Simplify the following expression : \[{\left( {\sqrt 5 + \sqrt 2 } \right)^2}\] ?

Hint: In order to solve and write the expression into the simplest form . To simplify this question , we need to solve it step by step . Here we are going to expand the part and perform some calculations to simplify the given equation by using algebraic identities ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ for any real numbers $a$ and $b$ .Also we will use the concept of operations on real numbers to get our required result .

Complete step-by-step answer:
If we see the question , we need to solve the given expression under the square root which is \[{\left( {\sqrt 5 + \sqrt 2 } \right)^2}\] .
We can apply the formula or we can say identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$
Substitute the values of $a$ and $b$ by comparing the algebraic identity to the expression given \[{\left( {\sqrt 5 + \sqrt 2 } \right)^2}\],
Plug in the values as $a = \sqrt 5 $and $b = \sqrt 2 $, we get-
\[
   \Rightarrow {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} \\
   \Rightarrow {\left( {\sqrt 5 + \sqrt 2 } \right)^2} = {\left( {\sqrt 5 } \right)^2} + 2 \times \sqrt 5 \times \sqrt 2 + {\left( {\sqrt 2 } \right)^2} \\
   \Rightarrow {\left( {\sqrt 5 + \sqrt 2 } \right)^2} = 5 + 2\sqrt {10} + 2 \;
 \]
As the operations can only be performed between the like terms only so , we get –
\[ \Rightarrow {\left( {\sqrt 5 + \sqrt 2 } \right)^2} = 7 + 2\sqrt {10} \]
This expression cannot be simplified further . Therefore , the required answer is \[7 + 2\sqrt {10} \]
So, the correct answer is “ \[7 + 2\sqrt {10} \] ”.

Note: Always try to understand the mathematical statement carefully and keep things distinct .
Remember the algebraic identities and apply appropriately .
Choose the options wisely , it's better to break the question and then solve part by part .
Cross check the answer and always keep the final answer simplified .
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$
This algebraic identity is used here to solve the given problem.