How do you rationalize the denominator and simplify $\dfrac{5}{{\sqrt {14} - 2}}?$

Hint:To rationalize the denominator of the given fraction, first find the conjugate part of the denominator of the given fraction. Then multiply and divide the fraction with the conjugate part. When multiplying you will face an expression equivalent to the algebraic identity of product of sum and difference of two numbers which is equals to difference of square of the two numbers and is expressed mathematically as follows: $(a + b)(a - b) = {a^2} - {b^2}$. Also conjugate part of the irrational number $a \pm \sqrt b $ is given as $a \mp \sqrt b $.

Formula used:
Algebraic identity of product of sum and difference of two numbers:
$(a + b)(a - b) = {a^2} - {b^2}$

Complete step by step answer:
In order to rationalize the denominator and simplify the given expression $\dfrac{5}{{\sqrt {14} - 2}}$, we will first find the conjugate part of the denominator, that is $\sqrt {14} - 2$.Conjugate part of $\sqrt {14} - 2$ is equals to $\sqrt {14} + 2$.Now, multiplying and dividing the given expression with conjugate part, in order to rationalize the denominator
$\dfrac{5}{{\left( {\sqrt {14} - 2} \right)}} \times \dfrac{{\left( {\sqrt {14} + 2} \right)}}{{\left( {\sqrt {14} + 2} \right)}} \\
\Rightarrow\dfrac{{5\left( {\sqrt {14} + 2} \right)}}{{\left( {\sqrt {14} - 2} \right)\left( {\sqrt {14} + 2} \right)}} \\ $
We can see that the denominator part is seems to be an algebraic identity given as,
$(a + b)(a - b) = {a^2} - {b^2}$
So using this identity to simplify the expression further, we will get
$\dfrac{{5\left( {\sqrt {14} + 2} \right)}}{{\left( {\sqrt {14} - 2} \right)\left( {\sqrt {14} + 2} \right)}} \\
\Rightarrow\dfrac{{5\left( {\sqrt {14} + 2} \right)}}{{{{\left( {\sqrt {14} } \right)}^2} - {2^2}}} \\
\Rightarrow \dfrac{{5\left( {\sqrt {14} + 2} \right)}}{{14 - 4}} \\
\Rightarrow \dfrac{{5\left( {\sqrt {14} + 2} \right)}}{{10}} \\ $
Now, taking out five common from the numerator part and the denominator part in order to simplify the fraction further, we will get
$\dfrac{{5\left( {\sqrt {14} + 2} \right)}}{{10}} \\
\therefore\dfrac{{\sqrt {14} + 2}}{2} \\$
Therefore $\dfrac{{\sqrt {14} + 2}}{2}$ is the rationalized and simplified form of $\dfrac{5}{{\sqrt {14} - 2}}$.

Note:Rationalization and simplification are two different terms. Rationalizing an expression means converting the denominator part of the given expression from an irrational number into a rational number, whereas simplifying an expression means writing the expression in the simplest possible way.