I wanted to share this post from a long time but somehow this couldn't be possible, but now here it is,
√16 < √23 < √ 25
4 < 4.7or 4.8 < 5 ( this is an approximation)
but √23 = 4.79583
• No doubts this method is easy than long Division and also giving answer to satisfactory manner, But it requires practice to become perfect. Try at least for twice or thrice with different values.
So this is our today'as tutorial of finding square root of any number without our traditional method.
Have a look at some best tutorials From Math's Buddies:
Now over to you, let me know how you have found this tutorial helpful to you, by commenting.
How to find square roots without long Division method
There is one method i.e. approximation method but this wouldn't lead us towards exact solution at all time,
" Confuse...??"
don't worry see this example.
Q. Find √23
Sol. using approx. method
A perfect square before 23 is 16 and just next perfect square is 25.
so this can be written as
√16 < √23 < √ 25
4 < 4.7or 4.8 < 5 ( this is an approximation)
but √23 = 4.79583
• And now our short method to evaluate square root of √23...
Next highest square root is √25
so,
let 23 = A
25 = B
- our formula of finding square root of will be
• √A = √ B + (A - B)/ 2×√B
√23 = 5 + (23-25)/ 2× 5
= 5 - 2/ 10
= 5 - 0.2
= 4.8
• let's have an another example for √45
so,
45 = A
49 = B
putting values into Formula
√A = √ B + (A - B)/ 2×√B
√45 = 7 + ( 45-49) /2×7
= 7 - 4/14
= 7 - 0.28571
= 6.71429
now by evaluating from calculating
√45 = 6.7082 ≈ 6.71429
so this Method is giving us nearly equals to exact values.
• let's have an another example for √45
so,
45 = A
49 = B
putting values into Formula
√A = √ B + (A - B)/ 2×√B
√45 = 7 + ( 45-49) /2×7
= 7 - 4/14
= 7 - 0.28571
= 6.71429
now by evaluating from calculating
√45 = 6.7082 ≈ 6.71429
so this Method is giving us nearly equals to exact values.
So this is our today'as tutorial of finding square root of any number without our traditional method.
Have a look at some best tutorials From Math's Buddies:
Now over to you, let me know how you have found this tutorial helpful to you, by commenting.
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