What is #sqrt192# in simplest radical form? Thanks!

Question

I need help checking my extra work for practice and wanted to know how to do #sqrt192# as well as the simplest answer in radical form. The answer I got was #8sqrt3# if this is not the correct answer, please let me know why. If it is, let me know that I am correct so I don't have to worry about failing my math test in a couple of weeks. Thanks!

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Answer 1 · 2016-10-18T23:31:17

#sqrt(192) = 8sqrt(3)#

Your answer is correct. Here's why...

If #a, b >= 0# then:

#sqrt(ab) = sqrt(a)sqrt(b)#

If #a >= 0# then:

#sqrt(a^2) = a#

Find the prime factorisation of #192# and identify square factors:

#192 = 2*2*2*2*2*2*3 = 2^6*3 = (2^3)^2*3 = 8^2*3#

So:

#sqrt(192) = sqrt(8^2*3) = sqrt(8^2)*sqrt(3) = 8sqrt(3)#