I am surprised at how small amount of information I could find on this subject when surfing the web. My teacher a while back showed us a clever little quick way to find the sqaure root of a number, does anyone know how to do this?{if so, please list} Thanks.

http://math.arizona.edu/~kerl/doc/square-root.html

 

follow this URL

The link Sama posted is good.

 

If you don't mind doing some long division, I always liked the divide and average method.

 

I.e. say you were trying to find the square root of 10, and you just guessed that it was 3. Now, divide 10 by 3 and you are left with 3.333333. Average the two results and you get 3.166667. Now, use that average as your next guess. Divide 10 by 3.16667 and you get 3.1578. Average the two again and repeat. This will converge to the square root (3.1622). It isn't fun to do it all by hand, but it can be done.

 

How much work it is depends on how accurate you need the result to be.

If you are feeling adventurous you could try this

 

[math] \Delta y \approx f'(x) \Delta x [/math]

 

Where

 

[math] y = f(x) = \sqrt x [/math]

 

then

 

[math] \Delta y = \sqrt {x + \Delta x } - \sqrt x = \sqrt {x + \Delta x} - y [/math]

 

so

 

[math] y + \Delta y = \sqrt { x + \Delta x } [/math]

 

now

 

[math] \Delta y \approx f'(x) \Delta x = \frac {1} {2} x ^ { - \frac {1} {2} } \Delta x = \frac {1} { 2 \sqrt x } \Delta x [/math]

 

let x = 25 because this is the nearest square root that is easily achieved mentally and [math] \Delta x = 2 [/math] because 25 + 2 = 27

 

then

 

[math] \Delta y \approx f'(x) \Delta x = \frac {1} {2 \sqrt {25} } * 2 = \frac {1} { \sqrt {25} } = \frac {1} {5} = .2 [/math]

 

thus

 

[math] \sqrt {27} = \sqrt { x + \Delta x } = y + \Delta y = \sqrt {25} + .2 = 5 + .2 = 5.2 [/math]

 

These are approximations CM..........

 

I just added this 'cause I thought it was cool when I learned it.........Calculus Bittinger!

Edited July 9, 200916 yr by buttacup

Nice...

 

just a question, how did you get [math]\Delta y = \sqrt {x + \Delta x } - \sqrt x[/math]?

where

 

[math] y = \sqrt x [/math]

 

because this is the question

 

so....

 

[math] y_2 = \sqrt {x + \Delta x } [/math]

 

and

 

[math] y_1 = \sqrt x [/math]

 

then

 

[math] y_2 - y_1 = \sqrt {x + \Delta x } - \sqrt x [/math]

 

and

 

[math] \Delta y = \sqrt {x + \Delta x } - \sqrt x [/math]

 

I do believe this may be a way used by calculators starting by finding the first closest squaring of a prime and then going through a couple iterations of this to find a good approximation.......

Thanks, didn't think of it that way

I remembered it was something like this:

Find the prime factors of a number, then group all 2's numbers, for example, to find the sq root of 144, its prime factors are 2, 2, 2, 2, 3, and 3. Then, for every two same numbers, that is equivalent to one of that number. For example, there are two 2, then, that number is equivalent to 2 only. 2, 2, 2, 2, 3, and 3 are equivalent to 2, 2, and 3. Then multiply those numbers. (2x2x3=12 12x12=144 Didn't know I would still remember it.)

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Merged post follows:

Consecutive posts merged

P.S. If you get an extra number (2, 2, 2, 2, 3, 3, "4"), put that number inside the radical sign. (2x2x3 \/"4"" or 12\/"4"") (just pretend \/""" is radical)

*facepalm*

I am surprised at how small amount of information I could find on this subject when surfing the web. My teacher a while back showed us a clever little quick way to find the sqaure root of a number, does anyone know how to do this?{if so, please list} Thanks.

 

Here's another trick that can work.

 

1) Take your original number (the one you want to find the square root of) - lets call it x1

2) Produce a new number thus: x2 = x1-(x1^2-x1)/(2*x1)

3) Produce a third number thus: x3 = x2-(x2^2-x1)/(2*x2) [iMPORTANT: Notice the second term in (x2^2-x1) (and only this term) is x1 - this is not a typo - it has to be the original number all the way through]

4) Repeat. The values of 'x' get closer to the square root of the original number. You can repeat as often as you want. The more you repeat the more accurate the answer, but you'll find that 4 or 5 repeats will be accurate enough for most purposes