# Definition
A root is an inverse of an exponent. For example:
* \\( 4^2 = 16; \sqrt{16} = 4 \\)
* \\( 5^2 = 25; \sqrt{25} = 5 \\)
* \\( 3^3 = 27; \sqrt[3]{27} = 3 \\)

# Simplifying Square Roots
You can simplify a root by factoring the root number with another number that
is rational. In other words, divide the root number by the perfect squares.  
After that, root the perfect square.

For example: **Simplify** \\( \sqrt{60} \\)!  
\\(  \sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15} \\)  
!---(
#--(
## 💡 Detailed explanation
)--#
1. We factorise 60. We choose the smallest factor with a perfect square (e.g. 4, 9, 16, 25).  
   We will get \\(4 \times 15 \\). To verify, \\( 4 \times 15 \\) must be equal to 60. And it is!
2. Then, root the perfect square, which in this case, is 4.  
   \\( \sqrt{4 \times 15} = 2\sqrt{15} \\)
)---!
More examples:
* \\( \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3} \\)
* \\( \sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5} \\)

!---(
#--(
## ❕ There are some "leftover" roots!
)--#
Some roots are **irrational numbers**. Irrational numbers are numbers that
cannot be represented by a ratio, aka a fraction.

We cannot simplify the roots that is irrational. So, it is **already in its
simplest form**.
)---!

# Root Operations
Firstly, some terminology!
1. Coefficient is the multiplier of a root. For example: \\( 3\sqrt{2} \\); 3 is the coefficient here, and 2 is the root.

## Addition
Root numbers with the same root can be summed together.

\\( a\sqrt{c} + b\sqrt{c} = (a + b)\sqrt{c} \\)

* \\( 3\sqrt{2} + 4\sqrt{2} = (3 + 4)\sqrt{2} = 7\sqrt2 \\)
* \\( \sqrt5 + 3\sqrt5 = (1 + 3)\sqrt5 = 4\sqrt5 \\)

## Subtraction
Root numbers with the same root can be subtracted together.

\\( a\sqrt{c} - b\sqrt{c} = (a - b)\sqrt{c} \\)

* \\( 7\sqrt3 - 4\sqrt3 = (7-4)\sqrt3 = 3\sqrt2 \\)

## Multiplication
Root numbers can be multiplied by any other root numbers.

\\( a\sqrt{c} \times b\sqrt{d} = (a \times b)\sqrt{c \times d} \\)

!---(
#--(
## 💡 Tip
)--#
If you multiply a root number with another root number that has the same root,
you can delete the root symbol, aka it becomes a regular number.  
For example:

\\( \sqrt5 \times \sqrt5 = 5 \\)

!---(
#--(
### ❓ Why?
)--#
Root is the inverse of exponent. If you multiply two numbers together with the
same value, it is an exponent. If two of them meet together, they will fight
and both of them dies (they will cancel out), which means it will be a regular
ol' number.

In this example:

\\( \sqrt5 \times \sqrt5 = \sqrt{5^2} = 5 \\)
* Here, \\( \sqrt{5} \\) was powered to 2, which cancels it out.
	* **Why do they cancel out?** We can expand this equation even more.  
	  \\( \sqrt5 \times \sqrt5 = \sqrt{5^2} = \sqrt{25} = 5 \\)
		* \\( \sqrt{25} \\) is equal to 5, so the answer is 5.

---

This also works for every other number.

---

### In conclusion of these tips:
* If a root is exponented, they will both cancel out. \\( \sqrt[x]{a^x} = a \\)
	* Because of that, \\( \sqrt{a} \times \sqrt{a} = a \\)

)---!
)---!

## Division
Roots can be divided with any other roots.

\\( a\sqrt{c} \div b\sqrt{d} = (a \div b)\sqrt{c \div d} \\)

* \\( 8\sqrt6 \div 4\sqrt3 = (8 \div 4)\sqrt{6 \div 3} = 2\sqrt2 \\)

# Rationalising Roots
If a fraction has an **irrational** root as the denominator, we should rationalise it.

!---(
#--(
## ❓ What? Why?
)--#
Usually in mathemathics, a fraction should have a rational number as the
denominator. If you remember, some roots are **irrational**, like \\( \sqrt2
\\), \\( \sqrt3 \\), \\( \sqrt 5 \\). They cannot be represented as integers or
a valid fraction.

You don't have to rationalise a root, but it is widely accepted that the
denominator is a rational number.
)---!

Let's learn how to rationalise a root.

## Simple Irrational Fraction
If you have a simple irrational fraction like this:

\\( \dfrac{1}{\sqrt2} \\)

You can rationalise it by multiplying it with its own denominator. If you
remember correctly, when you multiply a square root with itself, it will return
its own number, therefore a rational number!

\\( \dfrac{1}{\sqrt2} \cdot \dfrac{\sqrt2}{\sqrt2} = \sqrt{1\sqrt2}{2} \\)

The value is still (and must be!) the same, rationalising roots are basically
representing numbers in another way.

### More Examples
* \\( \dfrac{3}{\sqrt5} \cdot \dfrac{\sqrt5}{\sqrt5} = \dfrac{3\sqrt5}{5} \\)
* \\( \dfrac{2}{3\sqrt6} \cdot \dfrac{\sqrt6}{\sqrt6} = \dfrac{2\sqrt6}{3 \cdot 6} = \dfrac{2\sqrt6}{18} \\) (you don't need to multiply it with \\( 3\sqrt6\\))

## A little bit more complex fractions
If you have a fraction like this:

\\( \dfrac{5}{3 + \sqrt2} \\)

You can rationalise it by multiplying it with itself, but invert the signs. Like this:

\\( \dfrac{5}{3 + \sqrt2} \cdot \dfrac{3 - \sqrt2}{3 - \sqrt2} \\)  
\\( \dfrac{5 \cdot (3 - \sqrt2)}{(3 + \sqrt2)(3 - \sqrt2)} \\)

!---(
#--(
### ❓ Looks familiar?
)--#
If you look closely, \\( (3 + \sqrt2)(3 - \sqrt2) \\) is the same pattern as
\\( (a + b)(a - b) \\)! This pattern is one of the **special binominal
products**. We use this a lot in algebra.

So, remember this pattern:  
\\( (a + b)(a - b) = a^2 - b^2 \\)
)---!

Let's continue.

\\( \dfrac{5 \cdot (3 - \sqrt2)}{(3 + \sqrt2)(3 - \sqrt2)} = \dfrac{5(3 - \sqrt2)}{3^2 - \sqrt{2^2}} \\)

And if you remember correctly, \\( \sqrt{a^2} = a \\). So,

\\( \dfrac{5(3 - \sqrt2)}{3^2 - \sqrt{2^2}} = \dfrac{5(3 - \sqrt2)}{9 - 2}\\)  
\\( = \dfrac{5(3 - \sqrt2)}{6}\\)

And that's how you rationalise complex fractions.