11. 47.62 ? 12. (17)2 = ? Ans. 8464. 13. (1243) = ? 19. ()12? 20. .59 = ? 21. Involve 1.3 to the 6th power. 22. Raise 18 to the 5th power. 23. What is the difference between the square and the cube of 24. 24. What is the compound interest of $1.10 for 4 years, at 10 per cent.? 25. How many paving stones 13 inches square will be required to pave 100 rods of a street 3 rods in width? 26. How many dice measuring an inch each way may be made from a cubical foot of ivory, allowing for waste in the manufacture? EVOLUTION. 383. Evolution consists in finding the roots of numbers. 384. The root of a number is one of the equal factors which produce that number. The square root is one of the two equal factors; the cube root, one of the three equal factors; the fourth root, one of the four equal factors, and so on. 385. √ is the radical sign, and by itself signifies the square root, and with a figure above it, signifies the degree of the root indicated by the figure; thus, 27 signifies the third root of 27. The root may also be indicated by a fractional exponent; thus, 16 (read, 16 to the power)=√16 √16 = 2. SQUARE ROOT. 386. Table, showing the places occupied by the square of any number of units, tens or hundreds. 387. By the above we perceive that the square root of any whole number expressed by one or two figures, must be units; expressed by three or four figures, must be units and tens; expressed by five or six figures, must be units, tens, and hundreds. Hence, generally, if a number be separated into periods of two figures each, beginning with the units, the number of figures in the square root will be indicated by the number of periods. NOTE I. The left hand period may contain but one figure. NOTE II. The principle above elucidated applies also to decimal fractions; but every period in decimal fractions must contain two figures. 388. That the pupil may comprehend the method of extracting the square root of a number, we will multiply 64 by itself, i. e., square it, and keep the separate products, instead of reducing them and adding as in ordinary multiplication. 64 64 (604) × (60+ 4). By the above it will appear that a square whose root is com posed of tens and units, contains (1.) The square of the tens ; (2.) Twice the tens multiplied by the units; and (3.) The square of the units. 389. We will now extract the square root of 4096. square 36 (hundreds) from the 40 (hundreds), and to the remainder 4 (hundreds), bring down the next period, 96. This remainder (496) must contain two times the tens of the root multiplied by the units, plus the square of the units, or the product of two times the tens, plus the units, multiplied by the units. If it contained only two times the tens multiplied by the units, we should obtain the units' figure by dividing the remainder (496) by two times the tens. We make 2 X 6 (tens) = 12 (tens) the trial divisor, which is contained in 49 (tens) 4 times. We write 4 as the units' figure of the root, and also at the right of the 12 (tens), and have 124 for the true divisor. This we multiply by 4, and thus complete the square, obtaining at once, twice the product of the tens by the units, and the square of the units. If the root consists of more than two figures, having obtained the first two, we can consider them as tens in reference to the next figure, and proceed with them in all respects as above. Thus, suppose it be OPERATION. 4138.94 (64.33+ 36 124) 538 496 1283) 4294 12863) 44500 5911 required to extract the square root of 4138.94: find the first two places as before; bring down the next period, 94, and form a new trial divisor by doubling 64 (the root already found); find how many times this, considered as tens, is contained in 429 tens, for the third figure of the root. To obtain a fourth figure in the root, form another period by annexing two zeros, double 643, and so continue. From the above, we deduce the following RULE FOR EXTRACTING THE SQUARE ROOT OF A NUMBER. Point off the given number into periods of two figures each, by placing a dot over the units' figure and every alternate figure to the left in whole numbers, and to the right in decimals. Find the greatest square number in the left hand period, and write its root as the first term in the answer. Subtract the square number from the left hand period, and to the remainder bring down the next period for a dividend. Take twice the root already found for a trial divisor; rejecting the right hand figure of the dividend, divide it by the trial divisor; place the result, as the second term in the root, also at the right of the trial divisor, making a true divisor; multiply the true divisor thus obtained by the last term of the root, and subtract this product from the dividend; to the remainder bring down the next period for a new dividend. Double the terms of the root already found for a new trial divisor, and proceed as before. NOTE I. When a zero occurs in the root, annex a zero to the trial divisor, bring down another period, and proceed as before. NOTE II.. If a root figure proves too large, substitute a smaller, and repeat the work. NOTE III. When a remainder occurs after all the periods are brought down, the root may be more nearly approximated by annexing periods of zeros, and continuing the operation. NOTE IV. The square root of a common fraction may be obtained by extracting the root of both terms when they are perfect squares; when they are not, the fraction may first be reduced to a decimal. NOTE V.Mixed numbers may be reduced to the decimal form, or to improper fractions when the denominator of the fractional part is a square number. 390. The above rule may be illustrated by diagrams. Let A B C D (Fig. 1) represent a square court containing 4096 square feet, the length of whose side we wish to determine. Having found (Art. 389) that the greatest square of tens in 4096 is 3600, the root of which is 6 tens, we deduct 3600 from 4096, and have left 496 square feet, which are to be disposed on two sides of the square already found. The width of these additions we wish to ascertain. Fig. 2. 60 ft. M 60 ft. N By extending the lines a and b, we shall divide the addition into three parts, M, N, and O; M and N having for one side the tens of the root, and O being a square whose side is equal to the width of the side additions. If the 496 square feet equalled the feet in the side additions, M and N, the width of the additions would be determined by dividing 496 by twice the length of the square already found, 2X 60. Using this as the trial divisor, we obtain 4 as the width, which is the units' term of the root; but the entire length of the additions 60 + Fig. 3. 60 + 4 is two times the tens, plus the units, or 124 (Fig. 3), the product of which by 4, the units' term, is 496. There being no remainder, 4096 is found to be a square of which 64 is the root, and the length of the court is 64 feet. 391. EXAMPLES IN SQUARE ROOT. 1. What is the square root of 841 ? Ans. 29. Ans. 874. Ans. 3690. Ans. 503. Ans. 1006. Ans. 21.15. Ans. .075. Ans. .41109+. Ans. |
