How do I find the square root of a number which is not a perfect square?

How do I find the square root of a number which is not a perfect square?

'

----------

There are four ways, that I am aware of, to approximate the square root of an imperfect square.

Let's begin with the easiest one.

Method 1 :

Let x be the number whose square root we want to compute (x is referred to as the radicand).

First, we find the perfect square closest to x. Let this number be y.

The square root of x is then computed to be $$x−−√=y√+x−y2∗y√$\sqrt{x}=\sqrt{y}+\frac{x-y}{2\ast \sqrt{y}}$

Note : This method provides finite precision, often to only two decimal places. Using the closest perfect square is the key to finding acceptable solutions.

Computing a) 66−−√$66$ and b) 24−−√,$\sqrt{\mathbf{24}}\mathbf{,}$

a) 64 is the closest perfect square.

66−−√=$\sqrt{66}=$8+66−642∗8=8.125≈$8+\frac{66-64}{2\ast 8}=8.125\approx$8.124038$8.124038$

b) 25 is the closest perfect square.

24−−√=5+24−252∗5=4.9≈4.898979$\sqrt{24}=5+\frac{24-25}{2\ast 5}=4.9\approx 4.898979$

Method 2 :

Again, x is the radicand and y is our best guess such that y2$y^2$ is closest to x.

We proceed as follows -

1. Calculate y1$y_1$=xy$=\frac{x}{y}$
2. Find average of y$y$1$1$ and y. This becomes our new y, y$y$=y1+y2$=\frac{y_1+y}{2}$
3. Repeat steps 2 and 3 until you get your desired precision. Output y when done.

Note : Greater the number of times one iterates, greater the precision one achieves. Also, closer the initial approximated value, faster one approaches the true value of the square root of x.

Computing $$24−−√,$\sqrt{24},$

Let's start with y=5 as our initial guess, that is, 52=25∼24$5^2=25\sim 24$.

1. y1=245=4.8$y_1=\frac{24}{5}=4.8$
2. y=4.8+52=4.9$y=\frac{4.8+5}{2}=4.9$

Repeating steps 1 and 2,

1. y1=244.9=4.8979$y_1=\frac{24}{4.9}=4.8979$
2. y=4.8979+4.92=4.8989$y=\frac{4.8979+4.9}{2}=4.8989$

Although we could have continued, let's stop here. We get y=4.8989 which is a decent approximation of the actual square root of 24.

Method 3:

This method is what they call the Long Division Method for finding square roots.

Finding 23−−√$23$

Method 4:

x−−√=antilog(log(x)2)$\sqrt{\mathbf{x}}\mathbf{=}\mathbf{a}\mathbf{n}\mathbf{t}\mathbf{i}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{(}\frac{\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{(}\mathbf{x}\mathbf{)}}{\mathbf{2}}\mathbf{)}$

Note : It doesn't matter which logarithm function is used as long as the corresponding antilog function is used.

Computing 67−−√,$\sqrt{\mathbf{67}}\mathbf{,}$

To sum up, method 1 is the easiest of all the procedures listed. Method 2, while being tedious and iterative, is easy to use and can provide greater precision, reliably. Use method 3 if you can remember the steps involved accurately. Use method 4 only if you have a logarithm table and are familiar with using one.