We cannot here obtain the value of the unknown quantity x, as in art. 11., and the question ans simply to this, to find a

number which, multiplied by itself,

this number is 5; but it seldom

easy; hence arises this new nu

e 25. It is obvious that

that the solution is so

estion ; to find a numbe a product equal to a

ber, which, multiplied by itself, w proposed number; or, which is the same thing, from the second power of a number, to retrace our steps to the number from which it is derived, and which is called the square root. I shall proceed, n the first place, to resolve this question, as it is involved in the determination of the unknown quantities, in all equations of the second degree.

91. The method employed in finding or extracting the roots of numbers, supposes the second power of such as are expressed by only one figure to be known. See the nine primitive numbers with their second powers written under them respectively.

It is evident from this table, that the second power of a number expressed by one figure, contains only two figures; 10, which is the least number expressed by two figures, has for its square a number composed of three, 100. In order to resolve the second power of a number consisting of two figures we must attend to the method by which it is formed; for this purpose we must inquire, how each part of the number 47, for example, is employed in the production of the square of this number.

We may resolve 47 into 40 +7, or into 4 tens and 7 units; if we represent the tens of the proposed number by a, and the units by b, the second power will be expressed by

(a + b) (a + b) = a2 + 2 a b + b2 ;

that is, it is made up of three parts, namely, the square of the tens, twice the product of the tens multiplied by the units, and the square of the units. In the example we have taken, a = 4 tens or 40 units, and b = 7; we have then

Now in order to return, by a reverse process, from the number 2209 to its root, we may observe, that the square of the tens, 1600, has no figure, which denotes a rank inferior to hundreds, and that it is the greatest square, which the 22 hundreds, comprehended in 2209, contain; 22 lies between 16 and 25, that is, between the square of 4 and that of 5, as 47 falls between 4 tens or 40, and 5 tens or 50.

We find, therefore, upon examination, that the greatest square contained in 22 is 16, the root of which 4 expresses the number of tens in the root of 2209; subtracting 16 hundreds, or 1600, from 2209, the remainder 609 contains double the product of the tens by the units, 560, and the square of the units 49. But as double the product of the tens by the units has no figure inferior to tens, it must be found in the two first figures 60 of the remainder 609, which contain also the tens arising from the square of the units. Now, if we divide 60 by double of the tens 8, and neglect the remainder, we have a quotient 7 equal to the units sought. If we multiply 8 by 7, we have double the product of the tens by the units, 560; subtracting this from the whole remainder 609, we obtain a difference 49, which must be, and in fact is, the square of the units.

This process may be exhibited thus ;

We write the proposed number in the manner of a dividend, and assign for the root the usual place of the divisor. We then separate the units and tens by a comma, and employ only the two first figures on the left, which contain the square of the tens found in the root. We seek the greatest square 16, contained in these two figures, put the root 4 in its assigned place, and subtract 16 from 22. To the remainder we bring down the two other separating the last, which

figures, 09, of the proposed number, does not enter into double the product of the tens by the units, and divide the remainder on the left by 8, double the tens in the root, which gives for the quotient the units 7. In order to collect

into one expression the two last parts of the square contained in 609, we write 7 by the side of 8, which gives 87, equal to double the tens plus the units, or 2 a+b; this, multiplied by 7 or b, reproduces 6092ab+b2, or double the product of the tens by the units, plus the square of theits. This being subtracted leaves no remainder, and the operation shows, that 47 is the square root of 2209.

If it were required to extract the square root of 324; the operation would be as follows;

3,24 18
1

22,4 28

22,4

000

Proceeding as in the last example, we obtain 1 for the place of tens of the root; this doubled gives the number 2, by which the two first figures 22 of the remainder are to be divided. Now 22 contains 2 eleven times, but the root can neither be more than 10, nor 10; even 9 is in fact too large, for if we write 9 by the side of 2, and multiply 29 by 9, as the rule requires, the result is 261, which cannot be subtracted from 224. We are, therefore, to consider the division of 22 by 2 only as a means of approximating the units, and it becomes necessary to diminish the quotient obtained, until we arrive at a product, which does not exceed the remainder 224. The number 8 answers to this condition, since 8 x 28 = 224; therefore, the root sought is 18.

By resolving the square of 18 into its three parts, we find

and it may be seen, that the 6 tens, contained in the square of the units, being united to 160, double the product of the tens by the units, alters this product in such a manner, that a division of it by double the tens will not give exactly the units.

92. It will not be difficult, after what has been said, to extract the square root of a number, consisting of three or four figures; but some further observations, founded upon the principles above

laid down, may be necessary to enable the reader to extract the root of any number whatever.

No number less than 100 can have a square consisting of more than four figures, since that of 100 is 10000, or the least number expressed by five figures. In order, therefore, to analyze the square of any number exceeding 100, or 473, for example, we may resolve it into 470+ 3, or 47 tens plus 3 units. To obtain its square from the formula,

u2 + 2 a b + b2,

we make a 47 tens 470 units, b = 3 units, then

In this example, it is evident that the square of the tens has no figure inferior to hundreds, and this is a general principle, since tens multiplied by tens, always give hundreds. (Arith. 32.)

It is therefore in the part 2237, which remains on the left of the proposed number, after we have separated the tens and units, that it is necessary to seek the square of the tens; and as 473 lies between 47 tens, or 470, and 48 tens, or 480, 2237 must fall between the square of 47 and that of 48; hence the greatest square contained in 2237, will be the square of 47, or that of the tens of the root. In order to find these tens, we must evidently proceed, as if we had to extract the square root of 2237 only; but instead of arriving at an exact result, we have a remainder, which contains the hundreds arising from double the product of the 47 tens multiplied by the units.

We first separate the two last figures 29, and in order to extract the root of the number 2237, which remains on the left, we further separate the two last figures 37 of this number; the proposed number is then divided into portions of two figures, beginning on the right and advancing to the left. Proceeding with the. two first portions as in the preceding article, we find the two first. figures 47 of the root; but we have a remainder 23, which, joined to the two figures 29 of the last portion, contains double the product of the 47 tens by the units, and the square of the units. We separate the figure 9, which forms no part of double the product of the tens by the units, and divide 282 by 94, double the 47 tens; writing the quotient 3 by the side of 94, and multiplying 943 by 3, we obtain 2829, a number exactly equal to the last remainder, and the operation is completed.

93. In order to show, by what method we are to proceed with any number of figures, however great, I shall extract the root of 22391824. Whatever this root may be, we may suppose it capable of being resolved into tens and units, as in the preceding examples. As the square of the tens has no figure inferior to hundreds, the two last figures 24 cannot make a part of it; we may therefore separate them, and the question will be reduced to this, to find the greatest square contained in the part 223918, which remains on the left. This part consisting of more than two figures, we may conclude, that the number, which expresses the tens in the root sought, will have more than one figure; it may therefore be resolved, like the others, into tens and units. As the square of the tens does not enter into the two last figures 18 of the number 223918, it must be sought in the figures 2239, which remain on the left; and since 2239 still consists of more than two figures, the square, which is contained in it, must have a root which consists of at least two; the number which expresses the tens sought will therefore have more than one figure; it is then, lastly, in 22 that we must seek the square of that, which represents the units of the highest place in the root required. By this process, which may be extended to any length we please, the proposed number may be divided into portions of two figures from right to left; it must be understood, however, that the last figure on the left may consist of only one figure.