Algebraic Operations - Radicals, 7

Category: Algebra, Arithmetic

"Published in Newark, California, USA"

Perform the indicated operations

Solution:

Consider the given equation above

If you will multiply a radical with another radical with the same index, then the terms inside the radicals will be multiplied together. In this case, the given above equation can be written as follows

Apply the distributive property of multiplication over addition, as follows

Therefore, the final answer is

Algebraic Operations - Radicals, 6

Category: Algebra

"Published in Newark, California, USA"

Perform the indicated operations

Solution:

Consider the given equation above

The first thing that we have to do is to examine the radicals first if they can simplify or not. As a rule in Mathematics, all radicals must be simplified as much as we can.

At the first term, 162 is not a perfect 4th root. The factors of 162 are 81 and 2. 81 is a perfect 4th root.

At the second term, 32 is a not a perfect 4th root. The factors of 32 are 16 and 2. 16 is a perfect 4th root. Since x⁶ is not a perfect 4th root, then we can factor x⁶ into x⁴ and x².

At the third term, the denominator contains a radical. We need to eliminate the radical sign at the denominator by rationalization of the denominator. Multiply both the numerator and denominator by x² so that the denominator becomes a perfect 4th root which is x⁴. 

Hence, the given equation above becomes

Take the 4th root of the numbers inside the radicals that are perfect 4th root, we have  

Since all the terms inside the radicals are the same, then we can combine them and therefore, the final answer is

 

Algebraic Operations - Radicals, 5

Category: Algebra

"Published in Newark, California, USA"

Perform the indicated operations

Solution:

Consider the given equation above

The first thing that we have to do is to examine the radicals first if they can simplify or not. As a rule in Mathematics, all radicals must be simplified as much as we can.

At the first term, 16 is not a perfect cube. The factors of 16 are 8 and 2. 8 is a perfect cube.

At the second term, the denominator contains a radical. We need to eliminate the radical sign at the denominator by rationalization of the denominator. Multiply both the numerator and denominator by x² so that the denominator becomes a perfect cube which is x³. 

At the third term, the denominator contains a radical also. We need to eliminate the radical sign at the denominator by rationalization of the denominator. Multiply both the numerator and denominator by 2 so that the denominator becomes a perfect cube which is 8. 

Hence, the given equation above becomes

Take the cube root of the numbers inside the radicals that are perfect cube, we have 

Since all the terms inside the radicals are the same, then we can combine them and therefore, the final answer is