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When evaluating a problem where there is a square root, should I assume that the square root falls under the order part of bodmas?

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  • $\sqrt{\cdots} = (\cdots)^{\frac12},$ so yes, and treat the inside of a square root as its own group. – Stephen Donovan Jul 06 '22 at 19:18
  • Square roots and other functions typically come with brackets (in the case of square roots this can be shown by the vinculum - the horizontal line above). Evaluate what is inside the brackets and then apply the function – Henry Jul 06 '22 at 19:19
  • For those like me that hadn't ever met acronym BODMAS : https://www.twinkl.fr/teaching-wiki/bodmas ... – Jean Marie Jul 06 '22 at 20:21

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Paraphrasing Order of operations (Wikipedia),

The order of operations is

  1. parentheses
  2. exponentiation and root extraction
  3. multiplication and division
  4. addition and subtraction

The term root extraction refers to taking the $n$th root. In your case $n = 2$, so you can evaluate square roots as you would exponents.

Example:

\begin{align} & \quad 5 \times 2^2 + \sqrt{16 + 9} \\ &= 5 \times 2^2 + \sqrt{(16 + 9)}\\ &=5 \times 2^2 + \sqrt{25}\\ &= 5 \times 2^2 + 5\\ &=5 \times 4 + 5\\ &=20 + 5\\ &=25 \end{align}