Answer:
345
Explanation:
You want the square root of 119025 by the "division method."
Division method
This method develops one digit of the square root at each step of the process.
The steps are shown in the attachment, and are described below.
The first step is to mark off pairs of digits either side of the decimal point. The first digit of the root is the integer part of the square root of the leftmost pair of digits.
For this number, that is √11 ≈ 3.
This value is appended to the "divisor" as the units digit, and the product of this "divisor" and the found digit is subtracted from the "dividend". At the first stage, this means 3×03 is subtracted from 11.
As in long division, the next digit pair is brought down and appended to the right of the difference. Here, this means the next "dividend" will be 2 with 90 appended. The new divisor is double the quotient so far, so will be 6_, where the _ is filled with the next root digit.
290/6_ ≈ 4, so the next root digit is 4 and the product 4×64 is subtracted from the "dividend" to get 290 -256 = 34. When the next digit pair is appended, the new "dividend" 3425 and the new divisor is 68_.
3425/68_ ≈ 5, so the next root digit is 5 and the product 5×685 is subtracted from the "dividend" to get 0. Thus ends the algorithm. If the result were not zero, the next digit pair would be appended to the difference, and the "divisor" would become 690_.
The square root of 119025 is 345.
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Additional comment
The "Babylonian method" (Newton's iteration) develops the root fairly quickly. Starting with 300, the next "guess" is the mean of this value and the quotient 119025/300:
(300 + 119025/300)/2 = 348.375
The next iteration is ...
(348.375 +119025/348.375)/2 ≈ 345.0163...
Followed by ...
(345.0163 +119025/345.0163)/2 ≈ 345.000000387
This method will approximate the root better at each step. (The number of correct significant digits approximately doubles at each step.) It will terminate when the change in value is smaller than the precision limit of the calculator. For most calculators, that will be 6 iterations or fewer.
While this method develops a root quickly using a calculator, hand calculation becomes tedious.
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