Algebraic Operations - Radicals, 3

Category: Algebra, Arithmetic

"Published in Newark, California, USA"

Perform the indicated operations

Solution:

Consider the given equation above

The first 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. Although the given radical equations are all numbers, then still, we have to follow the principles of Algebra which is "like combines like". 

At the first term, 54 is not a perfect cube. The factors of 54 are 27 and 2. 27 is a perfect cube.

At the second term, 250 is not a perfect cube. The factors of 250 are 125 and 2. 125 is a perfect cube.

At the third term, we need to eliminate the radical sign at the denominator by rationalization of the denominator. Multiply both the numerator and the 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

 

Algebraic Operations - Radicals, 2

Category: Algebra, Arithmetic

"Published in Newark, California, USA"

Perform the indicated operations

Solution:

Consider the given equation above

The first 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. Although the given radical equations are all numbers, then still, we have to follow the principles of Algebra which is "like combines like". The second and the third terms of the given equation above can be rewritten and factored as follows

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

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

Simplifying Radicals, 4

Category: Algebra

"Published in Newark, California, USA"

Simplify

Solution:

Consider the given equation above

This equation is considered a difficult one because the index of a radical is 6. We need to think a number that is raised to the sixth power will give an answer of 576. If 2 raised to the sixth power, the answer is 64. If 3 raised to the sixth power, the answer is 729. We cannot use 3 because 729 is greater than 576.

If we divide 576 by 2, the answer is 288. Again, if we divide 288 by 2, the answer is 144. The factors of 576 are 4 and 144. 4 and 144 are perfect squares. We can rewrite the above equation as follows

As you noticed that the exponent of 2 is a multiple of 6 while the exponent of x is not. We need to factor and rewrite x into a multiple of 6 as follows

Take the sixth root of the terms with exponents that are multiples of 6, we have

The remaining terms inside the radical have even exponents. Since the index of a radical is 6, we need to further simplify the radical as follows

Therefore, the final answer is