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4. The roots of the equation

x 2 +5x= — 6,

are —2 and —3. Their sum is —5, and their product
is +6

Let these principles be applied to each of the examples
under " examples in the third form."

Fourth Form.

132. The fourth form is,

x 2 — 2px= — q ;
and by resolving the equation we find,
1st root, x=p-\-^/—q-\-p 2

2nd root, #— p — y^— q-j-p 2

Their sum is —2p.

In this form, as well as in the third, the quantity under
the radical being less than p 2 , its root will be less than p :
hence both the roots will be positive, and the first will be
the greatest.

If we multiply the two roots together, we have

EQUATIONS OP THE SECOND DEGREE 199

Hence we conclude,

1 st. That in the fourth form both the roots are positive.

2nd. That the first root is greater than the second.

3rd. That the sum of the roots is equal to the coefficient of
x in the second term, taken with a contrary sign.

4th. That the product of the roots is equal to the known
term in the second member, taken with a contrary sign.

EXAMPLES.

1 . The roots of the equation

x 2 —7x=-\2,

are + 4 and +3. Their sum is +7 and their pro-
duct + 12.

2. The roots of the equation

x 2 — 14a?=—24,

are +12 and +2. Their sum is +14 and their pro-
duct + 24.

3. The roots of the equation

x 2 — 20;r=: — 36,

are +18 and +2. Their sum is +20 and their pro-
duct + 36.

4. The roots of the equation

x 2 — 17#=— 42,

are +14 and +3. Their sum is +17 and their pro-
duct + 42.

Quest. — 132. In the fourth form, what are the signs of the roots 1
Which root is the greatest 1 What is the sum of the roots equal to 1
What is their product equal to ?

200 PIRST LESSONS IN ALGEBRA.

133. In the third and fourth forms the values of x some-
times become imaginary, and in such cases it is necessary
to know how the results are to be interpreted.

If we have q^>p 2 , that is, if the known term is greater
than half the coefficient ofx squared, it is plain that y — q-\-p 2
will be imaginary, since the quantity under the radical
will be negative. Under this supposition the values of x
in the third and fourth forms will be imaginary.

We will now show^that, when in the third and fourth
forms, we have q^>p 2 , the conditions of the question will be
incompatible with each other.

134. Before showing this it will be necessary to estab-
lish a proposition on which it depends : viz.

Jf a given number be decomposed into two parts and those
parts multiplied together, the product will be the greatest pos-
sible when the parts are equal.

Let 2p be the number to be decomposed, and d the differ-
ence of the parts. Then

p ^ as the greater part (page 80, Ex. 7.)

w

and p as the less part.

d 2
and p 2 — t~^' their product (Art. 40.)

Now it is plain that P will increase as d diminishes, and
that it will be the greatest possible when d—0 : that is,

pXp=p 2 is the greatest product.

Quest. — 133. In which forms do the values of x become imaginary 1
When will the values of x be imaginary ! Why will the values of x be
then imaginary 1

UF THE

UNIVERSITY

- OF

EQUATIONS OF THE SECOND DEGREE. 201

Now, since in the equation

x 2 — 2px — — q

Xp is the sum of the roots, and q their product, it follows
that q can never be greater than p 2 . The conditions of the
equation, therefore, fix a limit to the value of q, and if we
make q>p 2 , we express by the equation a condition which
by the values of x becoming imaginary. Hence we may
conclude that,

When the values of the unknown quantity are imaginary,
the conditions of the question are incompatible with each other.

EXAMPLES.

1. Find two numbers whose sum shall be 12 and pro-
duct 46.

Let x and y be the numbers.

By the 1st condition, #+y=12 ;

and by the 2d, xy=46.

The first equation gives

x=l2—y.
Substituting this value for x in the second, we have

I2y—y 2 = 46;
and changing the signs of the terms, we have
y 2 — I2y=. :— 46.

Quest. — 134. What is the proposition demonstrated in Article 134]
If the conditions of the question are incompatible, how will the values
of the unknown quantity be 1

202 FIRST LESSONS IN ALGEBRA.

Then by completing the square

y 2 -12y+36 = _46 + 36 = -10

which gives y = 6 + -y/— 10,

and y = 6— V — 1°;

both of which values are imaginary, as indeed they should
be, since the conditions are incompatible.

2. The sum of two numbers is 8, and their product 20 :
what are the numbers ?

Denote the numbers by x and y.
By the first condition,

x+y=8;
and by the second, xy=20.

The first equation gives

x=8— y
Substituting this value of x in the second, we have
8y-y*=20 ;
changing the signs, and completing the square, we have

y 2 -8y+16 = -4 ;
and by extracting the root,

y__4_j_^/Z74 an( j y— 4 — ^/— 4.

These values of y maybe put under the forms (Art. 106),

y=4+2-/^T and y=4— 2-/=T7

3. What are the values of x in the equation

#2+2*= — 10.

Am*. (— 1+3^H

EQUATIONS OF THE SECOND DEGREE. 203

Examples with more than one unknown quantity.

1. Given \ n n A^T to find x and y.
<a? 2 +y 2 = 100 J y

By transposing y in the first equation, we have

#=14— y ;

and by squaring both members,

* 2 = 196 — 28y+y 2 .

Substituting this value for x 2 in the 2nd equation, we have

196-28y+y 2 +y 2 =100;

from which we have

y 2 -i4y=-48;

and by completing the square,

3/ 2_ 14y+ 49_- 1 .

and by extracting the square root,

y-7=±' V / ~l~=+l or -1:
hence, y = 7+l=8, or y = 7 — 1=6.

If we take the larger value, we find x=6 ; and if we
take the smaller, we find x =8.

Verification.
For the largest value, y=8, the equation
#+y = 14 gives 6 + 8 = 14;
and o: 2 +y 2 =100 gives 36 + 64 = 100.

For the value y-6, the equation

#+y = 14 gives 8 + 6 = 14 ;
and # 2 +y 2 =100 gives 64 + 36 = 100.

Hence, both sets of values will satisfy the given equation.

204 FIRST LESSONS IN ALGEBRA.

2. Given \ n % ~~ .. I to find x and y.

i x 2 —y 2 = 45 ) *

Transposing y in the first equation, we have
x=3+y ;
and then squaring both members,

# 2 = 9 + 6y-|-y 2 .

Substituting this value for x 2 in the second equation, we
have

9 + 6y + y 2 — y 2 =45 ;
whence we have

6y-—36 and y=6.
Substituting this value of y in the first equation, we have

X — 6 = 3,

and consequently #=3 + 6 = 9.

Verification.
x—y=3 gives 9 — 6 — 3 ;
and x 2 — y 2 = 45 gives 81—36=45.

„ ^. t x 2 +3xy =22 )

3. Given < _ , _ . - ^ ' .."> to find a and y.

J x 2 +3xy+ 2y 2 = 40 J y

Subtracting the first equation from the second, we have
2y 2 = 18,
which gives y 2 = 9,

and y—^r^ or — 3.

Substituting the plus value in the first equation, we have
# 2 +9a;=22;

EQUATIONS OF THE SECOND DEGREE. 205

from which we find

x—-\-2 and #= — 11.

If we take the negative value, y= — 3, we have from the
first equation,

x 2 — 9*=22 ;
from which we find

#= + 11 and x=— 2.

Verification.

For the values y=+3 and x =+2, the equation

# 2 +3a:y=22

gives 2 2 +3x2x3 = 4 + 18=22 :

and for the second value, x = — 1 1 , the same equation

# 2 + 3;ry=22

gives (-ll) 2 + 3x -11x3 = 121—99=22.

If now we take the second value of y, that is, y= — 3,
and the corresponding values of x, viz, x = + 1 1 , and
a:=— 2 ; for ate + 11, the equation

x 2 +3xy=22
gives ll 2 +3xllX -3 = 121-99=22 ;

and for #=— 2, the same equation

a? 2 + 3#y=22
gives (-2) 2 +3x -2 X -3=4+18=22.

#*=y 2 (1)
4. Given

2 +y 2 +s 2 =21 (3)

( ocz=f (IK

< # +y +j? =7 (2) > to find x, y, and

( a: 2 +y 2 +^ 2 =21 (3) )

18

206 FIRST LESSONS IN ALGEBRA.

Transposing y in the second equation, we have
x+z=7-y (4) ;
then squaring the members, we have

x 2 +2xz+z 2 —49 — 14y-fy 2 .

If now we substitute for 2xz its value taken from the
first equation, we have

x 2 +2y 2 +z 2 =49 — Uy+y 2 ;

and cancelling y 2 in each member, there results

x 2 +y 2 +z 2 =49—14y.

But, from the third equation we see that each member of
the last equation is equal to 21 : hence

49-14y=21,

and 14y=49— 21=28.

28 «
hence, y=— =2.

' 14

Placing this value for y in equation (1) gives

xz=z4 ;

and placing it in equation (4) gives

x-\-z=5, and # = 5 — z.

Substituting this value of x in the previous equation, we
obtain

5z — z 2 =z4 or z 2 — 5z=— 4;
and by completing the square, we have
s2_5£ +6j2 5=2,5,

and z— 2,5 = ±-/23 = + l,5 or — *> 5 ;

hence, ^=2,5 + 1,5=4 or #=+2,5 — l,5=rl.

EQUATIONS OF THE SECOND DEGREE. 207

If we take the value

#=4, we find x=l :
if we take the less value

s=l, we find x=4.

r 3. Given • + ^xj+y = 19 > find and

and a: 2 + a?y-fy 2 = 133 J y

Dividing the second equation by the first, we have

x—-y/xy + y— 7

but x-\-^x~y-{- y=19

or a? + y=13
and substituting in 1 st equa. -\/x~y+ 13 = 19

or yxy^z 6

and by squaring a?y=36

From 2d equation, x 2 +xy +y 2 =\33

and from the last 3xy =108

Subtracting x 2 — 2xy-{-y 2 = 25

hence, x— y=± 5

but x-\-y= 13

hence a?=9 or 4 ; and y=4 or 9.

6. Given the sum of two numbers equal to a, and the
sum of their cubes equal to c, to find the numbers

By the conditions < m * m

208 FIRST LESSONS IN ALGEBRA.

Putting x =:s+z, and y=zs—z, we have

a— 2s. or s=z — ;

2 '

, I * 3 =^ 3 + 3s 2 ^+3^ 2 +* 3

( y 3 — s 3 — 3s%+3s# 2 — ^ 3 .

hence, by addition, x 3 -\-y 3 =z2s 3 -\-6sz 2 = c,

1 c—2s 3 . /~c~^2s 3 ~

whence ^= — and ^=±\/ — :

os V 05

nr_2s 3 I

*- ff± V67~ ; and y= szp \

c—2s 3

6s
or by putting for s its value,

a / / c — t-\ a /4c— a

12a '

Ac — a 3
\2a '

a*

and r-^vCTKrO^f^v/

Note. — What are the numbers when a =.-5 and c=35.
What are the numbers when # = 9 and c=243.

QUESTIONS.

1. Find a number such, that twice its square, added to
three times the number, shall give 65.

Let x denote the unknown number. Then the equation
of the problem will be

2x 2 +3x.=z65,
whence

3 f6T-\

9__ ___3_ 23
§~ T V

EQUATIONS OF THE SECOND DEGREE. 209

Therefore,

3 , 23 ■ , 3 23 13
x=z — 5, and x— ■ — = — -—.

4 4' 44 2

Both these values satisfy the question in its algebraic
sense. For,

2x(5) 2 +3x5=2x25-H5:=65 ;
13\ 2 . 13 169 39 130

and

/ 13^ , « 13 169 39 13U _ K

Remark. — If we wish to restrict the enunciation to its
arithmetical sense, we will first observe, that when x is
replaced by —x, in the equation 2x 2 -\-3x—65, the sign of
the second term 3a? only, is changed, because (— x) 2 =x 2 .

3 23

Therefore, instead of obtaining a?=s — — ±— , we should

find x= — db — , or #=— , and x =—-5, values which only

differ from the preceding by their signs. Hence, we may

13
say that the negative solution , considered indepen-

dently of its sign, satisfies this new enunciation, viz : To
find a number such, that twice its square, diminished by three
times the number, shall give 65. In fact, we have

»*(T)"-»*r=ir-¥=«-

Remark. — The root which results from giving the plus
sign to the radical, generally resolves the question both
in its arithmetical and algebraic sense, while the second
root resolves it in its algebraic sense only.

18*

210 FIRST LESSONS IN ALGEBRA.

Thus, in the example, it was required to find a number,
of which twice the square added to three times the number
shall give 65. Now, in the arithmetical sense, added means
increased ; but in the algebraic sense it implies diminution,
when the quantity added is negative. In this sense, the
second root satisfies the enunciation.

2. x\ certain person purchased a number of yards of cloth
for 240 cents. If he had received 3 yards less of the same
cloth for the same sum, it would have cost him 4 cents more
per yard. How many yards did he purchase ?

Let x— the number of yards purchased.

Then will express the price per yard.

If, for 240 cents, he had received 3 yards less, that is
x—3 yards, the price per yard, under this hypothesis, would

^.w^,,,^ g *******

this last cost would exceed the first by 4 cents. Therefore,
we have the equation

240 240

— o = 4 ;

whence, by reducing x 2 — 3x =180,
3 /9~

3+27

+ 180:

2 '
therefore #=15 and x= — 12.

The value x=15 satisfies the enunciation; for, 15 yards

24-0
for 240 cents gives , or 16 cents for the price of

1 D

one yard, and 12 yards for 240 cents, gives 20 cents for the
price of one yard, which exceeds 16 by 4.

EQUATIONS OF THE SECOND DEGREE. 211

As to the second solution, we can form a new enuncia-
tion, with which it will agree. For, going back to the
equation, and changing x into —x, it becomes

240 240 240 240

_— 4 ? or rz= 4 >

— x — 3 — x x x-\-3

an equation which may be considered the algebraic transla-
tion of this problem, viz : A certain person purchased a num-
ber of yards of cloth for 240 cents : if he had paid the same
sum for 3 yards more, it would have cost him 4 cents less per
yard. How many yards did he purchase ?

Ans. a?=12, and #= — 15.

3. A man bought a horse, which he sold after some time
for 24 dollars. At this sale, he loses as much per cent,
upon the price of his purchase as the horse cost him.
What did he pay for the horse ?

Let x denote the number of dollars that he paid for the
horse, #—24 will express the loss he sustained. But as

x

he lost x per cent, by the sale, he must have lost

F J 100

upon each dollar, and upon x dollars he loses a sum de-
noted by —^r ; we have then the equation

=x—24, whence x 2 — 100#=— 2400.

100

and x=50dzi/2500— 2400 = 50 ±10.

Therefore, x=60 and #=40.

Both of these values satisfy the question.

For, in the first place, suppose the man gave \$60 for the
horse and sold him for 24, he loses 36. Again, from the
enunciation, he should lose 60 per cent, of 60, that is,

212 FIRST LESSONS IN ALGEBRA.

of 60, or — , which reduces to 36 ; there-

100 ' 100

fore, 60 satisfies the enunciation.

Had he paid \$40, he would have lost \$16 by the sale ;

40
for, he should lose 40 per cent, of 40, or 40 x — rjr-, which

reduces to 16 ; therefore, 40 verifies the enunciation.

4. A man being asked his age, said the square root of
my own age is half the age of my son, and the sum of
our ages is 80 years : what was the age of each ?

Let x=z the age of the father.
y= that of the son.
Then by the first condition

and by the second condition

x+y=Q0.
If we take the first equation

and square both members, we have

4
If we transpose y in the second, we have
x=80—y:
from which we find

y=z— 2±<\/324=z 16;

by taking the plus root, which answers to the question in
its arithmetical sense. Substituting this value, we find
a? =64. A \$ Father's age 64

( Son's 16.

t < EQUATIONS OF THE SECOND DEGREE. 213

5. Find two numbers, such that the sum of their pro-
ducts by the respective numbers a and b, may be equal to
2s, and that their product may be equal to p.

Let x and y be the required numbers, we have the equa-
tions

ax-\-by—2s.
and %y=p.

2s — ax
From the first y=— - ;

whence, by substituting in the second, and reducing,
ax 2 — 2sx = — bp.
s 1

Therefore, x— — db — \/ s 2 —abp,

a a

and consequently,

s 1

y=—+—^/s 2 -abp.

This problem is susceptible of two direct solutions, be-
cause s is evidently > <y/s 2 —abp ; but in order that they
may be real, it is necessary that s 2 > or = abp.

Let a=b — l ; the values of x and y reduce to

x—szh-y/s 2 — p and y=LS^p-\/s 2 — p.

Whence we see, that the two values of x are equal to
those of y, taken in an inverse order ; which shows, that if
s-\- -\/s 2 — p represents the value of x, s — -y/s 2 — p will re-
present the corresponding value of y, and reciprocally.

This circumstance is accounted for, by observing that in
this particular case the equations reduce to

x-\-y=z2s.

c x+y = 2s, )
I ocy—p ; J

214 FIRST LESSONS IN ALGEBRA.

and then the question is reduced to, finding two numbers of
which the sum is 2s, and their product p, or in other words,
to divide a number 2s, into two such parts, that thett product
may be equal to a given number p.

Let us now suppose

2^14 and pz=z4S:
what will then be the values of x and y ?

Ans. < x
( v -

:8 or 6.
1 y—6 or 8.

6. A grazier bought as many sheep as cost him £60, and
after reserving fifteen out of the number, he sold the re-
mainder for £54, and gained 2s. a head on those he sold :
how many did he buy 1 Ans. 75.

7. A merchant bought cloth for which he paid «£T33 lbs.,
which he sold again at £2 8s. per piece, and gained by the
bargain as much as one piece cost him : how many pieces
did he buy 1 Ans. 15.

S. What number is that, which, being divided by the pro-
duct of its digits, the quotient is 3 ; and if 18 be added to
it, the digits will be inverted ? Ans. 24.

9. To find a number, such that if you subtract it from 1 0,
and multiply the remainder by the number itself, the product
shall be 21. Ans. 7 or 3.

10. Two persons, A and B, departed from different places
at the same time, and travelled towards each other. On
meeting, it appeared that A had travelled 18 miles more
than B ; and that A could have gone B's journey in 15f
days, but B would have been 28 days in performing A's
journey. How far did each travel ?

A (A 72 miles
Ans. -J

B 54 miles.

EQUATIONS OF THE SECOND DEGREE. 215

11. There are two numbers whose difference is 15, and
half their product is equal to the cube of the lesser num-
ber. What are those numbers 1 Ans. 3 and 18.

12. What two numbers are those whose sum, multiplied
by the greater, is equal to 77 ; and whose difference, multi-
plied by the lesser, is equal to 12 ?

Ans, 4 and 7> or fyTand y <\pZ.

13. To divide 100 into two such parts, that the sum of
their square roots may be 14. Ans. 64 and 36.

14. It is required to divide the number 24 into two such
parts, that their product may be equal to 35 times their dif-
ference. Ans. 10 and 14.

15. The sum of two numbers is 8, and the sum of their
cubes 152. What are the numbers ? Ans. 3 and 5.

16. Two merchants each sold the same kind of stuff;
the second sold 3 yards more of it than the first, and to-
gether they receive 35 dollars. The first said to the second,
other replied, " And I should have received 12^ dollars for
yours." How many yards did each of them sell ?

A C 1st merchant x= 15 x=5.

Ans. ? , or

*2nd „ y = 18 ° r y=8.

17. A widow possessed 13,000 dollars, which she divided
into two parts, and placed them at interest, in such a man-
ner, that the incomes from them were equal. If she had
put out the first portion at the same rate as the second, she
would have drawn for this part 360 dollars interest ; and if
she had placed the second out at the same rate as the first,
she would have drawn for it 490 dollars interest. What
were the two rates of interest ?

Ans. 7 and 6 per cent.

*

216 FIRST LESSONS IN ALGEBRA.

CHAPTER VII.
Of Proportions and Progressions.

135. Two quantities of the same kind may be compared
together in two ways : —

1st. By considering how much one is greater or less than
the other, which is shown by their difference ; and,

2nd. Bj considering how many times one is greater or
less than the other, which is shown by their quotient.

Thus, in comparing the numbers 3 and 12 together with
respect to their difference, we find that ] 2 exceeds 3 by 9 ;
and in comparing them together with respect to their quo-
tient, we find that 12 contains 3 four times, or that 12 is 4
times as great as 3.

The first of these methods of comparison is called Arith-
metical Proportion, and the second Geometrical Proportion.

Hence, Arithmetical Proportion considers the relation of
quantities with respect to their difference, and Geometrical
Proportion the relation of quantities with respect to their
quotient.

Quest. — 135. In how many ways may two quantities be compared
together 1 What does the first method consider 1 What the second 1
What is the first of these methods called 1 What is the second called 1
How then do you define the two proportions 1

ARITHMETICAL PROPORTION. 217

Of Arithmetical Proportion and Progression,

136. If we have four numbers, 2, 4, 8, and 10, of

which the difference between the first and second is equal
to the difference between the third and fourth, these num-
bers are said to be in arithmetical proportion. The first
term 2 is called an antecedent, and the second term 4, with
which it is compared, a consequent. The number 8 is also
called an antecedent, and the number 10, with which it is
compared, a consequent.

When the difference between the first and second is equal
to the difference between the third and fourth, the four num-
bers are said to be in proportion. Thus, the numbers

2, 4, 8, 10,

are in arithmetical proportion.

137. When the difference between the first antecedent
and consequent is the same as between any two adjacent
terms of the proportion, the proportion is called an arith-
metical •progression. Hence, a progression by differences, or
an arithmetical progression, is a series in which the succes-
sive terms continually increase or decrease by a constant
number, which is called the common difference of the
progression.

Thus, in the two series

1, 4, 7, 10, 13, 16, 19, 22, 25, . . .
60, 56, 52, 48, 44, 40, 36, 32, 28, . . .

Quest. — 136. When are four numbers in arithmetical proportion'?
What is the first called 1 What is the second called ! What is th©
third called 1 What is the fourth called !

19

218 FIRST LESSONS IN ALGEBRA.

the first is called an increasing progression, of which the
common difference is 3, and the second a decreasing pro-
gression, of which the common difference is 4.

In general, let a, b, c, d, e,f, . . . designate the terms
of a progression by differences ; it has been agreed to write
them thus :

a.b.c.d.e.f.g.h.i.k...

This series is read, a is to b, as b is to c, as c is to d, as d
is to e, Slc. This is a series of continued equi-differences,
in which each term is at the same time a consequent and
antecedent, with the exception of the first term, which is
only an antecedent, and the last, which is only a consequent.

138. Let r represent the common difference of the
progression

a . b . c . d . e . f . g . h, &c,

which we will consider increasing.

From the definition of the progression, it evidently follows
that

bz=a+r, c—b-\-r—a-\-2r, d=c + rz=za-\-3r;

and, in general, any term of the series is equal to the first
term plus as many times the common difference as there are
preceding terms.

Thus, let I be any term, and n the number which marks
the place of it : the expression for this general term is

l=a-{- (n — \)r.

Quest. — 137. What is an arithmetical progression 1 What is the
number called by which the terms are increased or diminished ! What
is an increasing progression 1 What is a decreasing progression 7
Which term is only an antecedent 1 Which only a consequent 1

ARITHMETICAL PROGRESSION. 219

Hence, for finding the last term, we have the following

RULE.

I. Multiply the common difference by one less than the
number of terms.

II. To the product add the first term: the sum will be the
.last term.

EXAMPLES.

The formula l—a+(n — \)r serves to find any term
whatever, without our being obliged to determine all those
which precede it.

1. If we make n=l, we have l=a; that is, the series
will have but one term.

2. If we make ?i=2, we have l=a+r ; that is, the series
will have two terms, and the second term is equal to the
first plus the common difference.

3. If a — 3 and r—2, what is the 3rd term ? Ans. 7.

4. If «=5 and r=4, what is the 6th term? Ans. 25.

5. If a=:7 and r=5, what is the 9th term? Ans. 47.

6. If a — 8 and rz=5, what is the 10th term ?

Ans. 53.

7. If a— 20 and r—\, what is the 12th term?

Ans. 64.

8. If a=40 and r=20, what is the 50th term ?

Ans. 1020.

Quest. — 138. Give the rule for finding the last term of a series when
the progression is increasing.

220 FIRST LESSONS IN ALGEBRA.

9. If 0=45 and r=30. what is the 40th term ?

Arts. 1215.

10. If a=30 and r—20, what is the 60th term?

Ans. 1210.

11. If a = 50 and r=10, what is the 100th term?

Ans. 1040.

12. To find the 50th term of the progression

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