How can you find the square root of a 3-4 digit number
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How can you find the square root of a 3-4 digit number

[From: ] [author: ] [Date: 11-10-24] [Hit: ]
2^2x7^2.This equals 169.But we want the square root of it.So we take 2^2x7^2 and divide the power in half.2^1x7^1.So 196 squared=14.......
Without using a calculator...

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Write down a factor tree for example:

196
/\
2 98
/\
2 49
/\
7 7
Now that you have your factor tree, you take the numbers that only 1 and itself can go into. In this case its one and seven. So we now have the equation:
2^2x7^2. (note: 2^2 means two squared and 7^2 means seven squared)
This equals 169.
But we want the square root of it.
So we take 2^2x7^2 and divide the power in half.
So it would be:
2^1x7^1. (note:2^1 means two to the power of one and 7^1 means seven to the power of one)
So you now multiply:
2^1x7^1= 14
So 196 squared=14.
You can check your work by multiplying 14x14.
14x14=196

WARNING
this method does not work if one of the powers is uneven. For example:
2^2x7^3
Because you can't divide 3 in half!

Hope this helped!^_^

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practice.
a trick I've picked up is realizing the squares of numbers divisible by 10. since it's just 100 times what they would be without the zero in front of it ie. 4^2=16 and 40^2=1600.
From there you have to know the difference of squares. ie 41^2= 40^2 + 40 + 41. It's easy if you're familiar with this rule. if I were to calculate 53^2 I would do 2500+ 101+ 103+ 105= 2809. Or 57^2 ----> 3600- 119 -117 -115= 3249.
if you can pick up the average of the numbers you can just multiply that by the quantity of numbers and add that or subtract from the multiple of 10.
Soon you pick up even niftier tricks ie:
55^2 = 60^2 - 115*5
55^2= 50^2 + 105*5
basically every 7.5 or 2.5 you just multiply by ten (5*2) and add to the square of multiples of 10.

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http://www.google.com/#q=square+root+cal…

http://www.homeschoolmath.net/teaching/s… (skip the guess and check part)

http://xlinux.nist.gov/dads/HTML/squareR…

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thats easy, just ask an asian :)
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