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The word radical is used in connection with other roots than square roots. Thus, 10 means the cube root of 10; that is, the number which when used as a factor three times gives 10. Similarly, 56 means the fourth root of 6, etc. All such numbers represent perfectly definite magnitudes, as did V2 in Fig. 74, yet we cannot express them exactly by means of decimals.
In general, the nth root of any number a is written Va, and this is known as a radical of the nth order. The number n is here called the index of the root, and the number a itself is called the radicand.
When no index is expressed, the index 2 is understood. Thus, V3 means 93.
These definitions apply also to algebraic expressions. Thus, V3 xy2 and V9 xłyż are both radicals, although 9 x y2 is a perfect square, thus making it possible to write V9 x-y2=3 xy.
In the following list, pick out those values that can be expressed exactly without radicals, and those that cannot. For each state the index and the radicand. 1. V7.
12. $50. 2. Vg.
[Hint.-8=(-2)3.) 13. 32. 3. V17.
14. V3 x2y2. 4. VI.
9. V 27. 15. V64 a6b4 5. V. 10. 5.
16. 98 9663 6. V.3
11. 16. 17. 5 xby. 140. Value of Radicals. Use of Tables. To get the value of a radical correct to two or more places of decimals usually calls for a rather long process, as we have seen in § 135, where we found the value of 550. For this reason, the radicals which are needed most in ordinary life (that is, those coming from square roots and cube roots) have been carefully worked out and placed together in a Table at the end of this book, on p. 314. For the sake of completeness, the squares and cubes of the numbers are also shown, thus making the table very convenient for all kinds of mathematical work. Just how to use the table is described on page 311, which the pupil should now read carefully. Below are a few illustrative examples :
EXAMPLE 1. Find v7 from the table.
SOLUTION. By the top number in the third column of p. 326 (Table) we see that V7=2.64575, this value being correct to 5 decimal places.
EXAMPLE 2. Find 7 from the table.
SOLUTION. By the top number in the sixth column of p. 326 we have v7=1.91293, this value being correct to 5 decimal places.
EXAMPLE 3. Find v70.
SOLUTION. By the top number in the fourth column of p. 326 we have V70=8.36660+. The sign + which we have placed over the last digit indicates that the number is correct only up to the last decimal place.
EXAMPLE 4. Find ♡70.
141. Artisan's Method for Finding Square Root. The square root of a number may be easily calculated correct to several places of decimals in case one knows its value correct merely to the first or second decimal place. For example, 'knowing that V2=1.414, suppose we wish to calculate V2 to a greater degree of accuracy. We first
divide the 2 by the 1.41, carrying out the work to several places of decimals, as indicated below.
Now, since 1.41 was known to be slightly less than the true root, it follows that our quotient, 1.41851, is slightly greater than the true root. In other words, the true root itself lies somewhere between these two numbers. So we now calculate the number which lies halfway between them (their average) by taking half their sum. Thus,
1.41851 2 2.82851
1.41426 This average, or 1.41426, gives us a very close approximation to the true root. In fact, our answer in the present instance is the correct value of V2 to four places of decimals. †
† In general, if the number whose square root is to be found is greater than one, the new result will be correct at least to twice as many decimal places as the first estimate. The first estimate need not be at all exact, however. If any arbitrary number is taken as the first estimate, the process will give accurate answers after a sufficient number of repetitions.
Similarly, this method may be used to secure a good approximation for the square root of any number in case the value of the root is already known correct to only one or two decimal places. If still greater accuracy is desired, simply repeat the same process.
Because of the convenience and simplicity of this process, especially in rapid calculations such as are made by engineers, mechanics, or other workmen, it is commonly known as the “ artisan's method.”
EXERCISES Using the tables, find the approximate values of each of the following quantities.
1. V5. 2. 150. 3. 5. 4. V 50. 5. 500. 6. V51.
[Hint. 51=5.1X10. So use the information given for 5.10 on page 322 of Table.]
7. 51. 8. V82. 9. Ý820. 10. 77.7 11. 77.71 12. V77.1 13. V771.
SOLUTION. 771=7.71X100. Therefore V771 is the same as V7.71 except that the decimal point in the root must be moved one place farther to the right. (See p. 311.) Now, V7.71=2.77669+, . so it follows that V771=27.7669+. Ans.
EXERCISES — APPLIED PROBLEMS In answering the following, use the tables on pp. 314-331.
1. Figure 75 represents a right-angled triangle, whose short sides AB and BC measure 3 inches and 2 inches respectively. What is the length of the long side AC (hypotenuse) ?
SOLUTION. Let x represent the length of AC. Then x2 = 32 + 22 = 9 + 4 = 13. Therefore, x = V13 = 3.60555+ inches AZ (Table). Ans.
Fig. 75. 2. A baseball diamond is a square 90 ft. on a side. What is the distance from home plate to second base?
3. The diagonal of a square is 12 feet long. What is the length of each side?
4. The dimensions of a certain rectangular field are 100 feet by 230 feet. In going from one corner to the opposite corner how much shorter is it to go by the diagonal than to go around?
5. If the area of a circle is 10 square inches, how long is the radius?
(Hint. See Ex. 25, p. 22. Take a=37.)
6. If the volume of a cube is 235 cubic inches, how long is the edge ?
7. If the volume of a sphere is 672 cubic inches, how long is the radius?
[Hint. See Ex. 28, p. 23. Take r=37.]
8. In the accompanying figure, how long should the radius of each semicircle
be in order that the entire area inclosed Fig. 76. may be 200 square feet ?