the third figure of the root. (1460 +6) 6=8796, which product, subtracted from 8796, gives a remainder of 0, and the square is completed. The addition of the 0 to the figures of the root already obtained gives a trial divisor of the same order as the remainder, or of the next lower order. The process gives fewer figures and less likelihood of error. If a given square number has decimal places, we point off by beginning at the decimal point, first separating the whole number as before, finally separating the decimal into periods from left to right. Ciphers may be annexed, if necessary, to complete any period. If a given number is not a perfect square, its approximate square root can be found to any desired number of places. The square root of a common fraction is best found by changing the fraction to a decimal and extracting the square root of the decimal to the required number of places. Exercise 75 Find the square root of: 1. 9216. 2. 67081. 3. 32761. 4. 182329. 5. 186624. 9. .717409. 13. .00002209. 6. 4202500. 10. 9617.7249. 14. .0001752976. 7. 49.434961. 11. 44994.8944. 15. .009409. 8. 9486.76. 12. .00119716. 16. .0000879844. Find, to three decimal places, the square root of: Find, to three decimal places, the value of: 33. How many rods in the side of a square field whose area is 2,722,500 square feet? 34. Simplify and extract the square root of x2 (10 x2+13) - 2 (2 x2 + 3 + 7 x2) x + (x2 + 1) (x − x2 + 1). 35. Simplify and extract the square root of х (x2 -2x-3) (x2-x-6) (x2 + 3x + 2). 36. Show that the required square root in the preceding example can be readily obtained by factoring and inspection. 37. If a, b, and c are the sides of a right triangle, and a lies opposite the right angle, we may prove by geometry that a2 = b2+ c2. Find the length of a in a right triangle in which b and c are 210 feet and 350 respectively. 38. Applying the principle given in the preceding example, find, to three decimal places, the distance from the "home" plate to the second base; the four base lines of a regulation baseball diamond forming a square 90 feet on each side. CHAPTER XIX THEORY OF EXPONENTS 239. The Index Laws for Positive Integral Values of m and 240. The extension of the practice of algebra requires that these laws be also extended to include values of m and n other than positive integral values only; and the purpose of this chapter is to so extend those laws. We shall, therefore: 1. Assume that the first index law (aTM × a2=am+”) is true for all values of m and n. 2. Define the meaning of the new forms that result under this assumption, m or n, or both, being negative or fractional. 3. Show that the laws already established still hold true with our new and broader values for m and n. THE ZERO EXPONENT If m and n may have any values, let n = 241. Any quantity with the exponent 0 equals 1. If m and n may have any values, let n be less than 0 and Hence, we define a-m as 1 divided by am. From this definition we obtain an important principle of constant use in practice: 242. Any factor of the numerator of a fraction may be transferred to the denominator, or any factor of the denominator may be transferred to the numerator, if the sign of the exponent of the transferred factor is changed. |