How do you get the square root of 7?

Answer 1

If you have a non-scientific calculator you can use the Newton-Raphson method.

Let f(x) = x2 - 7, and f'(x) = 2x

[f'(x) is the derivative of f(x) but you do not need to know that to use the N-R method.]

Make a guess at the square root of 7, and call is x0.

Then calculate

xn+1 = xn - f(x0)/f'(xn)

Provided you made a reasonable choice for the starting point, the iteration will very quickly converge to the true answer. Even if it is not so good:

Suppose you start with x0 = 5 (a pretty poor choice since 52 is 25, nowhere near 7).

Even so, x3 = 2.6457670442, which is less than 0.0006% from the true value.

Without any sort of calculator you are in for a lot of long multiplication.

First find an integers, i, such that i2 ≤ 7 < (i+1)2

Next, find integers j and j+1 between 10i and 10*(i+1) such that

j2 ≤ 700 < (j+1)2 note that there are two zeros added to the 7 at each step.

Then find integers k and k+1 between 10j and 10*(j+1) such that

k2 ≤ 70000 < (k+1)2.

Keep going until you reach the degree of precision required.

Here 22 = 4 < 7 < 9 = 32 so i = 2 which means sqrt(7) = 2.something

Now you want numbers between 20 and 30 whose squares bracket 700.

252 = 625 (small), 262 = 676 (still small), 272 = 729.

So 262 = 676 < 700 < 729 = 272 which means sqrt(7) = 2.6something.

The next step

2652 = 70225 (large), 2642 = 69696 (small). Therefore k = 264 and you now know that 2.64 < sqrt(7) < 2.65 and that sqrt(7) = 2.64something.

There is another method that resembles long division but this site is not really suited for descibing it and I d not know what it is called so cannot offer a link.