About Square Root

Square root of a number it can be represented by `sqrt(y)` . The square root of number is also written as exponent form `sqrt(y)` which is equal to y^1/2
                Square root of a number 25 which is equal to `sqrt(25)` =25^1/2=5.
                In this article we shall discuss about the some square roots of the numbers.

Square root table

                The following table shows the square root of the some numbers.
                Number (x)        square (x²)          Square root (x^1/2)
                1                              1                              1.000
                2                              4                              1.414
                3                              9                              1.732
                4                              16                           2.000
                5                              25                           2.236
                6                              36                           2.449
                7                              49                           2.646 
                8                              64                           2.828 
                9                              81                           3.000 
                10                           100                           3.162

Problem 1:

Find the Square root value of 150
Solution:
                Write prime factors for 150 = 2 * 3 * 5 * 5
                `sqrt(150)` = `sqrt(2 * 3 * 5 * 5)`
                                   = `sqrt(2 * 3 * 5^2)`
             `sqrt(2 * 3 * 5^2) ` we can written as (2 * 3 * 5²)^1/2
                (2 * 3 * 5²)^1/2      By using algebraic property (a*b)^m =a^m * b^m
                        (2 * 3)^1/2 * (5^2*1/2)
                5 (2 * 3)^1/2
                        Square root value of 150= 5 `sqrt(6)`

Problem 2:
Find the Square root value of 300
Solution:
                Write prime factors for 300 = 2 * 2 * 3 * 5 * 5
                `sqrt(300)` = `sqrt(2 * 2 * 3 * 5 * 5)`            
                                   = sqrt(2^2 * 3 * 5^2)
                `sqrt(2^2 * 3 * 5^2)` we can written as (2 * 3 * 5²)^1/2
                (2² * 3 * 5²)^1/2     By using algebraic property (a*b)^m =a^m * b^m
                        (3)^1/2 * (2^2*1/2 * 5^2*1/2)
                5 * 2 (3)^1/2
                        Square root value of 300=10 `sqrt(3)`

Problem 3:
Find the Square root value of 625
Solution:
                Write prime factors for 625 = 5 * 5 * 5 * 5
                `sqrt(625) ` = `sqrt(5 * 5 * 5 * 5)`   
                                   = `sqrt(5^2 * 5^2)`
                `sqrt(5^2 * 5^2)` we can written as (5² * 5²)^1/2          
                (5² * 5²)^1/2            By using algebraic property (a*b)^m =a^m * b^m
                        (5²)^1/2 * (5^2*1/2)
                5 * 5
                        Square root value of 625= 25

Problem 4:
Find the Square root value of 75
Solution:
                Write prime factors for 75 = 3 * 5 * 5
                `sqrt(75)` = `sqrt(3*5*5)`
                                   = `sqrt(3*5^2)`
                `sqrt(3*5^2)` we can written as (3*5²)^1/2 
                (3*5²)^1/2               By using algebraic property (a*b)^m =a^m * b^m
                        (3)^1/2 * (5^2*1/2)
                5 (3)^1/2
                        Square root value of 75= 5 `sqrt(3)`

Divide Complex Fractions

Complex fractions are having both the numerator and the denominator values. The complex fractions are also consists of number of rational fractions. The other names for the complex fractions are compound fractions. For example, `(1/3)/(2/3)` is called as the division complex fractions.  Complex fractions are also used for many of the arithmetic operations.

Example problem for divide complex fractions

The example problem for divide complex fractions is,

Problem 1: Divide the following complex fractions, `(2/3)/(3/8)` .

Solution:

Step 1: In the divide complex fractions, the first step is to we have to convert the given fraction into the following,

`(2/3)/(3/8)` = `(2)/(3)` `-:``(3)/(8)`

Step 2: In this step we have to multiply the fraction, so that we have to take reciprocals to the second fractions,

= `(2)/(3)` `xx` `(3)/(8)`

=`(2)/(8)`

=`(1)/(4)`

This is the required solution to the divide complex fractions.

Problem 2: Divide the improper fraction into the complex fractions, `(5 (3/4))/(2/4)` .

Solution:

Step 1: In the first step, we are convert the following into the proper complex fraction, we get,

`(23/4)/(2/4)` = `(23)/(4)` `-:` `(2)/(4)`

Step 2: In the next step, we are multiplying the complex fractions,

`(23)/(4)` `xx` `(4)/(2)`

=`(23)/(2)`

This is the required complex fractions.

Problem 3: Divide the improper fraction into the complex fractions, `(6 (2/8))/(5/8)` .

Solution:

Step 1: In the first step, we are convert the following into the proper complex fraction, we get,

`(50/8)/(5/8)` = `(50)/(8)` `-:` `(5)/(8)`

Step 2: In the next step, we are multiplying the complex fractions,

`(50)/(8)` `xx` `(8)/(5)`

=`(50)/(5)`

`This is the required solution for the complex fractions.`

Problem 4: Divide the following complex fractions, `(2/8)/(4/6)` .

Solution:

Step 1: In the divide complex fractions, the first step is to we have to convert the given fraction into the following,

`(2/8)/(4/6)` = `(2)/(8)` `-:``(4)/(6)`

Step 2: In this step we have to multiply the fraction, so that we have to take reciprocals to the second fractions,

= `(2)/(8)` `xx` `(6)/(4)`

=`(3)/(8)`

`(1)/(4)`

This is the required solution to the divide complex fractions.

Practice problem to divide complex fractions

Problem 1: Divide the following complex fractions, `(2/3)/(4/6)` .

Answer: `(2)/(3)`

Problem 2: Divide the improper fraction into the complex fractions, `(4 (2/4))/(1/4)` .

Answer: 32