**Radicals** is an opposite action from exponentiation. Just like exponentiation is repetitive multiplication, taking a root from a number is repetitive division.

For example, you know that $\ 2 ^ 2 = 4$. If you want to take second (also called square) root from number $4$ is number $2$. That means that second root from $4$ is equal $2$.

Analog, $\ 2 ^ 3 = 8$ so the third root (also called cube) root from number $8$ is equal $2$. Usually the second root from a is written without index, simply $\sqrt{a}$ .

The symbol of a radical is called the radical symbol, the number underneath it is called the argument of the radical or the *radicand*, and the number above is called the *index of a radical*.

Since radicals and exponents are opposite actions, they undo each other which can really simplify tasks.

What does it mean that they undo each other? That means that if you have a root with index $a$, and a number that has exponent $a$, the a-root from your number is equal to that number.

But, here is the catch. You know that $\ 2 ^ 2 = 4$ but it is also true that $\ (-2) ^ 2 = 4$.

What now? How do you chose what is the solution to the $\sqrt{4}$?

Well, they both are. Those are called double solutions because they both match the equality. Does this work on all other indexes? Not really.

For example: $\sqrt[3]{8}$

You know that $\ 2^3 = 8$, but $\ (-2)^3 = – 8$. That means that third root of eight will only have one solution, and that is number $2$. How do we generalize this? You will always have two solutions, negative and positive, if you have an even index and one solution if you have an odd index.

Even exponents -> double solution $\sqrt[a]{x^a} = \pm x = \pm x$ ($a$ even number)

Odd exponents -> single solution $\sqrt[b]{x^b} = x = x$ ($b$ odd number)

## Add Radicals

**Adding radicals** is very simple action. There is only one thing you have to worry about, which is a very standard thing in math. You can’t add radicals that have different index or radicand. The only thing you can do is match the radicals with the same index and radicands and add them together.

Summation is done in a very natural way so $\sqrt[3]{2} + \sqrt[3]{2} = 2\sqrt[3]{2}$

But summations like $\sqrt[3]{2} + \sqrt[4]{2725}$ can’t be done, and you simply leave it just the way it is.

**Example:**

$\sqrt[4]{a} + \frac{1}{2}\sqrt[5]{a} + 4\sqrt[4]{a} + \sqrt[3]{a} + \sqrt[3]{a} = 5\sqrt[4]{a} + \frac{1}{2}\sqrt[5]{a} + 2\sqrt[3]{a}$

## Subtract Radicals

The rules that apply to summation also apply to subtraction. Again, you have to be careful about what you are subtracting. Radicals you are subtracting must have **equal indexes** and **equal radicands**.

We also instinctively subtract radicals so: $\ 6\sqrt[5]{4} – \sqrt[5]{4} = 5\sqrt[5]{4}$,

or $\sqrt[7]{8} – 2\sqrt[7]{8} = – \sqrt[7]{8}$

$\sqrt[4]{a} – \frac{1}{2}\sqrt[5]{a} – 0.5\sqrt[4]{a} + \frac{3}{2}\sqrt[3]{a} – \frac{1}{2}\sqrt[3]{a}=$

$=(\sqrt[4]{a} – 0.5\sqrt[4]{a}) + (\frac{3}{2}\sqrt[3]{a} – \frac{1}{2}\sqrt[3]{a}) – \frac{1}{2}\sqrt[5]{a}$

$=(\sqrt[4]{a} – \frac{1}{2}\sqrt[4]{a}) + \frac{2}{2}\sqrt[3]{a} – \frac{1}{2}\sqrt[5]{a} $

$=\frac{1}{2}\sqrt[4]{a} + \sqrt[3]{a} – \frac{1}{2}\sqrt[5]{a}$

## Multiply Radicals

**Multiplying radicals** is a bit different. You can multiply if either your radicands are equal or your indexes are equal.

First, multiplications when the indexes of radicals are equal:

**Example 1:**

$\sqrt{6} \cdot \sqrt{2} = ?$

Solution:

$\sqrt{6} \cdot \sqrt{2} = \sqrt{6 \cdot 2} = \sqrt{12}$

**Example 2:**

$\sqrt{0.6} \cdot \sqrt{5} = ?$

Solution:

$\sqrt{0.6} \cdot \sqrt{5}$

$= \sqrt{\frac{6}{10}} \cdot \sqrt{5}$

$= \sqrt{\frac{3}{5}} \cdot \sqrt{5}$

$= \sqrt{\frac{3}{5} \cdot 5} \cdot \sqrt{3}$

And secondly, if you multiply two radicals that have the same radicands, but different indexes, you will get a radical that has the same radicand as those two, but index that is gain by multiplying indexes of factors.

**Example 3:**

$\sqrt[2]{5} \cdot \sqrt[3]{5} = ?$

Solution:

$\sqrt[2]{5} \cdot \sqrt[3]{5} = \sqrt[6]{5^5}$

## Divide Radicals

**Dividing radicals** is very similar to multiplying. You have to be careful: If you want to divide two radicals they have to have the same index.

If you have same bases but different indexes, the easiest way is to transform a radical into an exponent, but we’ll get to that later.

**Example 1:**

$\sqrt[3]{16} : \sqrt[3]{2} + \frac{4^3}{4} = ?$

Solution:

$\sqrt[3]{16} : \sqrt[3]{2} + \frac{4^3}{4} $

$= \sqrt[3]{\frac{16}{2}} + 4^{3 – 1} $

$= \sqrt[3]{8} + 4^2 = \sqrt[3]{2^3} + 16 = 2 + 16 $

$= 18$

## Rational exponents

Until now we met only whole exponents. But what if we encounter a fraction or a decimal in it? This is where exponents and radicals mix. There is a simple rule in their connection. That is:

$\sqrt[c]{a^b} = a^{\frac{b}{c}}$

In words, their connection is expressed as the fraction whose numerator is the power, and denominator is the index of a radical.

Using that we can now multiply radicals that have the same base but different indexes:

$\sqrt[b]{b} \cdot \sqrt[c]{a}$

$= b^{\frac{1}{b}} \cdot a^{\frac{1}{c}}$

$= (b \cdot a)^{\frac{1}{b} + \frac{1}{c}}$

**Example 1:**

$\sqrt[\frac{1}{2}]{8} \cdot 2^6 = ?$

Solution:

$\sqrt[\frac{1}{2}]{8} \cdot 2^6 $

$= 8^{\frac{1}{\frac{1}{2}}} \cdot 2^6 $

$= (2^3)^2 \cdot 2^6 $

$= 2^6 \cdot 2^6$

$ = 2^12$

**Example 2:**

$\sqrt[2]{7} \cdot 7^{\frac{3}{2}} + \sqrt[3]{7} \cdot \frac{1}{7^3} = ?$

Solution:

$\sqrt[2]{7} \cdot 7^{\frac{3}{2}} + \sqrt[3]{7} \cdot \frac{1}{7^3} $

$= 7^{\frac{1}{2}} \cdot 7^{\frac{3}{2}} + 7^{\frac{1}{3}} \cdot 7^{-3} $

$= 7^{\frac{1}{2} + \frac{3}{2}} + 7^{\frac{1}{3} – 3} $

$= 7^2 + 7^{-\frac{8}{3}}$

**Example 3:**

$\sqrt[4]{8} \cdot \frac{1}{2^{\frac{2}{3}}} + 2^{-\frac{1}{2}} + 0.4 \cdot 4^{0.2} = ?$

Solution:

$\sqrt[4]{8} \cdot \frac{1}{2^{\frac{2}{3}}} + 2^{-\frac{1}{2}} + 0.4 \cdot 4^{0.2} $

$= \sqrt[4]{2^3} \cdot 2^{-\frac{2}{3}} + 2^{-\frac{1}{2}} + \frac{4}{10} \cdot 4^{\frac{2}{10}}$

$= \ 2^{\frac{3}{4} \cdot 2^{-\frac{2}{3}}} + 2^{-\frac{1}{2}} + \frac{2}{5} \cdot (2^2)^{\frac{1}{5}} $

$= 2^{\frac{3}{4} – \frac{2}{3}} + 2^{-\frac{1}{2}} + \frac{2}{5} \cdot 2^{\frac{2}{5}}$

$= \ 2^{-\frac{1}{6}} + 2^{-\frac{1}{2}} + 2^{\frac{2}{5}}$

**Example 4:**

Solve and turn into a radical:

$\ 2^{-\frac{3}{2}} \cdot 2^{-3} = ?$

Solution:

$\ 2^{-\frac{3}{2}} \cdot 2^{-3} $

$= 2^{-\frac{3}{2} + (-3)} $

$= 2^{-\frac{9}{2}} $

$= \sqrt[2]{-9} $

$= \sqrt[2]{\frac{1}{2^9}}$

**Example 5:**

$\sqrt[4]{4^{-3}} \cdot 2^7 + 6^{-0.2} \cdot 36^2 = ?$

Solution:

$\sqrt[4]{4^{-3}} \cdot 2^7 + 6^{-0.2} \cdot 36^2 $

$= 4^{-\frac{3}{4}} \cdot 2^7 + 6^{-\frac{1}{5}} \cdot 36^2 $

$= (2^2)^{-\frac{3}{4}} \cdot 2^7 + 6^{-\frac{1}{5}} \cdot 6^4 $

$= 2^{-\frac{3}{2}} \cdot 2^7 + 6^{4 -\frac{1}{5}} $

$= 2^{7 – \frac{3}{2}} + 6^{\frac{19}{5}}$

$ = 2^\frac{11}{2} + 6^{\frac{19}{5}} $

$= \sqrt[2]{2^{11}} + \sqrt[5]{6^{19}}$

## Simplifying radicals

What does it mean to simplify a radical? Some radicals have arguments that can be factorized in a way that one factors exponent would be undone by a radicant.

**Example 1:**

$\sqrt[2]{8} $

$= \sqrt[2]{2 \cdot 4}$

$ = \sqrt[2]{2 \cdot 2^2} $

$= \sqrt[2]{2^2}\sqrt[2]{2}$

$ = 2\sqrt[2]{2}$

**Example 2:**

Simplify a radical:

$\sqrt[4]{32a^6b^{13}} = ?$

We are trying to match exponents from each factor to the index of the radical, and since we know that $\sqrt[b]{a} \cdot \sqrt[b]{c} = \sqrt[b]{a \cdot c}$, , numbers whose exponents match the indexes of radicals can come out as themselves. With this you have to be careful, you can only do this if those numbers are bound with multiplication or division.

$\sqrt[4]{32a^6b^{13}} $

$= \sqrt[4]{2^5a^{2 + 4}b^{12 + 1}} $

$= \sqrt[4]{2^{4 + 1}a^2a^4(b^3)^4b} $

$= 2ab^3\sqrt[4]{2a^2b}$

## Radical equations

**Radical equations** are equations in which the unknown is inside a radical.

**Example 1:**

$\sqrt{x} = 2$

(We solve this simply by raising to a power both sides, the power is equal to the index of a radical)

$\sqrt{x} = 2 ^{2}$

$ x = 4$

**Example 2:**

$\sqrt{x + 2} = 4 /^{2}$

$\ x + 2 = 16$

$\ x = 14$

**Example 3:**

$\frac{4}{\sqrt{x + 1}} = 5, x \neq 1$

Again, here you need to watch out for that variable $x$, he can’t be ($-1)$ because if he could be, we’d be dividing by $0$.

$\ 4 = 5\sqrt{x + 1}$

$\ 5\sqrt{x + 1} = 4 /: 5$

$\sqrt{x + 1} = \frac{4}{5} /^2$

$\ x + 1 = \frac{16}{25}$

$\ x = \frac{16}{25} – 1$

$\ x = -\frac{9}{25} \neq 1$

**Example 4:**

$\sqrt{x – 1} = \sqrt{3x – 3} /^2$

$\ x – 1 = 3x – 3$

$\ -2x = -2$

$\ x = 1$

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