FINDING SQUARE ROOTS AND CUBE ROOTS WORKSHEET

Problem 1 : 

Solve for x :

x²  =  121

Solution :

x²  =  121

Solve for x by taking square root on both sides.

x  =  ± √121

Apply the definition of square root.

Think: What numbers squared equal 121 ?

x  =  ± 11

So, the solutions are 11 and −11.

Problem 2 : 

Solve for x : 

x²  =  16/169

Solution :

x²  =  16/169

Solve for x by taking square root on both sides.

x  =  ± √(16/169)

Apply the definition of square root.

Think: What numbers squared equal 16/169 ?

x  =  ± 4/13

So, the solutions are 4/13 and −4/13.

Problem 3 : 

Solve for x : 

x³  =  729

Solution :

x³  =  729

Solve for x by taking cube root on both sides.

x  =  ³√729

Apply the definition of cube root.

Think: What number cubed equals 729 ?

x  =  9

So, the solution is 9.

Problem 4 : 

Solve for x :

x³  =  8/125

Solution :

x³  =  8/125 

Solve for x by taking cube root on both sides.

x  =   ³√(8/125)

Apply the definition of cube root.

Think: What number cubed equals 8/125 ?

x  =  2/5

So, the solution is 2/5.

Problem 5 : 

Find the square root of 0.16.

Solution :

√0.16 = √(16/100)

=√16/√100

= 4/10

= 0.4

Problem 6 : 

Find the cube root of 0.008.

Solution :

³√0.008 = ³√(8/1000)

=³√8/³√1000

= 2/10

= 0.2

Solve for x by taking cube root on both sides.

x  =   ³√(8/125)

Apply the definition of cube root.

Think: What number cubed equals 8/125 ?

x  =  2/5

So, the solution is 2/5.

Problem 7 : 

Find the cube root of -125.

Solution :

³√-125 = ³√(-5 x -5 x -5)

= -5

Problem 8 : 

Can you solve the equation x² = 27 and get a rational umber as solution ? Explain.

Solution :

No, we can not get a rational number as a solution. 

In the given equation we have "square" for "x".

So, we will be able to get a rational number as solution, only if we have a perfect square on the right side of the equation. 

Since we have 27 (not a perfect square) on the right side of the equation, we can not get a rational number as a solution. 

Problem 9 : 

Find the smallest number which when multiplied with 137592 will make the product a perfect cube. Further, find the cube root of the product.

Solution :

Decomposing 137592,

= 2 x 2 x 2 x 3 x 3 x 3 x 7 x 7 x 13

To group them as product of three same values, we need one 7 and two more 13's.

The numbers to be multiplied = 7 x 13 x 13

= 1183

So, the required smallest number to be multiplied is 1183.

Problem 10 : 

The volume of a cubical box is 13.824 cubic meters. Find the length of each side of the box.

Solution :

Volume of cubical box = 13.824 cubic meters

Let x be the side length of the cube.

x³ = 13.824

x = ³√13.824

= ³√13.824 x (1000/1000)

= ³√(13824/1000)

= ³√(3 x 3 x 3 x 8 x 8 x 8)/(10 x 10 x 10)

= (3 x 8) / 10

= 24/10

= 2.4

So, side length of the box is 2.4 meter.

Problem 11 : 

Multiply 26244 by the smallest number so that the product is a perfect cube. What is that number? Also find the cube root of the product.

Solution :

Decomposing 26244,

= 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3

= 2 x 2 x (3 x 3 x 3) x (3 x 3 x 3) x 3 x 3

To group as product of three same terms, there is one 3 and one 2 are insufficient. To make it as perfect cube one 3 and one 2. 

So, the smallest number to be multiplied to make it as perfect cube is 6.

Cube root after multiplying these values,

= ³√(2 x 2 x 2) x (3 x 3 x 3) x (3 x 3 x 3) x (3 x 3 x 3)

= 2 x 3 x 3 x 3

= 54

The answer is 54.

Problem 12 : 

Divide 88209 by the smallest number so that the quotient is a perfect cube. What is that number? Also find the cube root of the quotient.

Solution :

Decomposing 88209,

= 3 x 3 x 3 x 3 x 3 x 3 x 11 x 11

Grouping them as product of three same values,

= (3 x 3 x 3) x (3 x 3 x 3) x 11 x 11

There is two 11's extra. to make it as perfect cube these two 11's to be ignored. Then (11 x 11) that is 121 should be divided to make it as perfect cube.

After ignoring 121, making it as perfect cube

= 3 x 3

= 9

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