# How do you find the square root of 4783?

##### 2 Answers

#### Explanation:

As

Here first pair is

Now as we still have a remainder of

Now recall we had brought as divisor

We continue to do this till we have desired accuracy.

Hence

#### Explanation:

Here's a method using generalised continued fractions.

First the theory...

Suppose we can find numbers

#sqrt(n) = a+b/(2a+b/(2a+b/(2a+b/(2a+...))))#

Then we have:

#a+b/(a+sqrt(n)) = a+b/(a+color(blue)(a+b/(2a+b/(2a+b/(2a+b/(2a+...)))))) = sqrt(n)#

Multiplying both ends by

#a^2+color(red)(cancel(color(black)(asqrt(n))))+b = color(red)(cancel(color(black)(asqrt(n))))+n#

Hence:

#b = n-a^2#

In our example, first note that

#69^2 = (70-1)^2 = 70^2-2*70+1 = 4900-140+1 = 4761#

So putting

#b = n-a^2 = 4783-4761 = 22#

So:

#sqrt(4783) = 69+22/(138+22/(138+22/(138+...)))#

We can truncate after any number of terms to get a rational approximation for

#sqrt(4783) ~~ 69#

#sqrt(4783) ~~ 69+22/138 = 4772/69 ~~ 69.1594#

#sqrt(4783) ~~ 69+22/(138+22/138) = 69+22/(9533/69) = 659295/9533 ~~ 69.1592363#

A calculator tells me:

#sqrt(4873) ~~ 69.1592365487#