PROPERTIES OF SQUARE ROOTS

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. 

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 = 2^k, then solve for k. 

Solution : 

 (√4)⁷ ⋅ (√2)^-4 = 2^k

2⁷ ⋅ (2^1/2)^-4 = 2^k

2⁷ ⋅ 2^-2 = 2^k

2^{7 - 2} = 2^k

2⁵ = 2^k

k = 5

Problem 7 :

. Using properties of squares and square roots calculate:

100² − 98²

Solution :

= √(100² − 98²)

= √(100 + 98) (100 - 98)

= √198 (2)

= √(2 x 3 x 3 x 11 x 2)

= 2 x 3 √11

= 6√11

Problem 8 :

By what least number should 2028 be multiplied so that the product is a perfect square? Find the square root of the product so obtained.

Solution :

2028 is not a perfect square, by multiplying some numerical value by this we can change it as perfect square.

= √(2 x 2 x 3 x 13 x 13)

By grouping as pairs, we have 3 which doesn't have pair. Then 3 is the number to be multiplied to make it as perfect square.

Problem 9 :

By what least number should 3528 be divided so that the quotient is a perfect square? Find the square root of the quotient so obtained

Solution :

3528 is not a perfect square, by removing some numerical value from 3528 we can make as perfect square.

To find the number which should be removed, we have to decompose into product of prime factors.

= √(2 x 2 x 2 x 3 x 3 x 7 x 7)

Since 2 is the number which doesn't have pair, we have to remove 2. So, 2 is the number to be removed.

= √(2 x 2 x 3 x 3 x 7 x 7)

= 2 x 3 x 7

= 42

The square root of the quotient is 42.

Problem 10 :

Express 16 as sum of odd numbers.

Solution :

1 = 1

4 = 1 + 3

9 = 1 + 3 + 5

16 = 1 + 3 + 5 + 7

Problem 11 :

Find the least perfect square number which is divisible by each of the numbers 8, 12, 15 and 20.

Solution :

LCM of (8, 12, 15, 20)

LCM (8, 12, 15, 20) = 2 x 2 x 5 x 3 x 2

= 120

120 is divisible by 8, 12 and 20, but not divisible by 15.

120 x 2 ==> 240

To make this as perfect square,

2 x 2 x 2 x 2 x 5 x 5 x 3 x 3

= 3600

So, the required number is 3600.

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