PROPERTIES OF SQUARE ROOTS

1. If two or more numbers multiplied with square roots, you can take the square root once and multiply the numbers inside the square root. 

√m x √n = √(m x n)

2. If two numbers are in division with square roots, you can take the square root once and divide the numbers inside the square root. 

√m/√n = √(m/n)

3. One number can be taken out of the square root for every two same numbers multiplied inside the square root. 

√(m  x m) = m

√4 = (2 x 2) = 2

4. Square root of a number can be written as exponent 1/2. 

√m = m1/2

5. Addition and subtraction of two or more numbers with square roots can be performed with like radicands only. (Radicand is the number inside the square root)

For example, 9√2 and 4√2 can be added or subtracted. Because the numbers inside the square roots are same. 

9√2 + 4√2 = 13√2

9√2 - 4√2 = 5√2

6. If a square root is moved from side of the equation to the other side, it will become square. 

√x = a

x = a2

7. If a square is moved from side of the equation to the other side, it will become square root. 

y2 = b

y = √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 of a perfect square is always a rational number.

√4 = √(2 x 2) = 2

√49 = (7 x 7) = 7

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, 

√144  =  144

225  =  15

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

For example, √(-9), √(-12). 

Solved Problems

Problem 1 :

Simplify : 

√6  √3

Solution :

= √6  √3

= √(6  3)

= √(2  3  3)

= 3√2

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 : 

If a = 1/5, then find the value of a. 

Solution : 

a = 1/5

a = (1/5)2

a = 12/52

a = 1/25

Problem 6 :

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

Solution : 

 (√4)7 ⋅ (√2)-4 = 2k

27 ⋅ (21/2)-4 = 2k

27 ⋅ 2-2 = 2k

27 - 2 = 2k

25 = 2k

k = 5

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