extracting the square root of a number it is divided into periods of two figures, beginning at the units' place, above which a point is placed, as also on every alternate figure. The number of points show the number of figures in the root. Thus 243, because it has three figures, has two in the root, and the points show it also.

It follows also that 1 is the integer part of the square root of all numbers from 1 to 4. The square root of 2 is 1 with a fraction, the same applies to 3; a fraction is found in the root of every number which is not a perfect square, and the square root of such a number is the root of the preceding perfect square plus a fractional part. Thus the square of 20 is 4 and a fraction, that of 90 is 9 plus a fraction. It shall be shown that the roots of such quantities can only be approximately found, and these quantities are called surds or irrational quantities.

329. If we analyse what takes place when squaring a number we shall derive a formula or law which will enable us to extract the square root of numbers.

Every number of two or more figures allows of its being divided into two quantities. For example 27=20+7; 424=420+4; 8462584620+5. Let us express generally the first part by a, and the second by b, so that a+b represent all numbers of two or more figures; then squaring a+b we have

(a+b)2=(a+b) (a+b)=a×a+axb+b×a+bxb=a2+2×a

xb+b2.

330. Hence it follows that the square of a number of two or more figures contains the square of the first part (a2)+twice the product of the first part by the second (2x axb)+the square of the second part (b2).

According to this law the multiplication is thus carried on— 362 (30+6)=(30+6) (30+6.)

Also,

524=(520+4)=(520+4) (520+4).

=520 × 520+520 × 4+4 × 520+4×4.
=5202+2×520×4+42.

=274576.

331. Ex. 1. Let it now be required to extract the square

root of 576.

Since 576 is a number of three figures, and there are two in the root, as the points show. Therefore extracting the square root of the first period, or of 500, the root is found to be between 20 and 30, take 20 and subtract 202 (a2), or 400 from 500, the remainder is 100, to which the second period is added, and 176 contains twice the first part by the second + the square of the second part, then dividing the whole remainder by 2×20=40, the quotient is 4, which is the second part of the root; lastly we subtract 40×4 (2×a×b)+42(b2)=176, from the remainder 176 and 0 is left.

X

../576=20+4=24.

The proof of it is 242=576.

20
4 X 20

400

4

16

4 X 20

S

The annexed diagram shows plainly what we have just explained, here xs=576, xz= 20, ..xo 400, and vo+os+oy=176, zy is found to be 4, then oy+ov=2×20 × 4, and os is the square of 4 or 16, which completes the square xs.

We must observe that the first figure of the root (2) expresses tenths, twice its pro

duct is 40, or 4 tenths; so we might have divided 17 tenths by 4 tenths, instead of 176 by 40, and the quotient would have been the same.

Ex. 2. 7225 is the square of what number?

ab

8 and 9, therefore 8 is the first part of the root, and 82 or 64 being subtracted from 72, the remainder is 8, the next period 25 being brought down, we divide 82 by 2×8 or 16, and the quotient is 5, the second part of the root, we set 5 after 16 tenths to

=

fill up the units' place, for (16 tenths + 5 units) x 5 (2 × a +b)b=2a×b+b2, and subtracting 825 from 825 the remainder is 0, .. 7225=85. The proof of the operation is 852= 7225.

332. Having thus explained the extraction of the square root of two figures, it will not be difficult to find a root of three figures, for the two first are determined as before, and to find the third we consider the two first figures, as composing the first part of the square root.

Were we to extract the square root of a number of 7 or 8 figures, the root would consist of four, we should find the three first, as just mentioned, and then considering those three figures as the first part, the fourth would be the second part of the square root.

Ex. 3. Find the square root of 294849.

Here we find the number consisting of three periods, from the first period the square root is 5; the square of which being subtracted leaves 4. Bring down the next period 48, then 448 is the second part, divide 44 by 2×5 or 10, the double of the first figure in the root, we set the quotient 4, as the second figure in the root, then we subtract (2 × 50+4)×4, or 416, from 448, the remainder is 32. We then bring down the last period 49. Now we are to divide 3249 by 2× 54, or 108, twice the first figures of the square root, which we regard as 1080, the quotient is 3, which is the third figure of the square root, and we subtract 1080×3+32=3249 from the dividend 3249, the remainder is 0. .. 294849=543.

The proof is 5432=294849.

If after having brought down a period by the remainder, the double of the root was not contained in this dividend, the last figure being cut off, a cipher is set at the root, and also after the double of the root, and a new period is set down by the side of the other one, and proceed as before.

333. EXERCISES.

1. Extract the square roots of 676, 1689, 9801.

2. Find the square roots of 16129, 258064, 61340224. 3. What is the square root of 258064? 826281? 31360000?

334. The square roots of irrational quantities or surds cannot be exactly expressed, but we can get an approximate root, correct to as many decimals as we please, by affixing to the right hand of the last remainder as many periods of ciphers, as we require decimal figures in the root.

Thus, if it were required to extract the square root of 345, true to three places of decimals, or within .001 the operation is carried on as follows:

√345(=18.574

1

28)245

224

365) 2100

1825

3707) 27500
25949

37144) 155100

148576
6524

The square root of 345 is found as before; to the remainder 21 a period of ciphers is appended, and the quotient is found to be 5, the remainder is 275, to which a period of ciphers is added, and so on. Since three periods of ciphers are appended to the integral part, there are three decimal figures in the root, viz., as many decimal figures as periods of ciphers, which are pointed from the right hand.

In this and like cases where ciphers are appended, the root can never terminate, because no figure multiplied by itself can produce a cipher.

335. If the number from which the square root is to be extracted were a mixed number, the fractional part is converted into decimals, and periods of decimals are brought down as mentioned with regard to the ciphers; or if we had to extract the

square root of a decimal, the pointing is made from the units' place to the right hand.

Ex. 1. Required the square root of 28g.

√2828.625(=5.35023

25 103) 362 309

1065) 5350

5325

107002) 250000

214004

1070043) 3599600

3210129

389471

After what has been said the operation is easily understood:The square root has been extracted to five places of decimals. Ex. 2. What is the square of .2, true to .00001 or to 5 places? √.20 (=.44721

16

84) 400
336

887) 6400

6209

8942) 19100
17884

89441) 121600

The periods are perfectly similar to the cases we have examined; the first period is .20, and not .2, because .4 x .4+.04=.20=.2.

336. By what has been said of the squaring of vulgar fractions it is evident that the square root of such quantities is found by extracting the square root of both numerator and denominator. Thus the square root of is, since 3×3=9 and 5×5=25. But it generally happens that the numerator, or denominator, or both, are not square numbers, then divide the square root of their product by the denominator.