What is the square root of\(128 + \sqrt {262 - \sqrt {36} }\)
Solution
Correct Answer: Option D
Given expression
\(128 + \sqrt {262 - \sqrt {36} }\)
\(\begin{array}{l} = 128 + \sqrt {262 - 6} \\ = 128 + \sqrt {256} \end{array}\)
= 128 + 16
= 144
Now square root of the given expression\(= \sqrt {144}\)
= 12
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