Methods of computing square roots

The square root of a number $n$ is whatever number equals $n$ when multiplied by itself (squared). There are many ways to calculate square roots, from estimations to precise calculations.

The mathematical operation of finding a root is the opposite operation of exponentiation, and therefore involves a similar but reverse thought process.

Firstly, one needs to know how precise the result is expected to be. This is because square roots are usually irrational. For example, the square root of 28 is a number with infinite digits, and therefore it is impossible to express it exactly:$${\sqrt {28}}\approx 5.291502622129181....$$

Moreover, for some real numbers the square root is a complex number. For example, square root of -4 is a complex number 2i :$${\sqrt {-4}}=2i$$

In many cases there may be multiple valid answers. For example, square root of 4 is 2, but -2 is also a valid answer. One can verify that they are both valid answers by squaring each candidate answer and checking if you obtain 4 as the result of verification:$$2^{2}=2\times 2=4$$

$$(-2)^{2}=(-2)\times (-2)=4$$

Please note that calculating a square root is a special case of the problem of calculating N^th root.

Calculating

Most calculators provide a function for calculation of a square root.

	General Steps	Example
How to calculate a square root using a simple calculator.	First, make sure the operating space is clear. This is usually accomplished by clicking the C button a couple of times. Then type the number whose root you are trying to calculate. Then press the square root button (${\sqrt {}}$). The number you see on the screen is one of the answers. Remember, that often there are multiple valid answers, as explained above.	Press C button a couple of times. Type 16 Press ${\sqrt {}}$ button. The answer is 4. Keep in mind that -4 is also a valid answer.

Estimating

If the result does not have to be very precise, the following estimation techniques could be helpful:

Methodology	Example
Suppose you need to estimate the square root of some number $N$. Find some whole number $A$ such that $A^{2}$ is less than $N$, while $(A+1)^{2}$ is greater than $N$. Now we know that the square root of $N$ is somewhere between $A$ and $A+1$, so we can use $A+0.5$ as an approximate square root of $N$.	Suppose we need to estimate the square root of 2. $1^{2}=1$, which is less than 2, and $2^{2}=4$, which is more than 2. We now know ${\sqrt {2}}$ is somewhere between 1 and 2. Therefore, ${\sqrt {2}}$ is approximately 1.5.

You can get more accurate results by doing this technique again on the result. For example:

• $1.5^{2}$ is more than 2, so ${\sqrt {2}}$ is somewhere between 1 and 1.5. The halfway point of 1 and 1.5 is 1.25, so ${\sqrt {2}}$ is approximately 1.25.
• $1.25^{2}$ is less than 2, so ${\sqrt {2}}$ is somewhere between 1.25 and 1.5. The halfway point of 1.25 and 1.5 is 1.375, so ${\sqrt {2}}$ is approximately 1.375.
• $1.375^{2}$ is less than 2, so ${\sqrt {2}}$ is somewhere between 1.375 and 1.5. The halfway point of 1.375 and 1.5 is 1.4375, so ${\sqrt {2}}$ is approximately 1.4375.
• $1.4375^{2}$ is less than 2, so ${\sqrt {2}}$ is somewhere between 1.375 and 1.4375. The halfway point of 1.375 and 1.4375 is 1.40625, so ${\sqrt {2}}$ is approximately 1.40625.

With each step, the range of possible values that ${\sqrt {2}}$ can be is cut in half.

Babylonian method

The Babylonian method is a much faster way to estimate the square root of a number.

Methodology	Example
Suppose you need to estimate the square root of some number $N$. Start by guessing the result, as accurately as possible. The better this guess is, the faster the method is. We'll call this guess $X$. Then, calculate $(X+N/X)/2$. This will be a more accurate guess. You can set this as the new value of $X$ and repeat.	Suppose we need to estimate the square root of 2. We'll start with a guess of 1.5. The new guess will be $(1.5+2/1.5)/2$ = 1.4166666... Let's repeat for an even better guess. The new guess will be $(1.4166666+2/1.4166666)/2$ = 1.41421569...

References

• "Square Root".