How do you find out the square root of a decimal?
ex. 2.31
3 answers
You could use a calculator.
I need to do it without a calculator
This used to be taught to 6th graders back in the 1910's. With the advent of calculators and computers, the method got lost from the system. Some home schooling programmes still do it.
1. By modern calculator
Assuming you have a calculator, press 2.31 followed by the √ key.
2. By simple calculator
If your calculator does not have a √ key, you can do it by trial and error. Try to square 1.5 (=2.25), too small. Try 1.6² (=2.56), too big. So the square root is somewhere in between Keep trying and halving the gap, for example, try 1.55 in this case.
3. By simple calculator using Newton's method.
From a given estimate x, calculate a better estimate using
x1=x + (2.31-x²)/(2x)
Keep repeating the same procedure until the desired accuracy is achieved.
Example:
x=1.5
x1=1.5+(2.31-1.5²)/(2*1.5)=1.52
x2=1.52+(2.31-1.52²)/(2*1.52)=1.519868
which is already accurate to 6 figures after the decimal point.
4. Manual method.
It is rather difficult to describe the method without having an easy way to align numbers in the response. So I will refer you to a web-site that describes the method:
http://www2.qnet.fi/abehr/Achim/Calculators_SquareRoots.html
1. By modern calculator
Assuming you have a calculator, press 2.31 followed by the √ key.
2. By simple calculator
If your calculator does not have a √ key, you can do it by trial and error. Try to square 1.5 (=2.25), too small. Try 1.6² (=2.56), too big. So the square root is somewhere in between Keep trying and halving the gap, for example, try 1.55 in this case.
3. By simple calculator using Newton's method.
From a given estimate x, calculate a better estimate using
x1=x + (2.31-x²)/(2x)
Keep repeating the same procedure until the desired accuracy is achieved.
Example:
x=1.5
x1=1.5+(2.31-1.5²)/(2*1.5)=1.52
x2=1.52+(2.31-1.52²)/(2*1.52)=1.519868
which is already accurate to 6 figures after the decimal point.
4. Manual method.
It is rather difficult to describe the method without having an easy way to align numbers in the response. So I will refer you to a web-site that describes the method:
http://www2.qnet.fi/abehr/Achim/Calculators_SquareRoots.html
1 answer