RULE FOR EXTRACTING THE SQUARE ROOT OF A POLYNOMIAL. Arrange the terms according to the powers of some one letter; take the square root of the first term for the first term of the required root, and subtract its square from the given polynomial. sor. Divide the first term of the remainder by double the root already found, and annex the result both to the root and the diviMultiply the divisor thus increased by the last term of the root, and subtract the product from the last remainder. Proceed in the same manner to find the additional terms of the root. Ex. 1. Required the square root of a -2a3x+3a2x2—2ax3+x*. a*—2a3x+3a2x2-Qax3+x* | a2—ax+x2= the root. a1 -2a'x+3a2x2 | 2a2-ax = the first divisor. -2a3x+ a2x2 2a2x2-2ax3+x |2a3-2ax+x= the second divisor. 2a2x2-2ax3+x1| For verification, multiply the root a2-ax+x2 by itself, and we shall obtain the original polynomial. Ex. 2. Required the square root of a2+2ab+2ac+b2+2bc+c2. Ex. 3. Required the square root of 10x-10x-12x3+5x2+ 9x-2x+1. Ex. 4. Required the square root of 8ax3+4a2x2+4x‘+16b2x2 +16b+16ab❜x. Ans. 2x+2ax+4b". Ex. 5. Extract the square root of 15a'b'+a-6ab-20a'b +b+15a b'-6ab'. Ex. 6. Extract the square root of 8ab3+a1-4a3b+4b*. (276.) Method of extracting the square root of numbers. The preceding rule is applicable to the extraction of the square root of numbers. For every number may be regarded as an Algebraic polynomial, or as composed of a certain numher of units, tens, hundreds, &c. Thus, Also, 529 is equivalent to 500+20+9. 841 66 800+40+1. If, then, 841 is the square of a number composed of tens and units, it must contain the square of the tens, plus twice the product of the tens by the units, plus the square of the units. But these three terms are blended together in 841, and hence the peculiar difficulty in determining its root. The following principles will, however, enable us to separate these terms, and thus detect the root. (277.) I. For every two figures of the square there will be one figure in the root, and also one for any odd figure. The smallest number consisting of two figures is 10, and its square is the smallest number of three figures. The smallest number of three figures is 100, and its square is the smallest number of five figures, and so on. Therefore, the square root of every number composed of one or two figures will contain one figure; the square root of every number composed of three or four figures will contain two figures; of a number from five to six figures will contain three figures; and from 2n-1 to 2n figures must contain n figures. Hence, if we divide the number into periods of two figures, proceeding from right to left, the number of figures in the root will be equal to the number of periods. (278.) II. The first figure of the root will be the square root of the greatest square number contained in the first period on the left. For the square of tens can give no figure in the first right hand period; the square of hundreds can give no figure in the first two periods on the right; and the square of the highest, figure in the root can give no figure except in the first period on the left. or Ex. 1. Suppose we wish to find the square root of 529. 400+120+9. Here the three classes of terms are exhibited distinct from each other, and we might extract the root by the rule of Art. 275. But observe that in the number 529, since the square of the tens can not give a figure in the place of units or tens, it Now this period con must be contained in the first period 5. tains not only the square of the tens, but also a part of the product of the tens by the units. The greatest square contained in 5 is 4, whose root is 2; hence 2 must be the number of tens, whose square is 400; and if we subtract this from 529, the remainder 123 contains twice the product of the tens by the units, plus the square of the units. If, then, we divide this remainder by twice the tens, we shall obtain the units, or possibly a number somewhat too large. This quotient figure can never be too small, but it may be too large, because the remainder 129, besides twice the product of the tens by the units, contains the square of the units. We therefore complete the divisor by annexing the quotient 3 to the right of the 4, and then multiplying by 3. we evidently obtain the double product of the tens by the units, plus the square of the units. The entire operation may then be represented as follows: 5.29|23=the root 4 43 129 129. (279.) Hence, for the extraction of the square root of numbers we derive the following RULE. 1. Separate the given number into periods of two figures each, beginning at the right hand. 2. Find the greatest square contained in the left-hand period; its root is the first figure of the required root. Subtract the square from the first period, and to the remainder bring down the second period for a dividend. 3. Double the root already found for a divisor, and find how many times it is contained in the dividend, exclusive of its righthand figure; annex the result both to the root and the divisor. 4. Multiply the divisor thus increased by the last figure of the root; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. 5. Double the whole root now found for a new divisor, and continue the operation, as before, until the periods are all brought down. Ex. 2. Find the square root of 186624. The operation is as follows: 18.66.24 432 16 83 2 66 2 49 862 17 24 17 24. Ex. 3. Find the square root of 21086464. Ex. 4. Find the square root of 88078225. (280.) We have seen that the root of a fraction is equal to the root of its numerator divided by the root of its denominator. That is, the square root of a decimal fraction may be found in the same manner as a whole number, if we divide it into periods commencing with the decimal point. Ex. 5. Find the square root of 58.614336. Hence, also, if the square root of a number can not be found exactly, we may, by annexing ciphers, obtain the root approximately in decimal fractions. Ex. 7. Find the square root of 2. Ex. 8. Find the square root of 3. Ans. 1.4142136 nearly. Ex. 9. Find the square root of 10. The most expeditious method of extracting roots is usually by means of logarithms. See page 308. (281.) Method of extracting the cube root of a polynomial. We already know that the cube of a+b is a3+3a2b+3ab2+b3 If, then, the cube were given, and we were required to find its root, it might be done by the following method: When the terms are arranged according to the powers of one letter, we at once know, from the first term a3, that a must be one term of the root. If, then, we subtract its cube from the proposed polynomial, we obtain the remainder 3a2b+3ab' +b3, which must furnish the second term of the root. Now this remainder may be put under the form (3a2+8ab+b2)×b; whence it appears that we shall find the second term of the root, if we divide the remainder by 3a2+3ab+b. But as this second term is supposed to be unknown, the divisor can not be completed. Nevertheless, we know the first term 3a2, that is, thrice the square of the first term already found, and by means of this we can find the other part b, and then complete the divisor before we perform the division. For this purpose, we must add to 3a2 thrice the product of the two terms, or 3ab, and the square of the second term of the root, or b2. we derive the following Hence RULE FOR EXTRACTING THE CUBE ROOT OF A POLYNOMIAL. (282.) Arrange the terms according to the powers of some one letter, take the cube root of the first term, and subtract the cube from the given polynomial. Divide the first term of the remainder by three times the square of the root already found, the quotient will be the second term of the root. Complete the divisor by adding to it three times the product of the two terms of the root, and the square of the second term. Multiply the divisor thus increased by the last term of the root, and subtract the product from the last remainder. Proceed in the same manner to find the additional terms of the root. Ex. 1. Extract the cube root of a2+12a'+48a+64. a3+12a1+48a+64|a+4 = the root. 12a+48a+ 64/3a2+12a+16= the divisor. Having found the first term of the root a, and subtracted its cube, we divide the first term of the remainder, 12a3, by three times the square of a, that is, 3a2, and we obtain 4 for the sec ond term of the root. We then complete the divisor by adding to it three times the product of the two terms of the root, which is 12a, together with the square of the last term, 4 |