Finding Square Roots Manually

[deep0199]

What is the best way to manually find the square root of a positive rational number?

 

I understand that it will be an approximation, what what is the way to get the closest possible answer rounded to 2-3 decimal places.

[fuhrerkeebs]

You could just memorize the roots of a bunch of prime numbers, and just factor the number, pull out all pairs of two and just multiply the roots of the remaining primes...or you could use taylor expansion.

[Dapthar]

What is the best way to manually find the square root of a positive rational number?

It depends on what you mean by "best". If you want a simple' date=' and quick way to get a reasonably accurate estimate of a square root, try the first method on this page: http://mathforum.org/dr.math/faq/faq.sqrt.by.hand.html

 

If you want a more accurate approach, you could use the Binomial Theorem, and the details of how to use such a method are also on the aforementioned page, under the heading [b']Square Roots Using Infinite Series[/b].

[Ducky Havok]

If you only want it to a couple decimal places, I find this method useful.

 

[math]\frac{1}{2\sqrt{x}}dx+ \sqrt{x}[/math]

 

Here's an example: [math]\sqrt{24.8}[/math]. Pick the closest perfect square, in this case 25. So x=25, and dx=-.2. When you put the numbers in you get -.02+5, which is 4.98. When you put it in a calculator its really 4.97996, so it's pretty close.

[DQW]

Often, you could use a binomial approximation (as suggested previously) : choose the nearest perfect square to x; let it be n², where n is some convenient rational or known square root.

 

Let x = n² + d, where |d| << n²

 

Then [math]\sqrt{x} = \sqrt{n^2 + d} = n*\sqrt{1 + \frac{d}{n^2}} \approx n*(1+ \frac{d}{2n^2}) [/math]