Let us take the number 9. Nine divided by 3 equals to the divider 3 => 9/3 = 3, so 3.3 = 9 or 32 = 9. Let us take another number, 27 this time, 27 = 3.3.3 = 33. So we found that 9 and 27 are actually 3 with exponent 2 and 3.

Basically what radical is, is a function which finds a divider, of the argument, which upped on exponent gives us the argument. Sometimes this divider is not a rational number. The radical is actually the opposite function of an exponent. It even can be write down with the help of an exponent.

So in our case the square(2nd) root of 9 is 3, √9 and the third root of 27 is 3 = 327

If a is positive real number then the equation x2 = a has two solutions: x = +√a or x = -√a.

$\sqrt[2]{x}$ is $\sqrt{x}$

If a is real number then the equation x3 = a has only one solution => x = 3a. With the help of the equtions above we solve square and cubic equations.
A root can be write down with the help of an exponent, the following rule applies:

$x^{\frac{m}{n}}=\sqrt[n]{x^m}=(\sqrt[n]{x})^m$

If n is odd:
$\sqrt[n]{x^n}=x$

If n is even:
$\sqrt[n]{x^n}=|x|$ - the absolute value of x

Example: $\sqrt[3]{x^3}=x$ but $\sqrt[4]{x^4}=|x|$

$\sqrt[n]{a \cdot b}=\sqrt[n]{a}\cdot \sqrt[n]{b}$

Proof: let's have nab = (ab)1/n, which from the basic formula up of the exponent, comes to a1/n.b1/n, or nanb

$\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$

Proof: na/b = (a/b)1/n and from the basic equations of the exponent, comes to a1/n/b1/n, or na/nb

$\sqrt[n]{\sqrt[m]{a}}=\sqrt[n\cdot m]{a}$

Proof: if you have nma that equals to na1/m, which equals to (a1/m)1/n and from the basic equations of the exponent, comes to a1/(m.n), or n . ma

2nx ≥ 0     n is a natural number and x ≥ 0

If 0 ≤ x < y then nx < ny