303. As this rule, in high powers, is often found to be very laborious, it may be proper to observe, that the roots of certain compound quantities may sometimes be easily discovered: thus, in the last example, the root is 2a-3x, which is the difference of the roots of the first and last terms; and so on, for other compound quantities.

to

Hence, the following method in such cases; extract the roots of all the simple terms, and connect them together by the signs + or, as may be judged most suitable for the purpose; then involve the compound root thus found, to its proper power, and if it be the same with the given quantity, it is the root required. But if it be found to differ only in some of the signs, change them from + to or from +, till its power agrees with the given one throughout. However, such artifices are not to be used by learners, because the regular mode of proceeding is more advantageous to them; besides, a knowledge of those artifices which are used by experienced Algebraists, can only be acquired from frequent practice.

Ex. 3. Required the square root of a2+2ab+b2+ 2ac+2bc+c2.

Here, the square root of a2=

= a; the" square of be =b; and the square root c2=c. Hence, a+b+c, is the root required, because (a+b+c)2=a2+2ab+b2 +2ac+2bc+c2.

Ex. 4. Required the fifth root of 32x5-80x2 + 80x3-40x2+10x-1.

Ans. 2x-1.

Ex. 5. Required the cube root of a2-6x5+15x* +20x3-15x2+5x-1.

Ans. x2-2x+1.

Ex. 6. Required the fourth root of a 6a2x2-4ax+3x+.

4a3 x +

Ans. a-x.

Ex. 7. Required the square root of x+2x1y* +ys.

Ans. x+y.

4

Ex. 8. Required the square root of x3-2x+y++y®,
Ans. x-y.

Ex. 9. Required the cube root of a3-6a2x+12ax2

-8x3.

Ans. a-2x.

Ex. 10. Required the sixth root of x-6x5+15x4 -20x+15x3-6x+1.

Ans.x-1. Ex. 11. Required the fifth root of x10+15x8y2+ 96x*y*+270x+y+405x2y+243y1o. Ans. x2+3y2. Ex. 12. Required the square root of x2+2xy + y2 +6xz+6yz+9z2. Ans. x+y+3z.

§ III. INVESTIGATION OF THE RULES FOR THE EX

TRACTION OF THE SQUARE AND CUBE ROOTS OF NUMBERS.

304. It has been observed, (Art. 104), that, a denoting the tens of a number, and b the units, the formula a2+2ab+b2 would represent the square of any number consisting of two figures or digits; thus, for example, if we had to square 25; put a=20 and b= 5, and we shall find

a2=400 2ab=200 b2 = 25

(a+b)2=(25)2=625.

305. Before we proceed to the investigation of these Rules, it will be necessary to explain the nature of the common arithmetical notation. It is very well known that the value of the figures in the common arithmetical scale increases in a tenfold proportion from the right to the left; a number, therefore,

may be expressed by the addition of the units, tens, hundreds, &c. of which it consists; thus the number 4371 may be expressed in the following manner, viz. 4000+300+70+1, or by 4 x 1000+3x100+7×10 +1; also, in decimal arithmetic, each figure is supposed to be multiplied by that power of 10, positive or negative, which is expressed by its distance from the figure before the point: thus, 672.53=6x102+7x101 +2x10°+5X10-1+3 x 10-26 x 100+7x10+2

5 3

50 3

53

X1+ + =672+ + =672- Hence, if

10 100

100 100

100

the digits of a number be represented by a, b, c, d, e, &c. beginning from the left-hand; then,

A number of 2 figures may be expressed by 10a+b.

3 figures
by 100a10b+c.
4 figures by 1000a +100b+10c+d.

&c.

&c.

&c.

By the digits of a number are meant the figures which compose it, considered independently of the value which they possess in the arithmetical scale. Thus the digits of the number 537 are simply the numbers 5, 3 and 7; whereas the 5. considered with respect to its place, in the numeration scale, means 500, and the 3 means 30.

306. Let a number of three figures, (viz. 100a +10b+c) be squared, and its root extracted according to the rule in (Art. 299), and the operation stands. thus;

I.

10000a2+2000ab+10062 +200ac+20bc+c2

10000a2

200a +10b)2000ab+10062

2000ab+10062

(100a +10b+c

200a+20b+c)200ac+20bc+c2

200ac+206c+c2

II. Let a=2

b=3

c=

=1

and the operation is transformed into the following one;

40000+12000+900+400+60+1(200+30+1

40000

400+30) 12000-+900

400+60+1)400+60+1

400+60+1

III. But it is evident that this operation would not be affected by collecting the several numbers which stand in the same line into one sum, and leaving out the ciphers which are to be subtracted in the operation.

Let this be done; and let two figures be brought down at a time, after the square of the first figure in the root has been subtracted; then the operation may be exhibited in the manner annexed; from which it appears, that the square root of 53361 is 231.

307. To explain the division of the given number into periods consisting of two figures each, by placing a dot over every second figure beginning with the units, as exhibited in the foregoing operation. It

must be observed, that, since the square root of 100 is 10; of 10000 is 100; of 1000000 is 1000 ; &c. &c. it follows, that the square of a number less than 100 must consist of one figure; of a number between 100 and 10000, of two figures; of a number between 10000 and 1000000, of three figures; &c. &c., and consequently the number of these dots will show the number of figures contained in the square root of the given number. From hence it follows, that the first figure of the root will be the greatest square root contained in the first of those periods reckoning from the left.

Thus, in the case of 53361 (whose square root is a number consisting of three figures); since the square of the figure standing in the hundred's place cannot be found either in the last period (61), or in the last but one (33), it must be found in the first period (5); consequently the first figure of the root will be the square root of the greatest square number contained in 5; and this number is 4, the first figure of the root will be 2. The remainder of the operation will be readily understood by comparing the steps of it with the several steps of the process for finding the square root of (a+b+c)2 (Art. 299); for, having subtracted 4 from (5), there remains 1; bring down the next two figures (33), and the dividend is 133; double the first figure of the root (2), and place the result 4 in the divisor; 4 is contained in 13 three times; 3 is therefore the second figure of the root; place this both in the divisor and quotient, and the former is 43; multiply by 3, and subtract 129, the remainder is 4; to which bring down the next two figures (61), which gives 461 for a dividend. Lastly, double the last figure of the former divisor, and it becomes 46; place this in the next divisor, and since 4 is contained