 | James Bates Thomson - Arithmetic - 1862 - 436 pages
...Subtract the square of this figure of the root from tlie period under consideration ; and to the right of the remainder bring down the next period for a dividend. III. Double the root just found and place it on the left of the dividend for a partial divisor ; find how many times it... | |
 | John Flint (inspector of schools.) - 1862 - 152 pages
...period, and write its cube root as the first figure of the root. Subtract its cube from the left hand period, and to the remainder bring down the next period for a dividend. Multiply the square of the part of the root found by 300 for a trial divisor ; divide the dividend... | |
 | Horatio Nelson Robinson - Algebra - 1863 - 432 pages
...left hand period, and place its foot for the ßrst figure of the required root ; subtract this cube from the first period, and to the remainder bring down the next period for a dividend. III. At the left of the dividend write three times the square of the root already found, and annex two ciphers,... | |
 | Benjamin Greenleaf - 1863 - 338 pages
...square in the left-hand period, and place its root on the right ; subtract the square of this root from the first period, and to the remainder bring down the next period for a dividend. 4356 36 126 756 756 ELEMENTARY ALGEBRA. Divide this quantity, omitting the last figure, by double the... | |
 | Isaac Todhunter - Algebra - 1863 - 302 pages
...749 is contained in the first period; this 749 is the first figure in the root; subtract its square from the first period, and to the remainder bring down the next period. Divide this quantity, omitting the last figure, by twice the part of the root already found, and annex... | |
 | George Augustus Walton - Arithmetic - 1864 - 376 pages
...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,... | |
 | George Augustus Walton - Arithmetic - 1864 - 364 pages
...hand period, and write its root as the first term in the answer. Subtract the cube from the left hand period, and to the remainder bring down the next period for a dividend. Multiply the square of the root already found, considered as tens, by three for a trial divisor. Divide... | |
 | Francis Walkingame - 1865 - 222 pages
...the square number under the first period, and the root in the quotient ; subtract the square number from the first period, and to the remainder bring down the next period, and call that the resolvend. Thirdly, Double the quotient, and place it for a divisor on the left hand... | |
 | Joseph Ray - Algebra - 1866 - 250 pages
...the right, after the manner of a quotient in division. Subtract the square of the root from the left period, and to the remainder bring down the next period for a dividend. 3. Double the root already found, and place it on the left for a divisor. Find how many times the divisor... | |
 | Isaac Todhunter - Algebra - 1866 - 580 pages
...contained in the 1 2 6 J 7 5 6 first period ; this is the first figure in the 756 root; subtract its square from the first period, and to the remainder bring down the next period. Divide this quantity, omitting the last figure, by twice the part of the root already found, and annex... | |
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... and to the remainder bring down the next period for a dividend. 3. Place the double of the root already found, on the left hand of the dividend for a divisor. 4. Seek how often the divisor is contained... 
