If $\sqrt {\sqrt {2500} + \sqrt {961} } = {\left( x \right)^2}$, then x equals
(A). 81
(B). 3
(C). 6561
(D). 9

Hint: In this problem remember to use the idea that 2500 and 961 will be perfect squares of some numbers so try to get them in a simplified way, using this information can help you to approach the solution of the problem.

Complete step-by-step answer:
According to the given information we have equation $\sqrt {\sqrt {2500} + \sqrt {961} } = {\left( x \right)^2}$taking this as equation 1
 Since we know that square root of 2500 is given by



Since, the remainder is 0
Therefore, 50 is the square root of 2500
For square root of 961 is given by



Since, the remainder is 0
Therefore, 31 is the square root of 961
Now substituting the values in the equation 1 we get
$\sqrt {50 + 31} = {\left( x \right)^2}$
$ \Rightarrow $$\sqrt {81} = {\left( x \right)^2}$(equation 2)
For square root of 81 is given by



Since, the remainder is 0
Therefore, the square root of 81 is 9
Substituting the value in the equation 2 we get
$9 = {\left( x \right)^2}$
Now applying the square root operation both the sides we get
$\sqrt 9 = \sqrt {{{\left( x \right)}^2}} $(equation 3)
Square root of 9 is given by



Since, the remainder is 0
Therefore, the square root of 9 is 3
Substituting the value in the equation 3 we get
x = 3
Therefore, value of x is 3
Hence, option B is the correct option.

Note: In the above solution we used the term “operation” which can be explained as a method which includes an operand and an operator such as addition operation, subtraction option, and division operation as in the above solution we used the square root operation. The square root operation is unary operations which require only one operand whereas the operations which require more than one operand are called binary operations.