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6 answers

There is a method, akin to long division in form, to do it, but it's too complicated; there is Newton's method, to which I will give links in the sources; and there is smart guessing, which I will use in your example: square root of 150 * 20 = 3000.

First, note that 3000 = 30 * 100, so sqrt(3000) = sqrt(30) * sqrt(100) = 10 * sqrt(30). Now, we have to calculate sqrt(30).

Since 25 < 30 < 36, 5 < sqrt(30) < 6. Let's try 5.5: 5.5^2 = ... (hand calculating) = 30.25 . Bigger than 30.Let's try 5.4: 5.4^2 = 29.16. Smaller than 30.

Now we know that 5.4 < sqrt(30) < 5.5, leaning towards 5.5 (30.25 is nearer of 30 than 29.16). My next guess would be 5.48: 5.48^2 = 30.0304. Bigger than 30. 5.47^2 = 29.9209, smaller than 30.

Now we know that 5.47 < sqrt(30) < 5.48, or 54.7 < sqrt(3000) < 54.8. Not bad for about ten minutes of calculation.

Now, the same problem using Newton's method. Let _a_ be the number to which extract the square root, and x_0 be a decent first approximation of sqrt(a). Then, for all i >= 0, apply the formula

x_(i+1) = (x_i + (a/x_i))/2

The sequence x_0, x_1, x_2, etc., will approach quite quickly sqrt(a). Since you will work by hand, I suggest handling everything in fraction form, and only convert to decimal after 3 or 4 steps.

I will give the example of sqrt(3000) again:
a = 3000, x_0 = 50

x_1 = (50 + 3000/50)/2 = 55
x_2 = (55 + 3000/55)/2 = (55 + 600/11)/2 = 1205/22
x_3 = (1205/22 + 3000*22/1205)/2 = ...

As you see, the fractions become more and more unwieldy with each iteration. This is the reason I used guessing first.

Hope this helps.

2007-01-26 07:50:45 · answer #1 · answered by jcastro 6 · 0 0

Okeydoke - this is actually quite simple. Work by this idea:

(By the way - the 1/2 power means a square root. You can write cube roots, etc like that as well (1/3, etc).)

a = bc
(a)^1/2 = [(b)^1/2]*[(c)^1/2]

Therefore, you can split a square root of a composite number up into its factors. These factors can be a lot easier to find the square root of ('tis handy to memorize some common square roots -> 4^1/2 = 2, 16^1/2 = 4, etc). EX:

(8)^1/2 = [(2)^1/2][(4)^1/2] = 2[(2)^1/2]

Hope it helps!

2007-01-26 07:23:46 · answer #2 · answered by emsviper 2 · 0 0

150 x 20 = 3000
so you find the square root of 3000... and by the way the square root of 3000 is an irrational number (like pi) and it never stops.... this is because any square root that is not perfect is irrational.....like the square root of 37.

2007-01-26 07:20:27 · answer #3 · answered by Sheila 2 · 0 0

for this u must go through the book named VEDIC MATHEMATICS
this book has indian mythological origin and is very helpful in detemining products , squares , square roots , divisions upto 20th place after decimaland many more which takes just a few seconds
if u r not in india then u can find this book at www.mlbd.com
if u r in india , u can find it any good book store

the methods described in this book are shortest possible and effective for rocketting calculations
u can solve ur problem within 4-5 seconds then

2007-01-26 07:29:52 · answer #4 · answered by   2 · 0 0

That would be the square root of 3000. Try this link:
http://www.nist.gov/dads/HTML/squareRoot.html

It shows how to compute square roots by longhand.

2007-01-26 07:25:34 · answer #5 · answered by Anonymous · 0 0

Prime factorize the numbers

P = 2*5*3*5, R = 2*2*5
P*R = 2*2*2*3*5*5*5
√PR = √2^3*3*5^3
√PR = 2*5*√2*3*5
√PR = 10√30

2007-01-26 07:19:41 · answer #6 · answered by bequalming 5 · 0 0

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