# Math

Consider an example of adding two radical expressions: `sqrt(3)` and `sqrt(12)` .

If we attempt to simplify `sqrt(3) + sqrt(12)` , as in any calculation of a set of arithmetic operations, order of operation convention must be applied. Recall that radical is considered a grouping symbol: it must be evaluated first, before any other operation. Which means we cannot add `sqrt(3) ` and `sqrt(12)` easily and get an exact answer, because both of these radicals are non-terminating, non-repeating decimals.

However, since there is a property of radicals called "product rule",  which states that the product of radicals is a radical of a product, `sqrt(12)`  can be simplified:

`sqrt(12) = sqrt(4*3) = sqrt(4) * sqrt(2) = 2sqrt(3)`

Now we have to add `sqrt(3) ` and `2sqrt(3)` . This is possible because of the distributive property:

`1*sqrt(3) + 2*sqrt(3) = (1+2)sqrt(3) = 3sqrt(3)`

We apply the same property when we add polynomials and combine like terms. For example, `2x + 3x = (2 + 3)x = 5x` .

In this way addition of radical and polynomial expressions is similar. In fact, radicals such as `sqrt(3) and 2sqrt(3)` , and any other radical that can be simplified to become a multiple of `sqrt(3)` , are called like radicals.

Radicals that are not like cannot be combined, for example, one cannot simplify `sqrt(2) + 3sqrt(5)` , much like one cannot simplify `x + 3x^2` .

`3sqrt(45) - 2sqrt(20)`

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Radical expressions must be simplified before adding them together in order to ease the collection of like terms and ensure that the denominator is the same for all radical expressions that you wish to add together.

The process is similar to adding polynomial expressions because that is exactly what you are doing with the numerator: adding polynomial expressions.  The only thing that sets apart radical expression addition and polynomial expression addition is the existence of a denominator that you must contend with.

To illustrate with an example, I will add these two rational expressions together:

`f(x)=((x+2)(x-3))/(x+1) ` and `g(x)=(x^2+x+2)/(x-3)`

`p(x)=f(x)+g(x)`

First, simplify f(x):

`f(x)=(x^2-x-6)/(x+1)`

Next in order to make the denominators the same we must multiply the denominator of f(x) to both the numerator and denominator of g(x) and vice versa:

`f(x)=((x-3)/(x-3))((x^2-x-6)/(x+1))=(x^3-x^2-6x-3x^2+3x+18)/((x-3)(x+1))=(x^3-4x^2-3x+18)/((x-3)(x+1))`

`g(x)=((x+1)/(x+1))((x^2+x+2)/(x-3))=(x^3+x^2+2x+x^2+x+2)/((x-3)(x=1))=(x^3+2x^2+3x+2)/((x-3)(x+1))`

Now we can add the expressions together:

`p(x)=f(x)+g(x)`

`= (x^3-4x^2-3x+18)/((x-3)(x+1))+(x^3+2x^2+3x+2)/((x-3)(x+1))`

`=(2x^3-2x^2+20)/((x-3)(x+1))`