How do you factor and simplify the square root of 75?

Hint:
Recall that "square root of a" means $ {{a}^{\tfrac{1}{2}}} $ , which is also written as $ \sqrt{a} $ . For positive real numbers a and b, $ \sqrt{a} $ × $ \sqrt{b} $ = $ \sqrt{ab} $ (see Note below). Express 75 as a product of prime numbers and see if it can be simplified or not. The following rule of exponent $ {{a}^{\dfrac{p}{q}}} $ = $ \sqrt[q]{{{a}^{p}}} $ is useful.

Complete Step by step Solution:
We can write 75 = 3 × 5 × 5 = 3 × $ {{5}^{2}} $ , which is the required prime factorization.
Taking the square root, we get:
 $ \sqrt{75} $ = $ \sqrt{3\times {{5}^{2}}} $
We know that for positive real numbers a and b, $ \sqrt{a} $ × $ \sqrt{b} $ = $ \sqrt{ab} $ , therefore:
⇒ $ \sqrt{75} $ = $ \sqrt{3} $ × $ \sqrt{{{5}^{2}}} $
Using $ {{a}^{\tfrac{p}{q}}} $ = $ \sqrt[q]{{{a}^{p}}} $ , we have:
⇒ $ \sqrt{75} $ = $ \sqrt{3} $ × $ {{\left( {{5}^{2}} \right)}^{\tfrac{1}{2}}} $
⇒ $ \sqrt{75} $ = $ \sqrt{3} $ × $ {{5}^{\tfrac{2}{2}}} $
⇒ $ \sqrt{75} $ = $ \sqrt{3} $ × $ {{5}^{1}} $

⇒ $ \sqrt{75} $ = $ 5\sqrt{3} $, which is the required simplification.

Note:
If a and b are negative, then the rule $ \sqrt{a} $ × $ \sqrt{b} $ = $ \sqrt{ab} $ doesn't work. In that case, we have to consider complex numbers to make the radicals positive and multiply the imaginary unit separately. For example, $ \sqrt{-3} $ × $ \sqrt{-2} $ = $ \sqrt{3}\sqrt{-1} $ × $ \sqrt{2}\sqrt{-1} $ = $ -\sqrt{6} $ .
In general, the notation $ {{a}^{x}} $ is used to represent the value of the product a × a × a × a ... (x times). Here a is called the base (radix) and x is called the exponent / power (index). If $ {{a}^{x}} $ = b, then we say $ {{b}^{\tfrac{1}{x}}} $ = $ \sqrt[x]{b} $ = a, which is read as "x-th root of b is equal to a". The roots are also called radicals.
When representing the numbers, for radical numbers $ a\sqrt{b} $ means "a × $ \sqrt{b} $ ", whereas for fractions $ a\dfrac{b}{c} $ means "a + $ \dfrac{b}{c} $ ".