# Definisi
Sebuah akar merupakan kebalikan dari pangkat. Sebagai contoh:
* \\( 4^2 = 16; \sqrt{16} = 4 \\)
* \\( 5^2 = 25; \sqrt{25} = 5 \\)
* \\( 3^3 = 27; \sqrt[3]{27} = 3 \\)

# Menyederhanakan Akar Kuadrat
Kamu dapat menyederhanakan sebuah akar dengan memfaktorkan dengan angka lain
yang rasional. Dengan kata lain, membagi bilangan akar dengan kuadrat sempurna.  
Setelah itu, akar kuadrat sempurna itu.

Sebagai contoh: **Sederhanakan** \\( \sqrt{60} \\)!  
\\(  \sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15} \\)  
!---(
#--(
## 💡 Penjelasan terperinci
)--#
1. Kita faktorkan 60. Kita pilih faktor terkecil dengan kuadrat yang sempurna (cont. 4, 9, 16, 25).  
   Kita akan mendapatkan \\(4 \times 15 \\). Untuk membuktikannya, \\( 4
   \times 15 \\) harus sama dengan 60. Dan itu benar!
2. Lalu, akar kuadrat sempurnanya, dalam contoh ini, 4.
   \\( \sqrt{4 \times 15} = 2\sqrt{15} \\)
)---!
Contoh lebih banyak:
* \\( \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3} \\)
* \\( \sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5} \\)

!---(
#--(
## ❕ Ada akar "sisa"!
)--#
Beberapa akar adalah **bilangan tidak rasional**. Bilangan tersebut adalah
bilangan yang tidak bisa direpresentasikan dengan rasio atau pecahan.

Kita tidak dapat menyederhanakan akar yang tidak rasional. Jadi, akar itu **sudah
dalam bentuk paling sederhana**.
)---!

# Operasi Akar
Pertama-tama, beberapa terminologi!
1. Koefisien adalah pengali dari akar. Sebagai contoh: \\( 3\sqrt{2} \\); 3 adalah koefisiennya, dan 2 adalah bilangan akarnya.

## 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.