1. If two or more radicals are multiplied with the same index, you can take the radical once and multiply the numbers inside the radicals.

na x n√b = n(a x b)

2. If two radicals are in division with the same index, you can take the radical once and divide the numbers inside the radicals.

na/n√b = n(a/b)

3. One number can be taken out of a square root for every two same numbers multiplied inside the square root. And also, one number can be taken out of a cube root for every three same numbers multiplied inside the cube root and so on.

√4 = (2 x 2) = 2

3√8 = 3(2 x 2 x 2) = 2

4. A radical with index n can be written as exponent 1/n.

n√a = a1/n

√a = a1/2

3√a = a1/3

For example, 9√3 and 4√3 can be added or subtracted. Because the numbers inside the square roots are same.

9√3 + 4√3 = 13√3

9√3 - 4√3 = 5√3

6. If a radical with index n is moved from one side of the equation to the other side, it will become the exponent n.

n√x = a

x = an

7. If an exponent n is moved from one side of the equation to the other side, it will become a radical with index n.

yn = b

y = n√b

8. If the digit in one's place of a number is 2, 3, 7 or 8, then the number can not be a perfect square. So the square root of such numbers will be irrational.

For example, √23  =  4.795831.........

9. If a number ends with odd number of zeros, then, the square root of the number will be irrational.

For example, √3000  =  54.772255.......

10. The square root of a perfect square is always a rational number.

√4 = √(2 x 2) = 2

√25 = √(5 x 5) = 5

11. The square root of an even perfect square number is always even and the square root of an odd perfect square number is always is odd.

For example,

√64 = 8

√81 = 9

12. Square root of a negative number is considered to be an imaginary value.

For example, √(-2), √(-9).

## Solved Problems

Problem 1 :

Simplify :

√6  √15

Solution :

= √6  √15

= √(6  15)

= √(2  3  3  5)

= 3√(2  5)

= 3√10

Problem 2 :

Simplify :

√35 ÷ √7

Solution :

= √35 ÷ √7

= √(35/7)

5

Problem 3 :

Simplify :

3√425 + 4√68

Solution :

Decompose 425 and 68 into prime factors using synthetic division. √425 = √(5 ⋅ 5 ⋅ 17)√425 = 5√17 √68 = √(2 ⋅ 2 ⋅ 17)√68 = 2√17

3√425 + 4√68 :

= 3(5√17) + 4(2√17)

= 15√17 + 8√17

= 23√17

Problem 4 :

Simplify :

√243 - 5√12 + √27

Solution :

Decompose 243, 12 and 27 into prime factors using synthetic division. √243 = √(3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3) = 9√3

√12 = √(2 ⋅ 2 ⋅ 3) = 2√3

√27 = √(3 ⋅ 3 ⋅ 3) = 3√3

√243 - 5√12 + √27 :

= 9√3 - 5(2√3) + 3√3

= 9√3 - 10√3 + 3√3

= 2√3

Problem 5 :

Simplify :

√4 + 3√27 + 4√64

Solution :

√4 = √(2 ⋅ 2) = 2

3√27 = 3(3 ⋅ 3 ⋅ 3) = 3

4√625 = 4√(5 ⋅ 5 ⋅ 5 ⋅ 5) = 5

√4 + 3√27 + 4√64 :

= 2 + 3 + 5

= 10

Problem 6 :

Simplify :

3√4  3√16

Solution :

3√4  3√16

3√(4  16)

3√(4  4  4)

= 4

Problem 7 :

If 3a = 1/2, then find the value of a.

Solution :

3a = 1/2

a = (1/2)3

a = 13/23

a = 1/8

Problem 8 :

If (3√8)7 ⋅ (√2)-4 = 2k, then solve for k.

Solution :

(3√8)7 ⋅ (√2)-4 = 2k

27 ⋅ (21/2)-4 = 2k

27 ⋅ 2-2 = 2k

27 - 2 = 2k

25 = 2k

k = 5 Apart from the stuff given aboveif you need any other stuff in math, please use our google custom search here. Kindly mail your feedback to v4formath@gmail.com

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