Square roots?

Question

I remeber when I was younger, I could find square roots easily. I remember dividing first I believe. Well, I forgot, and when I try to learn again, I find the methods to be different.
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Can somebody point me to a place online to learn this? Or explain it to me?
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Thanks for your help.

Answer 1

Yes, it is a little like division. Try this simple description from the NIST web-site:

http://www.nist.gov/dads/HTML/squareRoot.html

Answer 2

4. 3 5 8
Example: {19.00000000000
4^2 = 16
-----
3 00
83 x 3 = 2 49
------
5100
865 x 5 = 4325
------
77500
8708 x 8 = 69664

Step 1:

If the number of digits to the left of the decimal point is odd, then
the first digit of the answer will be the largest number whose square
is less than the first digit of the original number. If the number of
digits to the left of the decimal point is even, then the first digit of
the answer will be the largest number whose square is less than the
first _two_ digits of the dividend. In this case, 19 is a 2-digit
number, so the first digit of the quotient is the largest number
whose square is less than 19, i.e. 4. Write that number above
either the first digit of the dividend (if odd) or the second digit of
the dividend (if even)

Step 2:

Subtract the square of the number in Step 1, then bring down two
more digits.

Step 3:

Double the current quotient (on top, ignoring any decimal point), and
put a blank space in front of it. Here 4 x 2 = 8. Put down 8_ x _ to
the left of the current remainder, in this case 300.

Step 4:

The next digit of the quotient will be the largest number that can
be put in both blanks so that the resulting multiplication problem
is less than or equal to the current remainder. In this case, the
number is 3, because 83 x 3 = 249, whereas 84 x 4 = 336 is too
high. Write this number above the second digit of the next two
numbers; in this case the 3 would go above the second 0. We now
have a quotient of 4.3.

Step 5:

As long as you want more digits, subtract the product from the
remainder (i.e. 300 - 249 = 51) and bring down the next two digits,
in this case 51 turns into 5100, which becomes the current
remainder. Now repeat steps 3 and 4.

If you ever get a remainder of zero, you've got a perfect square
on your hands.

http://mathforum.org/library/drmath/sets/mid_square_roots.html

Answer 3

If you mean doing it in your head ... iteration is the key. You try a number and square it and if it works out you keep it. If it's too small, make a bit bigger and so on. The trick is to get in the ball park by looking for nearby easy squares. If you are trying to get the square root of 65,721,456 you would first note that 8x8 is 64 and 1,000x1,000 is a million, so 8,000x8,000 is 64,000,000.

So your next try would be a bit bigger. A neat way of doing the next multiplication is to note that (x+a)*(x+a)=x^2 +2xa+a^2. Take x=8,000, so the thing becomes 64,000,000 + 16,000*a +a^2 is supposed to equal 65,721,456. So 16,000*a should be about 1.7 million. Well, take a=100. 16,000*100 = 1.6 million and 100^2= 10,000 so 8,100^2 is 65,610,000. Now it gets fiddlyer. You're trying to get another 110,000 by multiply 8,100 by 2 and a. You could use long division or guess a=7. 7*16,100 = 112,700 and 7^2 is 49, so 8,107^2 =65,722,749. Now it gets still fiddlyer. This is too big by 1,293, which is about .07 times 16,000 so anyway, you get to 8,107.07 and so on.

It is less fiddley and quicker if you are doing 3 or 4 digit numbers and the iteration is more exact if you do division.

Let's say you want the positive square root of a positive number M. Your first guess is x0.

x0^2=M0 M=(x0+a)^2 = M0 + 2*a*x0 + x0^2. Forget the last term as your approximation and you get M=M0+2*a*x0. Solving for a, a=(M-M0)/2*x0. To iterate, set x1=a, calculate M1 and a=(M-M1)/2*x1 and so on. Then x2=a, then x3=a. Then xn's will get quickly smaller unless you were silly in your choice of x0.

Answer 4

Do you mean using an iterative equation like:

X' = (N/X + X)/2

Where X is the square root (start with 1) and N is the number that we are trying to find the square root.

Answer 5

Ever used a calculator? A square root is the number you multiply twice to get the number you already have. Ex: The square root of 49 is 7 because 7 times 7 is 49.

Answer 6

you need to simplify sq. root 20 into 2 sq. root of five and sq. root of 8 into 2 sq. root of two for this reason, you need to ingredient out 2, which leaves 2(sq. root of five + sq. root of two)^2 via simplifying extra, 2(5+2square root of five * sq. root of two + 2) 2(7+ 2 sq. root of 10) for this reason the suited answer may well be: 14+ 4square root of 10

Answer 7

Here's a great page that explains a few different ways of finding roots by hand:

http://mathforum.org/dr.math/faq/faq.sqrt.by.hand.html

Answer 8

I can't explain it but, I can give an example The squre root of 144 is 12. 12 times 12 is 144

Answer 9

2x2 = 4
3x3 = 9
4x4 = 16
5x5 = 25
and so on.

you have to use the calculator for some number. like 20, 15 .....

Answer 10

kk
any multiplication # that can founnd by multiplying the same # twice is a square. these are some.
1x1=1
2x2=4
3x3=9
4x4=16
5x5=25
6x6=36
and so on. u just need to really think of wat ot could be close to and plug it in.