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+# 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.