root of arithmetical numbers can be found approximately by inspection. 70. EXAMPLES 1. Find approximately √19. 19 lies between 16 and 25. Therefore, √19 lies between √16 and √25, or between 4 and 5. 2. Find approximately √643. 643 lies between 400 and 900. Therefore, √643 lies between 400 and 900, or between 20 and 30. That is, it is 20 plus a number less than 10. 3. Find approximately √4678. 4678 lies between 3600 and 4900. Therefore, √4678 lies between √3600 and √4900, or between 60 and That is, it is 60 plus a number less than 10. The numbers to be added in any case will not change the first figure of the root found. That is, by inspection we can find exactly the first figure of the square root. ORAL EXERCISES State the first figure of the square root of each number: Since 102 = 100, we know 355. Pointing off into Periods. that the square root of any number greater than 1 but less than 100 is less than 10. Its integral part consists of one figure. Since 100210,000, we know that the square root of any number greater than 100 but less than 10,000 is greater than 10 but less than 100. That is, if the given number has 3 or 4 digits in its integral part, its square root will have 2 digits in its integral part. If larger numbers are given, the above reasoning can be repeated for 10002, etc., showing that in all cases if the number be pointed off into periods of 2 digits each (or possibly fewer in the left period), then each period will correspond to a digit of the root. Thus in 67/62/31, there are three periods, therefore there are three places in the integral part of the root. Since 67 lies between 64 and 81, the square root of 67 is approximately 8, and that of 676,231 is approximately 800. ORAL EXERCISES By the method above state an approximate square root of: 356. When the first digit of the square root has been found by inspection, the process may be continued thus: 5. Then 609 is greater than 80 r, or 502 is greater than r. ... 7 + decimal is greater than r, and it is possible that 7 = r. Trying, we find that 80.7 +72 = 609. That is, Therefore, r = 7. 220940 + 7 = 47. WRITTEN EXERCISES Use the above process to find the square roots of: 357. When once an approximate value, a, has been found for the root, an approximate value for the remainder, r, of the root can be found by means of the formula: (a+r)2 = a2 +2ar + r2. 1. Let n denote the number whose square root is sought, a denote the approximate root at any stage, and r the remainder of the root. 2. Then, n = (a +r)2 = a2 + 2 ar + r2. 5. Hence, a2 is greater than 2 ar, or, may be tried as an approximate value of r. 358. What precedes may be formulated into a process or working rule, thus: 1. Point off the number into periods of two figures each, beginning at units' place (at the decimal point). 2. By inspection find the largest integer whose square is not greater than the left period. (In Example A it is 9.) 3. Use this integer as the first digit of the root. Subtract its square from the left period. (In Example A this square is 81.) 4. Bring down the next period. (In Example A this makes 364.) Root (A) Number 92 84'64 81 18 364 182 364 5. Multiply the part of the root already found by 2. This number is called the trial divisor. (18 in Example A.) 6. Divide the remainder (omitting the right digit) by the trial divisor and use the digit found as the next digit of the root. (In Example A, 36÷ 182.) 7. Annex this digit to the trial divisor. This forms the complete divisor. (182 in A.) 8. Multiply the complete divisor by the digit of the root just found and subtract. NOTE. It may happen that the product to be subtracted is larger than the number from which it is to be subtracted. This indicates that the trial divisor produced too large a digit. Try the next smaller digit for the figure of the root last found. Root (B) 3 0.69 Number 9'41'.87'61 9 6 41 60 4187 606 3636 55161 612 9. Repeat the steps 4 to 8 until all of the Number 2'00. periods have been brought down. If the last remainder is zero, as in Example B, the process is ended, the given number is a perfect square, and its root has been found exactly. If the last remainder is not zero, as in C, the process may be continued as far as desired by supplying zeros. TEST. The square of the root, if complete, equals the given number. 359. When a number contains a decimal the decimal point of its root is placed between the figures furnished by the integral periods and those furnished by the decimal periods (as in C above). WRITTEN EXERCISES Find the square roots: 1. 361. 3. 625. 5. 2025. 7. 177,241. 9. 4,334,724. 2. 784. 4. 841. 6. 1936. 8. 120,409. 10. 4,888,521. Find the square roots to two decimal places: 360. To find the square roots of fractional numbers, either first reduce the fraction to a decimal or extract the square root of both numerator and denominator. 9. The area of an equilateral triangle of side s is known to be 2√3. Find the side of an equilateral triangle whose 18 area is 9 √3 sq. in. SUGGESTION: Solve the equation s2 √3 = 9 √3 for s. 10. The altitude of an equilateral triangle is 8√3. Find the side of an equilateral triangle whose altitude is 8 √3 ft. S√3 11. Each section, as ABC, of the bridge truss shown in |
