Ex. Find the square root of 565504. Pointing off the number, we find the first period to be 56. Now, the greatest square in 56 is 49, and the square next greater is 64. Hence the number lies between the square of 700 and 800; and, following the algebraical method, 700 I will be the first term of the root. The operation will stand thus— C 565504 (700+ 50+ 2 = 752 Or, omitting the useless ciphers, and bringing down one period of figures at a time, the operation will stand thus It will be observed that the second remainder, 51, is greater than the previous complete divisor, 48, and it might be supposed, therefore, that the second figure in the root should be 9 instead of 8. Now, the square of (a + 1) exceeds the square of a by 2 a + 1. Hence it follows that, so long as any remainder is less than twice the corresponding number in the root + 1, we may be certain that we have taken the figure of the root sufficiently large. Thus, since the remainder is less than 28 × 2 + 1 or 57, we may be certain that 8 is the correct figure and not 9. Square Root of a Decimal. 38. It is evident that the square of any number containing one, two, three, &c., decimal figures, will contain two, four, six, &c., decimal figures respectively; and, hence, conversely, every decimal considered as a square must contain an even number of decimal figures, and its square root must contain half this even number of figures. It will then be necessary to add a cipher when the given number of decimal figures is odd. Further, since decimals and integers follow the same system of notation, it is evident that if a dot be placed over the units' figure of the given number, the pointing off may be performed with regard to the integral part exactly as in integers, there being no necessity to point off the decimals, only taking care to bring them down in pairs, and putting a decimal point in the quotient when the first pair is brought down. And again, if an integer be given which is not a perfect square, we may, by affixing to the right of it a decimal point and an even number of ciphers, gradually approximate to the square root as nearly as we please. Ex. 4. Find the square root of 247 to four places of decimals. Instead of adding a decimal point and eight ciphers to the right of the given number, we will proceed in the ordinary way till we arrive at a remainder. Then putting a decimal point in the quotient, we shall add two ciphers to this and each successive remainder. 39. It will be shown hereafter that when n + 1 figures of a square root have been obtained by the ordinary method, n figures more may be obtained by dividing the remainder by the number formed by taking twice the quotient already obtained, provided that the whole number of figures in the root is 2 n + 1. Ex. Find the square root of 29 to six places of decimals. The square root required will evidently contain seven figures. We shall therefore find the first four figures by the ordinary method, and the other three by the above method. Thus― 40. We will next develop the method of finding the cube root of a quantity. Ex. 1. Find the cube root of a3 + 3 a2b + 3 ab2 + b3. = We know (Art. 33) that (a + b)3 a3 + 3 a2b + 3 ab2 + b3. Hence a + b is the cube root of a3 + 3 a2b + 3 ab2 + b3. We see then that, the quantity being arranged according to the powers of a, the first term a of the cube root is the cube root of the first term of the given quantity; and if this term a3 be subtracted, there remains 3 ab + 3 ab2 + b3. We see again that if this remainder be divided by 3 a2, its first term gives b the second term of the root, and, further, if it be divided by b, we get 3a2 + 3 ab + b2 as a quotient. If we wish therefore to arrange the whole process in a way similar to ordinary division, it is evident that we must write as a divisor 3 a2 + 3 ab + b2, in order that after multiplication by the quotient figure b we may obtain a quantity which when subtracted shall leave no remainder. The operation will then stand thus— 3a2 + 3 ab + b2 a3 + 3a2b + 3 ab2 + b3(a + b a3 3 a2b + 3 ab2 + b3 3a2b + 3 ab2 + b3 We call 3 a2 the trial divisor, because by means of it we search for the second term of the cube root. Having obtained this second term, we then form the complete divisor 3a2 + 3 ab + b2. Ex. 2. Find the cube root of 8 x3-36 x2y + 54 xy2 – 27 y3. 8x3-36x3y +54 xy 27 y3(2x-3y 12x18xy+9 y2 8x3 − 36 x2y + 54 xy2 – 27 y3 EXPLANATION.-We find the cube root of 8x3 to be 2x. This is then the first term of the quotient, and corresponds to a in the previous example. We now require 3 a2 for a trial divisor. This, of course, = 3(2x)= 12x2. Subtracting the first term of the given quantity and dividing the first term of the remainder by this trial divisor, we obtain 3 y for By the quotient. This forms the second term of the root, and corresponds to b in the last example. We now easily obtain 3 ab and b2. Thus, 3 ab 3 (2x) (-3y) 9 y2. 62 18xy, and b2= (-3y)2 Hence, the complete trial divisor, corresponding to 3 a2 + 3 ab+b2 in the last example, = 12 x2 18 xy + 9 y3. Multiplying now by 3y the quotient, we obtain -36xy+54 xy 27 y3, which subtracted leaves no remainder. Hence, 2 x 3 y is the cube root. |