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How do you simplify #sqrt32#?

1 Answer

May 17, 2015

The answer is #4sqrt2#.

Let's decompose #32# in its prime factors :

#32 -: 2 = 16#,
#16 -: 2 = 8#,
#8-:2 = 4#,
#4-:2 = 2#,
#2-:2=1#.

That gives us #32 = 2^5#.

Therefore, #sqrt32 = sqrt(2^5) = sqrt(2*2*2*2*2) = sqrt4 * sqrt4 * sqrt2 = 2*2"sqrt2 = 4sqrt2#.

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