Adding and Subtracting Radical Expressions:

To start we can remind ourselves how to simplify and work with like terms, the strategies for simplifying polynomials can be used to simplify sums and differences of radicals. For simplifying polynomials an example could be (3x + 5x =), since the variables are the same (x) then we can add the coefficients which equals to (8x). For radicals, what must be the same in order to simplify each other is the radicand and index, such as, (4√2 + 3√2) can be simplified to (7√2). Another example could be (5√2 + 3√18 – 4√8) Although it may seem like the radicands are different, you can simplify the radical (3√18) into (9√2) and (4√8) into (8√2) then we can simplify that into (5√2 + 9√2 – 8√2 = 6√2). Also, it is important to notice that the indexes are the same, which is 2 in this case, if not, you cannot simplify the radicals such as (3√3 + 4∛3) because even though they have the same radicand the indexes are different at 2 and 3. That being said, once we have like terms (radicand and index) we can simplify them by combining the co-efficient, and ultimately rewriting the equation as the most simplified result as it can be. To add, knowing the prime factors of numbers and how to simplify radicals are really important as it is basically the foundation of doing these types of questions like if you don’t know how to find the prime factors of a number you cannot fully simplify the radical making it really difficult to complete a question.

We can try a question that is a little harder but still builds with the knowledge and rules we know.

Simplify (√20x⁵y⁴ – √125xy²)  First we know the indexes are the same so that’s good but the radicands are not, so the next step is to simplify 20 and 125 which makes the expression into (2√5x⁵y⁴ – 5√5xy²), next are the variables, for (2√5x⁵y⁴) 2 groups of 2 of x⁵ can move to the co-efficient because the index is 2 which leaves 1 x remaining in the radicand, for y⁴ 2 groups of 2 can move with the co-efficient which is all the y’s because 2 x 2 = 4 and it is y⁴ (y⋅y⋅y⋅y). We can apply the same with (5√5xy²), in this case, only 1 group of 2 of y can move next to the co-efficient because it is y², the reason for being groups of 2 is due to the index (√ no number means 2) if it was ∛ then it would be groups of 3. After simplifying the radicand’s it brings it to (2x²y²√5x – 5y√5x) As you can see the radicands are the same so we can simplify to ( (2x²y² – 5y)√5x) ) TIP: If you are not sure if you did it right, just expand the simplified radicals and see if you got the original expression.

One harder question is (5√8x³ + 4y√75y³ – 2√27y⁵ – 3x√50x + 4∛3y)

First, simplify the numbers in the radicand

Then the variables

Combine like terms

Simplified!

Along with all this, we can also identify the values of the variable after simplifying and there are some rules, one rule is how the radicand cannot be negative if the index is even, for example, 5√x + 3√x – 4√x = 4√x, since the index is 2 the radicand cannot be negative, meaning each radical is defined for x ⩾ 0, or else it would be called “undefined” or “non-real number”. However, if the index is odd like 5∛x then it can be defined as x ∈ R (any real number). If it is something like q∜p³q, then to be defined, p⩾0 q⩾0 or p⩽0 or q⩽0, because if both p and q are negative in the radicand then the outcome would be positive which is why a negative could be possible in the radicand even with an even index as long as it comes out as a positive. Another scenario when there could be negatives in the radicand with an even index is if a question were to be 2√2a², this would be (a ∈ R) because of the exponent of 2, for example, if a = -2, it would become 2√2 (-2)(-2), which equals to positive 4 (2√2(4)) -> 2√8 -> 4√2 which would work and can be defined as (a ∈ R) because both a negative or positive can replace (a). But it would not work if the exponent in the radicand was odd because the end result of the radicand would be odd and that is not allowed. Overall, the end result of the radicand before must be positive (greater or equal to 0) if the index is even for it to be defined, and if the index is odd then the variable is equal to all real numbers.