How do I find a square root of a number, without a calculator?
Oh sorry, Its math
How do I find the estimate of a square root of a number that is not a perfect square without using a calculator.
Newton's Method
https://www.school-for-champions.com/algebra/square_root_approx.htm#.WqalredG3IU
for more exhaustive derivation:
https://math.mit.edu/~stevenj/18.335/newton-sqrt.pdf
This site has a pretty good explanation.
https://www.homeschoolmath.net/teaching/square-root-algorithm-example.php
It is based on the fact that (a+b)^2 = a^2 + 2ab + b^2
extending the logic, you can work out (as I did many moons ago) how to extract cube roots using the fact that
(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
have fun.
Ok my name is Alexis as well, so, yeah, I’m a different person. You just find out what times itself would equal that number that you’re finding the square root of. For example, the square root of 9 is 3, because 3 times itself is 9. The square root of 16 is 4, because 4 times itself is 16. And the square root of 25 would be 5, cause 5 times itself is 25. I think you get the idea. Hope I could help
To find the square root of a number without a calculator, you can follow these steps:
1. Start by estimating the square root. Choose two perfect square numbers between which the given number lies. For example, if you want to find the square root of 20, you might estimate that it lies between 4 (2 squared) and 9 (3 squared).
2. Take the average of the two chosen numbers. In this case, the average of 4 and 9 is 6.5.
3. Square the average obtained in step 2. In this example, 6.5 squared is 42.25, which is higher than 20.
4. Adjust the estimate. Since the squared value is higher than the actual number, we need to choose a smaller number between 4 and 6.5. Let's choose 5.
5. Repeat steps 2 to 4. Take the average of 5 and 6.5, which is 5.75. Square this value to get 33.0625.
6. Continue these iterations until you find a square close to the given number. In this case, the square root of 20 is approximately 4.47.
You can further refine the estimate by repeating the process for more decimal places, depending on the desired level of accuracy.