School Algebra Volume 2 (Paperback)

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1913 Excerpt: ... tu? + M8. We see that there will be two integral places in the root. The first period, 42, contains t8. The greatest cube in 42 is 27, and the cube root of 27 is 3. Hence t = 3. The remainder, 15,875, resulting from subtracting the cube of the tens, will contain 3t2u + 3tu? + u8. Each of these three parts contains tt as a factor. Hence the 15,875 consists of two factors, one of which is u, and the other is 312 + 3 tu + u2. The largest part of this second factor is St2. If the 158 hundreds of the remainder is divided by 3 i2 = 3 x 302, or 27 hundreds, the quotient will be approximately u. The second factor can now be completed by adding to the 2700 the sum of 3 x (30 x 5), or 450, and 52, or 25. If this factor, 3175, is now multiplied by 5, the result is 15,875, which completes the cube of 35. There being no remainder, 42875 = 35. To check the work, 35s = 42,875. 130. Cube Root of Larger Numbers. The me.thod of 129 can be applied to numbers of more than two periods, by considering the part of the root already found as so many tens with respect to the next figure of the root. For example, find the cube root of 57,512,456. 57 512 456(386 27 Therefore the cube root is 386. 131. Cube Root of Decimals. If a cube root has decimal places, the cube will have three times as many. Thus, if 0.11 is the cube root of a number, the number is 0.11 x 0.11 x 0.11 = 0.001331. Hence, if a given number contains a decimal, we separate it into periods of three figures each, beginning at the decimal point and proceeding toward the left for the integral part, and toward the right for the decimal. The last period of the decimal must contain three figures, zeros being annexed when necessary. Since there can be only one integral place, the decimal point is placed after the 5, ...