Examples of perfect square roots of decimal numbers

Before exposing some exercises that can serve as an example to the correct way in which every operation of perfect square root that has as a decimal number must be solved, perhaps the best thing is to revise briefly the own definition of this operation, in order to be able to understand each one of the posed exercises, within its just mathematical context.

Perfect square root of a decimal number

In this sense, we will begin by saying that Mathematics has defined the operation of perfect square root of a decimal number as the procedure by means of which a decimal number, which by nature has a perfect square, that is to say, is the product of a perfect square, acts as a root in a square root, in order to find the number that elevates it to that square, hence some authors have pointed out that this operation could also be understood as the inverse of an operation of potentiation of decimal base.

Steps to determine the perfect square root of a decimal number

However, despite the fact that it is a square root, and that it is essentially a simple operation, due to the nature of its radicando, i.e., the decimal number condition of its radicando, a series of steps must be followed in order to ensure that the correct result is achieved, whenever an operation of this type is involved. Next, each one of them:

1. Firstly, once the settlement operation has been proposed, each one ofits elements must be reviewed in order to understand the procedure to be followed.
2. Found that it is a radication of index 2, that is to say, of square root, and of radicando decimal, the first thing that must be done then is to suppress the comma of this number, to be able to handle it, while the operation lasts, as an integer.
3. The square root of the integer obtained after the comma was deleted is then calculated.
4. Once you have the square root of the integer, you must locate the comma, which will always be as many spaces as half of the decimal elements that the original decimal radicand had.
5. Once the comma is located, the final result of the operation is expressed.
6. If one wanted to verify the result of the operation carried out, it would then be enough to take the number obtained and submit it to a procedure of potentiation,where it is the base, and on the other hand the element that served as the radicando is the exponent. In other words, to square the decimal root that has been obtained. The result of this potentiation must give the number that serves as radicando.

Examples of how to solve perfect square roots of decimal numbers

However, the best way to complete an explanation about this type of operation, in which we seek to determine the perfect square root of a decimal number, may be the exposition of some concrete examples,in which we can see in a practical way how each of the steps, indicated by Mathematics, are applied in the operation. Next, each one of them:

Example 1

Solve the following square root operation: √0,36 =

The first thing to do then is to delete the comma from the decimal radicand. In this case, the decimal has a zero in its entire part, fact that in reality does not represent any additional procedure, because when removing the comma is like a zero to the left, therefore without any value, so it can also be translated:

√0,36 → √36

Once this number has been found, the square root obtained will be resolved:

√36 = 6

The answer is 6, since if I squared this number it will result in 36. Now, it is time to put the comma in the result. Therefore, the number of places to be counted from right to left in the root obtained will be half of this number,in this case it is equal to 1:

6 → 0,6

The comma is placed, even if some places must be completed with zero, either in incomplete units, as in whole parts. Once this is done, all that remains is to express the operation as resolved:

√0,36 = 0,6

Example 2

Solve the following square root operation: √2,25 =

In this case, the comma of the number that has served as the radicando will also be deleted,in order to be able to continue the radication operation as if it were an integer:

√2,25 → √225

It is then resolved the operation of radication, trying to find out what is the number that when elevated to its square gives as result 225:

√225 = 15

Once the answer has been found, it is time to place the comma in the root obtained. For this, we will count how many decimals the radicando decimal originally had. The mitas of this value will be the number of spaces that will be counted from right to left in the root before placing the comma, even if it must be completed with zeros, which is not necessary in this case, where the original radicando had only two decimals, and therefore in the root will be counted only one space before placing the comma:

15 → 1,5

Once this is done, the final answer of the operation is expressed:

√2,25 = 1,5

Example 3

Resolve the next operation: √0,0064 =

In the same way, one should begin, as in any operation of perfect square root of a decimal number, taking the radicand of the root, and suppressing its comma, to be able to handle it as an integer. In this case, in addition to the numbers other than zero, there are three zeros to the left, which will also disappear,because once the comma is suppressed -precisely because it is to the left- they lose all value:

√0,0064 → √64

Once this is done, the square root is resolved:

√64 = 8

Once this answer is obtained, it will be time to locate the comma, which will be as many spaces as half of the value represented by the totality of decimals that the original radicando has had. In this case, the decimal radicand had four decimals, so then the comma should be arranged to two spaces from the right in the root obtained, so these places should be completed with zeros:

8 → 0,08

Once this is done, it is enough to express the operation as resolved:

√0,0064 = 0,08

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