root value

How to caluclate root of 2 Note by Sai Venkata Raju Nanduri
6 years, 5 months ago

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We can use this method to calculate the root of 2 or any other non-square integer.

http://en.wikipedia.org/wiki/Methodsofcomputingsquareroots

After 5 steps computing using the CASIO fx-500MS calculator, I got $\sqrt{2} \approx 1.414213562$

- 6 years, 5 months ago

u are correct

- 6 years, 5 months ago

One method I learned was use the binomial theorem. (1+1)^(1/2)

- 6 years, 5 months ago

But, how can you calculate 2^(1/2)? I didn't know about that method. Can you tell me please?

- 6 years, 5 months ago

- 6 years, 5 months ago

$2^{\frac {1}{2}}$ is the same as $\sqrt{2}$, which is the same as $\sqrt 2$. Infact, $\sqrt[n]{x}$ is the same as $x^{\frac {1}{n}}$.

- 6 years, 5 months ago

Yes I knew it before I post. But are there any other methods to calculate a power of $\frac{1}{n}$ with n integer?

- 6 years, 5 months ago

Yes very good

- 6 years, 5 months ago

however if you know the basic of number theory and if you know the theorem of "continued infinite fractions" then you could find square root of any integer as a form of continued fraction tending to infinity....

- 6 years, 5 months ago

Le V is correct....

its 1.414

- 6 years, 5 months ago

take a calculator.the type like this (2)^1/2.then press "=".you will get the answer.

- 6 years, 5 months ago

by using taylor - macloren method

- 6 years, 5 months ago

around 1.414

- 6 years, 5 months ago

read NCERT of class 8 chapter 6

- 6 years, 4 months ago

read NCERT of class 8 chapter 6

- 6 years, 4 months ago

One of my favorite ways is to solve the diophantine equation x^2-2y^2=1 for positive integers x and y. The larger the integers, the better x/y approximates the square root of 2. You can find solutions using Brahmagupta's chakravala (cyclic) method, which is an elegant algorithm for finding solutions to Pell equations.

- 6 years, 4 months ago

you are 17...so well by ur standard, Let x be the nearest perefct square near it, i.e. 1... f(x) = x, i.e. 1; so f'(x) = nx^(n-1) putting n=1/2, f'(x) = (x^(-1/2))/2 = 1/(sqrt{1}) now f(x+delta(x)) = f(x) + f'(x)delta(x) = 1 + 1/21 = 1 + .5 = 1.5(approx) but to find the actual value, use the division method as u did in the earlier classes VII or VIII but for the larger no. such as 345 or 234, this method is quite awesome...

- 6 years, 5 months ago

Use scientific calculator.

- 6 years, 5 months ago