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the binomial theorem.

Let us now apply these principles in the two following
examples, in which the coefficients are omitted : —

(a+b) 6 . . . a 6 +a b b + aW+a?P + a 2 W + afr+b\
(a — b) 6 . . . a 6 — a 5 b+a±b 2 —a 3 b 3 +a 2 ¥—ab 5 + b 6 .

As the pupil should be practised in writing the terms with-
out the coefficients and signs, before finding the coefficients,
we will add a few more examples.

Quest. — 88. Which is the leading letter of the binomial 1 What is
the exponent of this letter in the first term 1 How does it change in the
terms towards the right 1 What is the exponent of the second letter in
the second term 1 How does it change in the terms towards the right 1
What is it in the last term 1 What is the sum of the exponents in any
term equal to?

BINOMIAL THEOREM. 129

1. (a+b) 3 . . a 3 + a 2 b+ab 2 + b 3 .

2. (a— by . . « 4 — a 3 b-\-a 2 b 2 — ab 3 +b\

3. (a+b) 5 . . a 5 +a*b+a 3 b 2 +a 2 b 3 +ab±+b 5 .

4. (a -b) 1 . . a I -a G b + a 5 b 2 -a*b 3 -\-a 3 b*-a 2 b 5 +ab 6 - b 7 .

Of the Coefficients.

89. The coefficient of the first term is unity. The co-
efficient of the second term is the same as the exponent of
the given power. The coefficient of the third term is found
by multiplying the coefficient of the second term by the
exponent of the leading letter, and dividing the product ley
2. And finally— If the coefficient of any term be multiplied
by the exponent of the leading letter, and the product divided
by the number which marks the place of that term from the
left, the quotient will be the coefficient of the next term.

Thus, to find the coefficients in the example

(a—bf.:. aT-atb + aW-aW + aW-aW+ab^V

we first place the exponent 7 as a coefficient of the second
term. Then, to find the coefficient of the third term, we
multiply 7 by 6, the exponent of a, and divide by 2. The
quotient 21 is the coefficient of the third term. To find the
coefficient of the fourth, we multiply 21 by 5, and divide
the product by 3 : this gives 35. To find the coefficient of
the fifth term, we multiply 35 by 4, and divide the product
by 4 : this gives 35. The coefficient of the sixth term,
found in the same way, is 21 ; that of the seventh, 7 ; and
that of the eighth, 1. Collecting these coefficients, we
have

(«-*)' =

u'-'7a%—2\a i b 2 —Z5(i i b' i +^a 3 b i —2\a i b 6 +7a¥~b\

130 FIRST LESSONS IN ALGEBRA.

Remark. — We see, in examining this last result, that the
coefficients of the extreme terms are each unity, and that
the coefficients of terms equally distant from the extreme
terms are equal. It will, therefore, be sufficient to find the
coefficients of the first half of the terms, from which the
others may be immediately written.

EXAMPLES.

1. Find the fourth power of a-{-b.

Ans. a±+4a 3 b+6a 2 b 2 + 4ab 3 +b*.

2. Find the fourth power of a — b.

Ans. a ±—4a 3 b+6a 2 b 2 —4ab 3 -{-b±.

3. Find the fifth power of a+b.

Ans. a 5 +batb+lQa 3 b 2 +\Qa 2 b 3 +5aV+b 5 .

4. Find the fifth power of a — b.

Ans. a 5 — 5a±b+l0a 3 b 2 —\0a 2 b 3 +5ab±—b 5 .

5. Find the sixth power of a-{-b.

Ans. a 6 +6a 5 b+15a*b 2 +20a 3 b 3 -\-l5a 2 b±+6ab 5 + b 6 .

6. Find the sixth power of a—b.

Ans. a 6 — 6a 5 b+l5a*b 2 —20a 3 b 3 +l5a 2 b*—6ab 5 + b 6 .

7. Let it be required to raise the binomial 3a 2 c— 2bd to
the fourth power.

It frequently occurs that the terms of the binomial are
affected with coefficients and exponents, as in the above

Quest. — 89. What is the coefficient of the first term'? What is the
coefficient of the second 1 How do you find the coefficient of the third
term 1 How do you find the coefficient of any term 1 What are the
coefficients of the first and last terms ? How are the coefficients of
terms equally distant from the extremes 1

BINOMIAL THEOREM. 131

example. In the first place, we represent each term„of the
binomial by a single letter. Thus, we place

3a 2 c=:x, and — 2bd=y,

we then have

{x-{-yY=:x ]: +4x 3 y-\-Qx 2 y 2 -\-4xy 3 -\-y A .

But, x 2 =9a 4 c 2 , x 3 =27a e c 3 , * 4 = 81a 8 c 4 ;

and y 2 = 4b 2 d 2 , f=z - 8b 3 d 3 , y 4 = I6b*d*.

Substituting for x and y their values, we have

(3a 3 c-2bd)± = (3a 2 cy+4(3a 2 c) 3 (-2bd) + 6 (3a 2 c) 2 (-2bd) 2
-f4(3« 2 c) (-2bd) 3 + (-2bd)S

and by performing the operations indicated,
(3a 3 c-2bdy = 81 a 8 c i -216a G c 3 bd + 216a 4 c 2 6 2 d r2 -96a 2 c6 3 d 3
-f 166 4 d 4 .

8. What is the square of 3a — 6b ?

Ans. 9a 2 — 36ab+36b 2

9. What is the cube of 3x — 6y ?

Ans. 27x 3 — l62x 2 y-i-324xy 2 —2l6b 3 .

10. What is the square of x — y ?

Ans. x 2 — 2xy-\-y 2 .

11. What is the eighth power of m-\-n 1

Ans. m s +8m 1 n + 28m 6 n%56m 5 n 3 -{- 70mW+56m 3 n 5 .
+ 28m 2 n 6 -\-8mn I + n s .

12. W r hat is the fourth power of a— 3b 1

Ans. a 4 — I2a 3 b+54a 2 b 2 — I08ab 3 +Slb*.

13. What is the fifth power of c — 2d 1

Ans. c 5 — \0cH+4Qc 3 d 2 — 80c 2 d 3 + 80cd*—32d 5 .

14. What is the cube of 5a — 3d 1

Ans. I25a 3 —225a 2 d+l35ad 2 — 27d 3

132 FIRST LESSONS IN ALGEBRA.

Remark. The powers of any polynomial may easily be
found by the Binomial Theorem.

15. For example, raise a-\-b-\-c to the third power.
First, put .... b+c=d.

Then, . (a+h + c) 3 = (a + d) 3 =a 3 +3a 2 d+3ad?+d?.
Or, by substituting for the value of d,

(a+b+c) 3 = a 3 + 3a 2 b+3ab 2 + b 3

3a 2 c + 3b 2 c-\- 6abc

+ 3ac 2 +3bc 2

+ c 3 .

This expression is composed of the cubes of the three
terms, plus three times the square of each term by the first
powers of the two others, plus six times the product of all
three terms. It is easily proved that this law is true for any
polynomial.

To apply the preceding formula to the development of
the cube of a trinomial, in which the terms are affected
with coefficients and exponents, designate each term by a
single letter, then replace the letters introduced, by their values,
and perform the operations indicated.

From this rule, we find that

(2a 2 — 4ab + 3b 2 ) 3 =:8a 6 — 48a 5 b+ I32a i b 2 — 208a 3 b 3
+ I98a 2 b±-l08ab 5 +27b*.

The fourth, fifth, &c, powers of any polynomial can be
found in a similar manner.

16. What is the cube of a— 2b-\-c 1

Ans. a 3 — 8b 3 +c 3 — 6a 2 b + 3a 2 c+12ab 2 +l2b 2 c+3ac 2
— 6bc 2 —l2abc.

*

£XTR.\CT ON OF THE SQUARE ROOT. 133

CHAPTER V.

Extraction of the Square Root of Numbers. Forma-
tion of the Square and Extraction of the Square
Root of Algebraic Quantities. Calculus of Radicals
of the Second, Degree.

90. The square or second power of a number, is trie
product which arises from multiplying that number by itself
once : for example, 49 is the square of 7, and 144 is the
square of 12.

fl 1 . The square root of a number is that number which,
being multiplied by itself once, will produce the given num-
ber. Thus, 7 is the square root of 49, and 12 the square
root of 144: for, 7x7 = 49, and 12 X 12=144.

02. The square of a number, either entire or fractional,
is easily found, being always obtained by multiplying this
number by itself once. The extraction of the square root
of a number is, however, attended with some difficulty, and
requires particular explanation.

Quest. — 90. What is the square, or second power of a number? —
91. What is the square root of a number 1

12

134 FIRST LESSONS IN ALGEBRA*

The first ten numbers are.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10;

and their squares,

1, 4, 9, 16. 25, 36, 49, 64, 81, 100;

and reciprocally, the numbers of the first line are the square
roots of the corresponding numbers of the second. We
may also remark that, the square of a number expressed by a
single figure, will contain no figure of a higher denomination
than tens.

The numbers of the last line, 1, 4, 9, 16, &c, and all
other numbers which can be produced by the multiplication
of a number by itself, are called perfect squares.

It is obvious that there are but nine perfect squares among
all the numbers which can be expressed by one or two
figures : the square roots of all other numbers expressed
by one or two figures, will be found between two whole
numbers differing from each other by unity. Thus 55,
which is comprised between 49 and 64, has for its square
root a number between 7 and 8. Also 91, which is com
prised between 81 and 100, has for its square root a number
between 9 and 10.

03. Every number may be regarded as made up of a
certain number of tens and a certain number of units.
Thus 64 is made up of 6 tens and 4 units, and may be ex
pressed under the form 60 + 4.

Quest. — 92. What will be the highest denomination of the square
of a number expressed by a single figure X What are perfect^ squares %
How many are there between 1 and 100 1 What are they 1

EXTRACTION OF THE SQUARE ROOT. 135

Now, if we represent the tens by a and the units by b,
we shall have

a+b = 64,
and (a+&) 2 = (64) 2 ;

or a 2 +2ab + b 2 = 4096.

Which proves that the square of a number composed of
tens and units, contains the square of the tens plus twice the
product of the tens by the units, plus the square of the units.

94. If, now, we make the units 1, 2, 3, 4, &c, tens, or
units of the second order, by annexing to each figure a ci-
pher, we shall have

10, 20, 30, 40, 50, 60, 70, 80, 90, 100,

and for their squares,

100, 400, 900, 1600, 2500, 3600, 4900, 6400, 8100, 10000.

From which w r e see that the square of one ten is 100, the
square of two tens 400 ; and generally, that the square of
tens will contain no figures of a less denomination than hun-
dreds, nor of a higher name than thousands.

Ex. 1. — To extract the square root of 6084.

Since this number is composed of more than
two places of figures, its roots will contain 60 84

more than one. But since it is less than 10000,
which is the square of 100, the root will contain but two
figures : that is, units and tens.

Now, the square of the tens must be found in the two

Quest. — 93. How may every number be regarded as made up ] What
is the square of a number composed of tens and units equal to? —
94. What is the square of one ten equal to 1 Of 2 tens 1 Of 3
tens'! &c.

136 FIRST LESSONS IN ALGEBRA.

loft-hand figures, which we will separate from the other two
by putting a point over the place of units, and a second over
the place of hundreds. These parts, of two figures each, are
called periods. The part 60 is comprised between the two
squares 49 and 64, of which the roots are 7 and 8 : hence,
7 is the figure of the tens sought ; and the required root is
composed of 7 tens and a certain number of units.

The figure 7 being found, we
write it on the right of the given 60 84 78

number, from which we separate 49

it by a vertical line: then we 7x2 = 14
subtract its square, 49, from 60,

1184
1184

which leaves a remainder of 1 1 , 0~

to which we bring down the two

next figures 84. The result of this operation, 1184, con-
tains twice the product of the tens by the units, pluslthe square
of the units.

But since tens multiplied by units cannot give a product of
a less name than tens, it follows that the last figure, 4, can
form no part of the double product of the tens by the units :
this double product is therefore found in the part 118, which
Ave separate from the units' place, 4.

Now if we double the tens, which gives 14, and then di-
vide 118 by 14, the quotient 8 is the figure of the units, or
a figure greater than the units. This quotient figure can
never be too small, since the part 118 will be at least equal
to twice the product of the tens by the units : but it may be
too large ; for the 118, beside-s the double product of the
tens by the units, may likewise contain tens arising from
the square of the units. To ascertain if the quotient 8 ex-
presses the units, we write the 8 on the right of the 14,
which gives 148, and then we multiply 148 by 8. Thus,
we evidently form, 1st, the square of the units; and,
2nd, the double product of the tens by the units. This

EXTRACTION OF THE SQUARE ROOT. 137

multiplication being effected, gives for a product 1184, a
number equal to the result of the first operation. Having
subtracted the product, we find the remainder equal to :
hence 78 is the root required.

Indeed, in the operations, we have merely subtracted
from the given number 6084, 1st, the square of 7 tens, or
70 ; 2nd, twice the product of 70 by 8 ; and, 3d, the square
of 8 : that is, the three parts which enter into the composi-
tion of the square 70 + 8? or 78 ; and since the result of
the subtraction is 0, it follows that 78 is the square root of
6084.

05. Remark. — The operations in the last example have
been performed on but two periods, but it is plain that the
same reasoning is equally applicable to larger numbers, for
by changing the order of the units, we do not change the
relation in which they stand to each other.

Thus, in the number 60 84 95, the two periods 60 84
have the same relation to each other as in the number
60 84 ; and hence the methods used in the last example
are equally applicable to larger numbers.

96. Hence, for the extraction of the square root of
numbers, we have the following

RULE.

I. Separate the given number into periods of two figures
each, beginning at the right hand : — the period on the left will
often contain but one figure.

II. Find the greatest square in the first period on the left,
and place its root on the right, after the manner of a quotient

Quest. — 95. Will the reasoning in the example apply to more than
two periods 1

12*

138 FIRST LESSONS IN ALGEBRA.

in division. Subtract the square of the root from the first
period, and to the remainder bring down the second period for
a dividend.

III. Double the root already found, and place it on the left
for a divisor. Seek how many times the divisor is contained
in the dividend, exclusive of the right-hand figure, and place
the figure in the root and also at the right of the divisor.

IV. Multiply the divisor, thus augmented, by the last figure
of the root, and subtract the product from the dividend, and to
the remainder bring down the next period for a new dividend.
But if any of the products should be greater than the dividend,
diminish the last figure of the root.

V. Double the whole root already found, for a new divisor,
and continue the operation as before, until all the periods are
brought down.

\$7. 1st Remark. If, after all the periods are brought
down, there is no remainder, the proposed number is a per-
fect square. But if there is a remainder, you have only
found the root of the greatest perfect square contained in
the given number, or the entire part of the root sought.

For example, if it were required to extract the square
root of 665, we should find 25 for the entire part of the
root, and a remainder of 40, which shows that 665 is not
a perfect square. But is the square of 25 the greatest per-
fect square contained in 665 1 that is, is 25 the entire part
of the root ? To prove this, we will first show that, the
difference between the squares of two consecutive numbers, is
equal to twice the less number augmented by unity.

Quest. — 96. Give the rule for extracting the square root of numbers.
What is the first step 1 What the second 1 What the third 1 What
the fourth 1 What the fifth 1

EXTRACTION OF THE SQUARE ROOT. 139

Let . . a=z the less number,

and . . a-f-1 = the greater.

Then . (a+l) 2 =za 2 +2a+ 1,

and . . (a) 2 z=:a 2 .

Their difference is =. 2a-\-l as enunciated.

Hence, the entire part of the root cannot be augmented,
unless the remainder exceeds twice the root found, plus
unity.

But 25 x2-fl=:51>40 the remainder : therefore, 25 is
the entire part of the root.

98. 2nd Remark. — The number of figures in the root
will always be equal to the number of periods into which
the given number is separated.

EXAMPLES.

1. To find the square root of 7225. Ans. 85.

2. To find the square root of 17689. Ans. 133.

3. To find the square root of 994009. Ans. 997.

4. To find the square root of 85673536. Ans. 9256.

5. To find the square root of 67798756. Ans. 8234,

6. To find the square root of 978121. Ans. 989.

7. To find the square root of 956484. Ans. 978.

8. What is the square root of 3^372961 ? Ans. 6031.

9. What is the square root of 22071204 ? Ans. 4698.

10. What is the square root of 106929? Ans. 327.

11. What is the square root of 12088868379025 ?

Ans. 3476905.

Quest. — 98. How many figures will you always find in the roott

140

FIRST LESSONS IN ALGEBRA.

99. 3rd Remark. — If the given number has not an exac*
root, there will be a remainder after all the periods are
brought down, in which case ciphers may be annexed,
forming new periods, each of which will give one decimal
place in the root.

1. What is the square root of 36729 ?

In this example there are
two periods of decimals,
which give two places of
decimals in the root.

3 67 29
1

191,64 + .

2 9 267
261

38

1

629

381

382 6
3832 A

24800
22956

[ 184400
153296

31104

Rem.

2. What is the square root of 2268741 ?

Ans. 1506,23 + .

3. What is the square root of 7596796 1

4. What is the square root of 96 ?

5. What is the square root of 153 1

6. What is the square root of 101 ?

Ans. 2756,22 + .

Ans. 9,79795 + .

Ans. 12,36931 + .

Ans. 10,04987+.

Quest. — 99. How will you find the decimal part of the root 1

EXTRACTION OF THE SQUARE ROOT. 141

7. What is the square root of 285970396644 1

Arts. 534762.

8. What is the square root of 41605800625 ?

Arts. 203975.

9. What is the square root of 48303584206084 ?

Arts, 6950078.

Extraction of the square root of Fractions.

1 OO. Since the square or second power of a fraction is
obtained by squaring the numerator and denominator sepa-
rately, it follows that the square root of a fraction will be
equal to the square root of the numerator divided by the
square root of the denominator.

oft d

For example, the square root of — is equal to — : for

a a a 2

T X T Z=Z 1F'

1 . What is the square root of — ?

9

2. What is the square root of — ?

64

3. What is the square root of — 1

81

256

4. What is the square root of

361

5. What is the square root of — ?

64

Ans.

2'

Ans.

3

Ans.

8

Ans.

16
19'

1

Ans.

2"

Quest. — 100. If the numerator *and denominator of a fraction are
perfect squares, how will you extract the square root 1

142 FIRST LESSONS IN ALGEBRA.

4096 64

6. What is the square root of ^tt™ ? Ans. -p^pr.

m *»n. • x. c 582169 , , 763

7. What is the square root oi „ n n 1 Ans. ——-.

* 956484 978

101. If neither the numerator nor the denominator is a
perfect square, the root of the fraction cannot be exactly
found. We can, however, easily find the approximate root.
For this purpose^

Multiply both terms of the fraction by the denominator,
which makes the denominator a perfect square without altering
the value of the fraction. Then, extract the square root of
the numerator, and divide this root by the root of the denomi-
tor ; this quotient will be the approximate root.

3

Thus, if it be required to extract the square root of — t

1 5
we multiply both terms by 5, which gives — .

We then have

t/T5=z 3,8729+ :

hence, 3,8729+ ~* 5 = ,7745+ = Ans.

7

2. What is the square root of — 1 Ans. 1,32287 + .

14

3. What is the square root of — 1 Ans. 1,24721 + .

4. What is the square root of 1 1— 1

16

Ans. 3,41869+.

Quest. — 101. If the numerator and denominator of a fraction are not
perfect squares, how do. you extract the square root t

EXTRACTION OF THE SQUARE ROOT. 143

13

5. What is the square root of 7— ? Ans. 2,71313 + .

36

15

6. What is the square root of 8— 1 Ans. 2,88203+.

49

5

7. W r hat is the square root of — - 1 Ans. 0,64549+.

3

8. What is the square root of 10— ?

Ans. 3,20936+.

102. Finally, instead of the last method, we may, if we

Change the vulgar fraction into a decimal, and continue the
division until the number of decimal places is double the number
of places required in the root. Then, extract the root of the
decimal by the last rule.

Ex. 1. Extract the square root of — to within ,001.

This number, reduced to decimals, is 0.785714 to within
0,000001 ; but the root of 0,785714 to the nearest unit, is

,886 ; hence 0,886 is the root of — to within ,001.

14

2. Find the \/2— to within 0,0001.

V 15

Ans. 1,6931 + .

3. What is the square root of — 1 Ans. 0,24253 + .

7

4. What is the square root of — ? Ans. 0,93541 + .

8

5

5. What is the square root of — 1 Ans. 1,29099 + .

o

Quest. — 102. By what other method may the root be found 1

144 FIRST LESSONS IN ALGEBRA.

Extraction of the Square Root of Monomials.

103. In order to discover the process for extracting the
square root, we must see how the square of the monomial
is formed.

By the rule for the multiplication of monomials (Art. 35),
we have

(5a 2 b 3 c) 2 =5a 2 b 3 c X 5a 2 b 3 c — 25a i b G c 2 ;

that is, in order to square a monomial, it is necessary to
square its coefficient, and double each of the exponents of the
different letters. Hence, to find the root of the square of a
monomial, we have the following

RULE.

1. Extract the square root of the coefficient,
II. Divide the exponent of each letter by 2.

Thus, ^64aW=8a 3 b 2 for 8a 3 b 2 x8a 3 b 2 =z64a e b*.

2. Find the square root of 625a 2 b 8 c 6 . Ans. 25ab 4 c 3 ,

3. Find the square root of 57 6a 4 b 6 c 8 . Ans. 24a 2 b 3 c i ,

4. Find the square root of 196x 6 y 2 z 4 . Ans. \\x 3 yz >

5. Find the square root of 441 a 8 b e c 10 d m .

Ans, 2la 4 b 3 c 5 d s

6. Find the square root of 784a 12 b u c 1G d 2 .

Ans. 28a%~c 8 d,

7. Find the square root of 8la s b 4 c 6 .

Ans. 9a 4 b 2 c 3 .

Quest. — 103. How do you extract the square root of a monomial 1

EXTRACTION OF THE SQUARE ROOT. 145

104. From the preceding rule it follows, that when a
monomial is a perfect square, its numerical coefficient is a
perfect square, and all its exponents even numbers. Thus,
25a 4 b 2 is a perfect square, but 98a5 4 is not a perfect square,
because 98 is not a perfect square, and a is affected with
an uneven exponent.

In the latter case, the quantity is introduced into the cal-
culus by affecting it with the sign -y/ , and it is written

thus :

i/98ab\

Quantities of this kind are called radical quantities, or irra-
tional quantities, or simply radicals of the second degree.
They are also, sometimes called Surds.

These expressions may often be simplified, upon the prin-
ciple that, the square root of the product of two or more factors
is equal to the product of the square roots of these factors ; or,
in algebraic language,

-y/abcd . . . =V a • V° • V c • V ^ • • •

This being the case, the above expression, y98a6 4 , can
be put under the form

V49Mx2a= V49Fx ^Z

Now a/496 4 may be reduced to 7b 2 ; hence,

In like manner,

^/45a 2 b 3 c 2 d— y/^oJWc 2 X 5bd= 3abc -y/lM,

y/864a 2 b 5 c ll = yflteaWc 10 X 6fc= 12ab 2 c 5 i/Wc.
13

146 FIRST LESSONS IN ALGEBRA.

The quantity which stands without the radical sign is
called the coefficient of the radical. Thus, in the expressions

7b 2 -y/2a, 3abc */5bd, 1 2ab 2 c 5 -yJWc,

the quantities 7b 2 , 3abc, I2ab 2 c 5 , are called coefficients oj

Hence, to simplify a radical expression of the second
degree, we have the following

RULE.

I. Separate the expression into two parts, of which one shall
contain all the factors that are perfect squares, and the other
the remaining ones.

II. Take the roots of the perfect squares and place them
before the radical sign, under which leave those factors which
are not perfect squares.

105. Remark. — To determine if a given number has
any factor which is a perfect square, we examine and see
if it is divisible by either of the perfect squares

4, 9, 16, 25, 36, 49, 64, 81, &c ;

and if it is not, we conclude that it does not contain a fac-
tor which is a perfect square.

Quest. — 104. When is a monomial a perfect square] When it is
not a perfect square, how is it introduced into the calculus 1 What are
quantities of this kind called 1 May they be simplified ] Upon what
principle 1 What is a coefficient of a radical 1 Give the rule for reducing
radicals. — 105. How do you determine whether a given number has a
factor which is a perfect square 1

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