers are

ARITHMETICAL PROGRESSION.

IV.]

Put n = 1 ; then the

23.

n=15; then

put

=5

term

first

the last term =61.

Sum=— (first term +

last term)

Example 24 may be solved in the same

put

78=2);

= -^x66=495.

-way.

term

= - + &:

then the last term

= - + 6;

Putn=l; then

25.

11

the

first

a

a

:.

sum=| (first

The

26. .:

series

= 2a-

term + last tenii)=|

i,

a

f^^ +2b)

.

n .terms \

.

ia--, 6a--, a

a

/S=(2a + 4a + 6a + ... tore terms)

/I

3

\a

a

-l-H

EXAMPLES.

5 1

X

a

IV. b.

h

...

to

J

Pages

35, 36.

3.

Hereo + 2d=4a, and a + 5d=17; henee a=2, d=3.

4.

Herea + d=—

31

a\?,M=-^, a + (re-l)(i=-—-;

d=- J 6.

13

1

,

Denote the instalments by

,

so that

a = 8, ra=59. o,

a + d, a + 2ci

;

then sum of 40 terms = 3600; o

and sum of 30 terms = 5

of

3600=2400.

o

.-.

20(2a + 39d) = 3600, and 15(2a + 2M) = 2400; .-.

7.

2a + 39d=180, 2a + 29(Z = 160.

Denote the numbers by a and

Then a + Z=-5-, and the sum

of the

I,

and the number

of

sum=2m+l; 13

.".

and the number

of

2m + l = m(a + Z)=-^m, whence m = 6; means

is 12.

means by 2m.

means = 2mx -jr— =m(a+Z).

But

this

12

[CHAP.

ARITHMETICAL PROGRESSION. The

9. first

term

Hence

is

^

,

1—x

S= ^

1-x

and

.

Lt^^ !-«

_5_-

1-x

difference

is

therefore an A. P. whose

—

:P^

l~x

| {2a + 6(J} =49, that

^{2a + 16d} = 289,

Similarly

and

{^j^ + ^^f = ^p^^

We have

10.

Iz^

series is

is

that

{2

+ (» - 3) V-J.

a + 3d=7. a + Sd=n.

is

Thusa=l, d=2. Let

11.

ic

be the

common difference then = x + (q-\)y, c=x + {r-l)y; (ti-r)a + (r-j))b + ('ja-q)c = 0, term, y the

first

a=x + {p-l)y, :.

since the coeflSoients of x

Here

12. that

and y

|

;

h

{2ffi

will

both be found to vanish.

+ (^-l)d} =2;

2a + (2)-l)d=?2.

is,

2a + (j-l)(?=

Similarly

a=^ + ^-i-- + l.

d=-2fi + lV

Whence

\P

—

qj

i

P

13. Assume for the integers a -3d, these is 4a; thus 4(i=24 and a =6. .-.

that

is,

P

q

/I

_._^^^|2p^2,_2_2

IM

a-d, a + d, a + 3d; the sum

of

(6-3d)(6-d){6 + d)(6 + 3d) = 945, 9 (2 - d) (2 + d) (6 - d) (6 + d!) = 945.

14. Assume for the integers o-3d, a-d, a + d, a + Sd; thus from the part of the question a=5; and from the second

first

(5-3d)(5 + 3d) (5-d)(5 + d)

2

=3;

^

whence d=l.

15. Here a + (p-l)d=j, and a + (g-l)d=ji; whence d= - 1, a=p + q-l. Thus the TO* term=^ + j-I + (TO-l)(-l)=^4.2_m. 17. of

sum

Putting n=r, the sum of r terms - 1) terms is 2 (»• - 1) + 3 (» - 1)2.

{r

is

2r + 3r^; putting n=r-l. the difference gives the ?•* term.

The

ARITHMETICAL PROGRESSION.

IV.]

«" n{2a+n-l.d) ra(2a+m-l .d)=m{2a+n-l d); whence 2a=d. m'''term _ a + (TO-l)d _l + 2(m-l) 2m-l n»i> term ~ a + (n-l)d~ 1 + 2 (re -1) ~ 2re-l

thatis,

.

rpjj

19.

number

m{2a+m^l .d) ^r^

We have

18.

13

m

Let be the middle term, d the common difference, and 2^ + 1 the of terms ; then the pairs of terms equidistant from the middle ;tcrm

are

m-d, m + d; m-2d, m + 2d; m-3d, Thus the

m,

+ 3d;

m-{p-l)d,

m + {p-l)d.

result follows at once.

20.

See the solution of Example 17 above.

21.

Let the number of terms be

Denote the

a + d, a + 2d, a + 3d,

a,

Then we have

2re.

by

series

the equations

a + {2n-l)d.

:

|{2a+(re-l)2d}

= 24

~{2{a + d) + {n-l)2d}

(1),

= 30

(2),

(2n-l)i=10i

From From

and and

(1) (3)

22.

(4)

In each

(3).

nd=e

(2), re

=4, and the number

set the

middle term

is

(4).

of terms is 8.

5 [Art. 46i Ex.

1].

Denote the first set of numbers by 5-d, 5, 5 + d; then the second will be denoted by 5 - (d - 1), 5, 5 + (d - 1) hence {5-d){5 + d) _7 (6-d)(4 + (iJ~8' d;=2 or -16. whence

set

;

_

The

latter value Is rejected.

23.

In the

being the (r+

first

1)"'

case the

term,

is

In the second case the

common

x+ r""

T (2y

.,

mean

m+1 (n

difference is

——

5-

n+1

:

and the

— x)

is

•

2x +

r{y~2x) ra+1 re+1

+ l)a! + r(2i/-a:) = 2(ra + l)a; + r(y-2a;), ry = {n + X-r)x. .-.

r""

mean,

14

[CHAP.

GEOMETBICAX PEOGEESSION. 24.

^{2a + {p-l)d} =^{2a + {q-l)d},

Here

.:

{2a-d)p+p^d={2a-d)q + qH; {2a-d){p-q) + {p^-q')d=0. 2a-d + (p + q)d=0, 2a + {p + q-l)d=Q.

or

^{2(i +

.-.

that

is, tlie

sum otp + q terms

EXAMPLES. o(r8-l)

„„

9a(r3-l)

20.

-Tnr-

21.

ar*=81, or=24;

r-1

24, 25.

The

assume

V.

Pages

a.

41, 42.

j-=| and o=16.

.-.

Use the formula

25,

+ 2-l)4=0;

„ , „ ••r°+l=9;'-=2.

'

22.' 23.

Ex.

(i)

is zero.

s

= ^-^. r-1

solutions of these two questions are very similar.

for the three

numbers -,

a,

r'

'

ar; then

r

'

ft

a=

and the numbers are -

6,

,

6, 6r.

T

(^x6^ + (6x6r) + ^^

Again,

3

that

—i-3r=:10,

is,

T

27.

Let/ denote

the

first

.:

28.

from

r=3

whence

term, x the

a=fxP-\

x6r"j

= 156; 1

or ^, o

common

h=fxi-';

ratio

;

then

c^fx'-K

oS-'"6'"-Pc»-«=/«-'^T-m'-«a;a>-i)w-'-m8-i)o--3))+(r-i)(ji-«)=fOa.o=

Here

the

first

^±-=4, and j^=192. equation o = 4 (1 - r)

64(l-r)'

1-ya

,„„

;

In

-xasxar=216:' wnence

hence

=192, or (l-r)==3(l + r+r=),

that

2r='-5r+2=0, whence r=2 or ^

is,

The first of these values is inadmissible in the other value gives a =2.

EXAMPLES. 1.

.-.

V.

subtraction,

4

a'^'^-na'^

31

.

5+i6 + 6i + 256+64^256 8 ,16

4

2

,

45, 46.

+na^^-\ + (n-l)a"-i + na"i

15

7

3

4^16

4 3 „

Pages

na".

1-a

^=l +

,

^=1 + 4 +16+64''' 256'^

,1111,

o =2-

= 1 + 2 + 4 + 8 + 16+ 3_

.

an infinite geometrical progression

b.

S=l + 2a+Sw'+ a + 2a«+ aS= S{l-a) = l + a+a^+

=-

By

15

GEOMETRICAL PROGRESSION.

V.J

S=l + Sx+5x^ + 73i? + 9x*+ x + dx^ + 5x^ + 7x^+ xS= {l-x)S=l + 2x + 2x^ + 2x^ + ix* + 2x _! + « _ .•.

By

subtraction,

""

l-a;~l-a!' 2

1

"2^=

''2^^'

n-1 n 12 2'^¥'^2^'^ + 2i=r + 11 ^ + 2^^'" 2" 3

2'''

2"

—

1 "•"2

1

ii iL_Q_A 2"" 2™'

2""

^=1+1+1 + 1+

5.

"2is.

subtraction,

n

4

1 2^~''''^2'''P'''

By subtraction,

By

3

S = l + 2 + 25'^25 +

4.

i+?+5+ 2+4+8^

2^^^'^^'^2^i'''

= 1+2=3-

[CHAP.

GEOMETRICAL PROGRESSION.

16

S = l + 3x + 6x^+10x^+ xS= x + 3x'+ 6a;'+ (l-x)S = l + 2x + 3x'+ ia^+

6.

/.

By

subtraction,

b

that

is,

are equal

r 9. .-.

By

= ap^,

(n+ 1)'" term of

The sums

8.

-17

are

first series

"^

r-

= -s— — r;

•'•

~

'

= (2?i+l)"' term

and

~ r^-1

'

''I

of second series.

respectively;

and since these

o=a(r + l) = a + ar. ^

1

r^

ratios of the two progressions; then and 6=iig*; hence ^ = 2^

common

Jjetp and q be the

7.

S=l + (l + S)r)-(l+6 + 62)r2+(l + 6 + 62+J5))-3 + r-l-(l + S)j-2 + (l+,6 + 62)r3 +

rS=

subtraction,

{l~r)S=l + ir + iV + b^r^+ 10.

= irbr'

a + ar+ar'' = 70

Wehave

(1);

4a + 4ar''=10ar

from

11. at

l

r=2

(2),

We

shall first

any term, say the

{n

(2);

or -.

shew that the sum of an infinite G.P. commencing + iy\ is equal to the preceding term multiplied by

—r

1-r In this particular example, the value of

1-r 1

1

the

first series is

r

-, so that r=j.

is

=

Again

a+ar=5, hence a=4. 12. S=(x + x^ + x^+...) the second in A. P. 13. in G.P.

14.

+ {a+2a+3a+...);

S={x^ + x* + x^+...) + lxy+xh/^+ a?y^+...); here both

S = (a + 3a + Sa+

...)

+ ^---+

— -...

in G.P.,

series are

;

the

first series is

in A. P.,

of

two

infinite series in

J

the second in G.P.

15.

The

series

may be

expressed as the

16.

The

series

may be

expressed as the difference of two infinite series

mG.P.

sum

G.P.

^]

GEOMETRICAL PROGRESSION.

= - = -;

B.eT6 -

17.

hence

b''

= ao,

17

c^=bd, ad=bc.

Thus

{b-c)^+{c-a)''+{d-bf=V-2bc + c^ + c''-2ca + a^ + d'-2bd + b^ =a^-2bc + d^ = a^-2ad + d''={a-d)\

that

^~- = 2 J^;

Here

18.

+ bf = 16ab,

or a' - liab

+ 62=0;

^|y_ 14^1^+1=0.

is,

19.

so that {a

Giving to r the values 1, 2, 3,...n, we have S=3.2 + 5.22 + 7.2'+ + (2ra + l)2»; .-.

+ {2n-l)2''+(2re + l)2"+i.

3.22 + 5.23 +

2Sf=

Subtracting the upper line from the lower,

S=(2re + l)2'»+i-3.2-(2.22 + 2.2S +

= (2ra + 1) 2"+! - 6 - ^ ^^"~ = (2ra+ 1) 2"+i - 6 - 2 20.

The

series

.

Y

+ 2.2")

^^

2^+1 + 8 = M

.

2"+2 - 2"+! + 2.

ial + a + ac + a^e + aV + a^e" +

= (l + ac + aV+ ton = {l + a){l + ac + aV+

terms)

...

to

to 2n terms

+ a (1 + ac + a^cS +

to

n terms)

m terms)

_( l+a)(a"c''-l) ~ ac-1

21.

We have S„=

we obtain the values

=-Ar

=

22.

—

-

r-l

We have

— a Ir^ '

—

1)

_..

,

and by putting in succeEsionM=l,

of Sj, Sj, Sj,

\{r-l)

+ (r'^-'i.) + {r^-l) + ...

ir+r'+r^+... to

re

to

m

termsj-

terms -ml

[

j

5'i=-i- = 2;

S.,

1--

= -^=3; 13

2

'S3=-^-4,

1-4 4

&0.;

^

:Sj,=

1i)

=p + l. +

l

sum = 2 + 3 + 4+... topterms=|{4 + (i)-l)}=|(p+3)". 2^ H. A. K.

3, 5,...

Thus the required sum

...

'

[CHAP.

THE PROGRESSIONS.

18 23.

Wehave

Now

(l-r™)"

l+r+r^+r^ + that

is positive;

-^_y

•

- 2r^+r^'^>0, or l + r2»>2)». r- 2r™+r«"'-'>0 or r+r^"" i>2r'»;

1

is,

Sinularly r(l-r'»-')2>0; that

is,

r^ (1 + 1-™-*)^ > 0, that is r^ - 2r"'

and generaUy

r"'^-"

> 0,

rP+r2™-P>2r™.

that is

Now i+r + ?-=+rS+r"'+ and

+i-°"=

is therefore

greater

+r*"

= (l + r2'») + (r+r2"-i) + (j^ + '-="~^) + + r™, that is greater than .2r'" + 2r™ +

+'^. than

(2OT+l)r™. .-.

(2m + 1) r"

< \rl^

Multiply both sides by

that

,

(2m + 1) r™ (1 - r)

is

< 1 - ?-2^+^

thus

j-™+',

< r"+Ml - J'™"^^)-

(2m + 1) j-^""*! (1 - r)

w+l

Put 2m + l=n, then

Br''(l-r))-9a62=0;

+ 362(x-3a)=0; whence x = 3o, y = b^-3aK

4x-4=(^«+-iJ-4 = (>--iJ;

44.

.-.

2V^^=v/a-4-1

^"'^ ,

•

-TV, .V, Thus the expression

=

^

\

/

EXAMPLES. Vm. 4.

The produot = (x + w)

=

w wehave

5.

fi 6.

V.

{x

_^ 1

H. A. K.

^2

\

~ ~2~

Pages

'

81, 82.

+ w^) =x^+(o>+ oi^ x + u^ = x'' - x + 1.

V^

+ v'^ = 3 + = S^-^-^ _jj_

-^^^^±^>^The Ihe expression ~ 3

b.

«-l

i

2 ^5

<

+ ^'>/5)\ V^18-20

.

33

[CHAPS.

SURDS AND IMAGINARY QUANTITIES.

34

+ (3 = (3 + 2i) (2 +45i) - -

7.

The expression

8.

The expression =

9.

The expression

=

- 2i) (2-5i)

+ if-ix-i)^

^^^

2(3ia2 + i')

29

=

30^^

2(3is2 + i3) ^^2

+ 1'

j^^— = —^^ =

10. Theexpression=

12.

_

2i(3x2-l)

^TiT

+1

'

Sa^-l

-^

•

l)4n+3

11.

8^

29

^^-r,

^^::^, i

_ 2(6 + 10f) ^ _

iiax

{a+ix)^-{a-ix)''

(x

"

25)

(

^

^

(-V^)^»+3 = (_l)4>.+Sxj^— = (-l)x(^/^l)' = (-l)x(-^/-l)=^/-l•

= (9 + iOi) + (9 - 40i) + 2 ^81 - IBOOi^ = 18 + 2.^/1681 = 100.

The square

Examples IB

may

to 18

be solved by the method of Art. 105, or by inspec-

tion as follows.

J^T= -5 + 2 J^M=-9 + 4:+2,J^^^i=U^+2y.

13.

-5 + 12

14.

-11-60

V^=- 11-2 n/- 900 =

15.

_ 36 + 25 -

J-36x25 = (5 - J^M)^

-il + sJ^S = -47+2j^i8

= (-48 + 1 + 2 16.

2

V^^ = (1+n/^^'-

-8v'^=0-2 J^^=i-i-2 J -4.xi = {2- ,J^f.

17.

a'-l + 2aJ^l = (a+J^\

18.

4a5-2(a2_62)^rT=(a + 6)2_(a-5)2_2(a2-62)

V^l

= {{a + J)-(a-6)V^}'iq 19.

+ 5i _ We we have 3^_^.s/3-iV2 _

20.

(

3

+ 5i)

,

behave

^^

.

12-2i2

14

—- = —

1+i =

19 i + 3i) _ - 9+

(^3-V2)(2^3 + i;V2) _ 8-^6

2v/3-is/2 ^

(2

^_^.^

:

1

(l ^

i)(l + i) l i2 2i +=^^-^5 '-= + = +

l-z^

.

=i.

2

+ i)^_l+i^ + 2i_ 2i _ 2i (3 + i) _ 6i + 2i^ _ 3i-l 3_j 3_j 9_j2 3_i 10

(l

22.

23.

Theexpression=

(7;^);-/°-.;f {a + ib){a~ib)

=

gi^ +b a^

THE THEORY OF QUADRATIC EQUATIONS.

VTII., IX.J

24.

Wehavel + u2=-fc,;

thae

{l

35

+ t^y = {-a)*=a'>=a.

25. Wehave l-o} + J^ = (l + o) + oi^)-2u=0-2i,i=-2oi. Similarly l + (a-ur'=-2uy'.

The product 26. Since the expression =

27.

The

4u'=4.

is

1-(o* = 1-(d and l-u'' = l -w^, (1

- u)^ (1 - u^f

2 + 5ai + 2oj2 = 2

(1

+ u + 0)2) + 3u = 3a>, and

solution of the second part

(3m)«

= 729ai6 = 729.

is similar.

28. The factors are equal to 1 - u+.u^ the product of each pair is 2". Ex. 25.

and

1

-

w^^

+u

alternately,

+ y^ + z^-3xyz = {x+y+z) (x^+y^+z^-yz-ex-xy) = {x+y + z)(x + mj + by'z)(x + i^y + \

= ^3 + js.

x^+y''+z^=ai?+(y+zY-2yz = (a + hf+(a + l)f-2(a?-ab + V) = &ah.

(2)

x^ + y^+z^=3i? + (y+z)(y^+z>-yz)

(3)

= ^ + {yJrz){(y + zf~^z} = (a + 6)3_(o + 6)j(a + 6)2_3{a2-a5 + 52)} = 3(a3 + 63).

EXAMPLES. 13.

(1) ,

Again in

a.

Pages

Ax''+Bx + C=0 are

If the roots of

Now in

IX.

real,

4a^-i (a" ~b^-c^)=ib^ + 4:C^, a

(2),

88, 89, 90.

then

is positive.

U{a-b)^-4:{a-b + c){a-b-c)

= 16(a-6)=-4(a-6)2 + 4c==12(a-6)2 + 4c2, 14,

B^-iAG

positive quantity.

Applying the

a positive quantity.

test for equal roots to the equation

x^-2mx + 8m-15 = 0,

m°=8m-15;

wehave 15,

If the roots are

equal (1

16.

is

(m -

5)

(m - 3) = 0.

+ 3m)' = 7 (3 + 2m)

or Qm"

-

8m - 20 ^ 0,

(9m + 10)(m-2) = 0.

that is

that is

that

On

reduction

we have (m + l)a;^-6a;(m+l)=ox(m-l)-c(m

-1),

(m + l)a;2_{j(m + l) + a(m-l)}a; + c(m-l) = 0.

The required condition

is

obtained by equating to zero the coefficient of

3—2

x.

THE THEORY OF QUADRATIC EQUATIONS.

36

17. If the roots of perfect square.

Ax''+Bx+G=0

In(l), 4c2-4(c + o-6)(c-a +

are rational,

(2),

be

aB=-; whence

1

1

a^ (« + ^) («' + ^' "

~a5V~oj

^y

/"

V;8"gy

(a^-^)^

~

+ ^=

b^-2ac — « -j

_a= + |S2_62_2a£___c^_62j-2ac

a7^«= a*/3^ (a^ + ,33) =

a*^' +

,

a^

a

a

20-

a perfect square.

(

3=~-, a + '^

on

be a

+ b^c^ - 4a6c« - 6a-' -ah + 26^) = c' {9a* + 2ia^i + lOa^b^ - 8a6' + b*) = c^ (Sa" + iab - 6^)2 = a perfect square.

(3a2

In Examples 18 to 20 we have

19.

B^-iAC must

6)

= ic^ - 4(;2 + 4 (a - bf = 4 (a - 6)^ In

[CHAP.

a>^^

(g

+

~

/3)°

(g

«/3)

a'

a^

- p)^ _

fg

+ j3)° Ua + ^)^ - 4a{(ac-M)2-2(a(J-Jc)2}2, that

is,

that

is,

that

6c)*,

{ac-id)^>(ad-bc)^; (ac-bd + ad- be) is a positive quantity;

(ac -bd-ad + be) + b){c- d) (a -b)(c + d),

is,

(a

.-.

- UY> (ac - bdy - 4 (ae - id)" {ad -bo)^+i{ad-

(ac

Hence

a"

~ V and

c^

or (a« -

If) (c"

- d') must be

- d^ must have the same

EXAMPLES.

IX.

Page

ff.

positive.

sign.

95.

and 2 may be solved by application of the formula of

Questions 1 Art. 127. 1.

m- 1 + 3 = 0,

Here

m=- 2.

whence

Or thus: the given equation may be written 2x{y + l) + y^+my-3=0; hence y + 1 must be a factor of y'+my -3; that ia,y=-i must satisfy the equation

y^+my-3=0.

2.

Here the condition gives-12-

3.

The condition that the

should be real

is

— =0, whence m'=i9. — + m^ 25

roots of

Ax'^-{'B-C)xy-Ay'^=(S that (B - Cf-^iA? should be a positive quantity:

this con-

dition is clearly satisfied. 4.

common + 2)-(a;2+ya; + 5') =

Since the equations are satisfied by a (a;2+^a;

Also by eliminating the absolute term,

we

5.

(1)

we

When

get

x=^^'

the condition

,

and from

is fulfilled,

(2)

(1).

+ 3') = o

z=-

(2;.

^^'~^'^ .

the equations

h? +mxy + ny^=0 and l'x^ + m'xy + n'y^=Q must be satisfied by a common value of the ratio x y. :

From

these equations

we have by

_ mn' - m'n

whence 6.

(nl'

cross multiplication

^y nV - n'l

- n'l)' = {mn' - m'n)

we must have

obtain

g'(a;2+px + 2)-}(a;2+p'a;

From

root,

Im'

^ -

{Im'

I'm

- I'm).

Applying the condition of Art. 127, we have

6- 4aP- 12- 2a2-p2=0.

THE THEORY OF QUADRATIC EQUATIONS.

IX.]

41

2/ -ma; is a factor of ax^+ihxy + hy^, then tins last expression when 2/ = ma;; that is, a + 2hm+bm,^=0. Similarly iimy + xisa, factor of a'x^ + 2h'xy + h'y^, we must have

7.

If

vanishes

a'm'-2h'm + b' = 0.

From

these equations,

we have by

cross multipUcation

1 _ m _ ~ -2(bh' + a'h) 2{b'h+ah') ~ aa" {aa' - bby= - 4 {ah' + b'h) (a'h + bK). mi'

whence

W

x^-x(Zy + 2)+2y^-Zy~Z5 = 0; Here 8. whence solving as a quadratic in x, 2ar=32/ + 2±v'(32/ + 2f-4(2j/2-3j/-35) = 32/+2±(2/ + I2). Giving to y any real value, we find two real values for x: or giving to x any real value we find two real values for y. 9.

Solving the equation 9a^ + 2a (y- 46)+;/^-

ratic in a,

we have 9a;=-

(2/

-46)±7(j/ -46)2-9

20y+ 244=0

as a quad-

(3/2-20!/ + 244>

= -(y-46)± V-8(j/2-llt/ + 10) = -(2/-46)±^/-8(2/-l)(y-10). 1

Now the quantity under the radical is only positive when y Ues between and 10; and unless y Ues between these limits the value of x wUl be

imaginary.

Again

j/2+2?/ (a;-10)

whence

Thus in

+ 9a;2-92ar+244=0;

y=-{x-10)± J{x- lO)^ - (9x2 - g2x + 244) = -{x-10)J= J -8 {x-6) {x-3}. order that y may be real x must He between 6 and

8.

x^{ay + a')+x{by + 10. Wehave solving this equation as a quadratic in x,

b')+cy + c'=0;

2 {ay + a')

x=

-

+ b')±J{by + b'f-4:{ay + a')

(by

{cy

+ &).

order that x may be a rational function of y the expression under the radical, namely (62-4oc)!/2 + 2(66'-2ac'-2a'c)y + 6'2-4a'c', must be the square of a linear function of y; {bb' - 2ac' - 2a'c)^ ={b^- 4ac) (6'2 - 4a'c') hence

Now in

we have + a'^a^ - ac'bV - a'cbV + 2aa'cc'=4,aa'cc' -acb'"- a'c'b^ o'%2 - 2aa'cc' = ac'bb' + a' ebb' - acb'^ - a'c'b' a^c'^

Simplifying a'c'^ .-.

;

.-.

{ac>

- a'cf={aV -

EXAMPLES. 1.

(a-i - 4)

(ic-^

+ 2) =

;

X.

a.

a'b) {be'

-

Pages

whence - = 4 or - 2.

b'e).

102, 103.

;

[CHAP.

MISCELLANEOUS EQUATIONS.

42 2.

(a;-2-9)(a;-2-l)=0; whence

3.

(2s4 -

4.

(3a;4

5.

(a:»-3)(a;'*-2)=0.

7.

Putting

y= x/o> we have

8.

Putting

y= \/ YZT

9.

(Ss^-l)

-

1) (a;4

- 2) =

2) (2a;4

_

whence

;

=

1)

i=9 \/a;

or 1.

=-

or

2.

1

whence ijx=-^ or

;

11

i

i_

(2a;i

'

'^^

7y + -

= -^-;

satisfied

whence

= 0.

y=ar

whence "-"""' 2/=^ " 3' !>-2 or

2/

-5.

or

a modified form of the given equation.

11.

(3>^-9)(3»'-l)=0; whence 3=^=9 or

12.

(t.5'-l) (5»-5)=0; whence 5*=|=5-i, and 5»^=5.

13.

22=^8

14.

8.22»^-65.2=: + 8 = 0; that

2"=+*

+ 1 = 0;

2^

1)2

that

16.

Putting y = J2x, we have

.

is (8

.

2»'-l) (2»'-8) 2==

3

+ 4=0.

= 0;

= 2'. 2"^

59

v

a;

= 1. 1

--5 = 10;

whence y = ^ot - 30.

(x-7)(a; + 5)(a;-3)(^ + l) = 1680; (!