PRACTICE PROBLEMS ON SOLVING SQUARE ROOTS

Problem 1 :

Solution :

Problem 2 :

The square root of 

0.09 + 2 x 0.21 + 0.49

is

Solution :

= square root of 0.09 + 2 x 0.21 + 0.49

Problem 3 :

Find the value of 

Solution :

Problem 4 :

Find the value of 

Solution :

Problem 5 :

Find the value of √45796, hence find the quotient of 

Solution :

√45796 = √214 x 214

= 214

√4.5796 = √(45796/10000)

= √(214 x 214)/(100 x 100)

= 214/100

= 2.14 ----(1)

√457.96 = √(45796/100)

= √(214 x 214)/(10 x 10)

= 214/10

= 21.4 ----(2)

Problem 6 :

√(0.04 x 0.4 x a) = 0.004 x 0.4 x √b

then find the value of a/b

Solution :

√(0.04 x 0.4 x a) = 0.004 x 0.4 x √b

√0.04 x √0.4 x √a = 0.004 x 0.4 x √b

Problem 7 :

If 3√5 + √125 = 17.88, then find the value of √80+6√5

Solution :

3√5 + √125 = 17.88

3√5 + √(5 x 5 x 5) = 17.88

3√5 + 5√5 = 17.88

8√5 = 17.88

√5 = 17.88/8

√5 = 2.235

√80 + 6√5 = √(5 x 2 x 2 x 2 x 2) + √(5 x 5 x 5)

= 2 x 2 √5 + 5√5

= 4√5 + 5√5

= 9√5

= 9(2.235)

= 20.115

Problem 8 :

If 3a = 4b = 6c and a + b + c = 27√29 then find the value of

Solution :

3a = 4b = 6c

3a = 6c then a = 6c/3 ==> 2c

4b=6c then b = 6c/4 ==> 3c/2

a + b + c = 27√29

2c+(3c/2)+c = 27√29

9c/2 = 27√29

c = (27√29) x (2/9)

c = 6√29

Squaring on both sides,

(a + b + c)² = (27√29)²

a² + b² + c² + 2ab + 2bc + 2ca = (27√29)²

a² + b² + c² = 729(29) - 2(ab + bc + ca)

= 21141 - 2(2c (3c/2) + (3c/2)c + c(2c))