EXAMPLES. 1. Extract the square root of 7569. Ans. 87. 2. Extract the square root of 76807696. Ans. 8764+. Ans. 295917+. 3. Extract the square root of 87567. 223. Since the square of a vulgar fraction is obtained by multiplying its numerator by itself for a new numerator, and its denominator by itself for a new denominator, it follows, that to extract the square root of a fraction whose terms are perfect powers, we have only to extract the root of the numerator and also of the denominator. Thus, ✔, since the square root of 9 is 3, and that Q. How is the square root of a vulgar fraction extracted? Why is this? 224. Should the numerator of the fraction be an imperfect power, we extract its square root approximately, and place the result over the root of the denominator. Thus, to extract the square root of 2. Q. How is the square root extracted when the numerator is an imperfect square? How far may the operation be extended? EXAMPLES. 1. Extract the square root of 88. 2. Extract the square root of 23. Ans. 3. Extract the square root of 1⁄2 of 7. Ans. 225. Should the denominator of the fraction be an imperfect power, we multiply the two terms of the fraction by the denominator. By this operation, the denominator is made a perfect square, and the root of the fraction may be extracted as in the last article. Thus, extract the square root of g. Multiplying numerator and denominator by 5, the denominator becomes 25, the square root of which is 5. Extracting the square root of 15 to three places of decimals, it becomes 3.872. The root sought is 3.872. 5 OPERATION. }= 3x5 5x5 15 - 25 ✓15=3.872 3.872 Q. How is the square root of a fraction extracted when its denominator is an imperfect power? Is the value of the fraction changed by this operation? Why do you multiply by the denominator? 226. Since a decimal may be expressed as a vulgar fraction by placing 1 as its denominator, with as many ciphers annexed as there are decimal places, we may extract the square root of a decimal by the principles just explained. Should the given decimal contain an uneven number of places, a cipher must be annexed to make them even, and by this means the denominator will be made a perfect square. The square root of the decimal may then be extracted as if it were a whole number, care being taken to point off as many decimal places in the root as there are periods in the given decimal. Thus, extract the square root of .5. The number of decimal places being odd, we annex a cipher, making the decimal .505. Extracting the root of 50, it is .7 By annexing two more ciphers, the decimal becomes .5000-5000, the root of which is .707. By continuing this = 100009 OPERATION. .50 (.707+ 49 .1407) 10000 9849 151 operation we may approximate as near as we please to the root of. Q. How may the square root of a decimal fraction be extracted? EXAMPLES. 1. Extract the square root of .25. 2. Extract the square root of .42573. Ans. .5. Ans. Ans. Ans. 4.683. Ans. .736. Ans. .073. Note. By reducing the vulgar fractions in articles 224 and 225 to equivalent decimals, their square root might be extracted as has just been explained for decimal fractions. EXTRACTION OF THE CUBE ROOT. 227. THE extraction of the cube root of a number consists in finding a number which, multiplied by itself twice, will produce the given number. Thus, the cube root of 27 is 3, since 3× 3× 3=27. We have seen (Art. 219) that there are only nine perfect cubes expressed by one, two, or three figures, viz: 1, 8, 27, 64, 125, 216, 343, 512, 729, the cube roots of which are the first nine numbers, 1, 2, 3, 4, 5, 6, 7, 8, 9. Hence, the cube root of every number composed of less than four figures cannot contain more than one figure, that is, units. The root of an imperfect cube will be found between two numbers which differ from each other by one, and whose cubes are the next greatest and smallest cubes, between which the given number is comprised. Thus, 130 is comprised between 125 and 216, the cube roots of which are 5 and 6. The cube root of 130 will therefore be between 5 and 6. Q. In what does the extraction of the cube root of a number consist? What is the cube root of 27? Why 3? How many perfect cubes may be expressed by one, two, or three figures? What are their roots? How many figures will the cube root of a number composed of fewer than four figures contain? What will the root be? Between what numbers will the root of an imperfect cube be found? 228. To explain the method of extracting the cube root of a number, let us see of what parts the cube is composed The cube of the number 16 is the product of 16 or its equal 10+6 by itself twice. Performing the operation as in Art. 121, we find the result as follows: OPERATION. 10+6 10+6 10×6+62 102+10×6 102+2(10×6)+62=square of (10+6) 10+6 102×6 +2(10×63)+63 103+2(102 × 6)+ 10×62 103+3(102×6)+3(10×62)+63=cube of (10+6). The first multiplication gives the square of 10+ 6, as found in Art. 121. Multiplying again by 10+6, we obtain the cube of 10+6. But upon examining this result we find that it is com posed as follows: 1st, 103=cube of the tens. 2d, 3(102 × 6)=three times the square of the tens multiplied by the units. 3d, 3(10×62)=three times the square of the units multiplied by the tens. 4th, 63-cube of the units. And as the same may be proved for every other such number, we conclude, that the cube of a number composed of tens and units contains the cube of the tens, plus three times the square of the tens multiplied by the units, plus three times the square of the units multiplied by the tens, plus the cube of the units. Thus, the cube of 25 is as follows: |