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

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!

↳Redirected from "Suppose that I don't have a formula for #g(x)# but I know that #g(1) = 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
1 Answer
George C.
Oct 18, 2016

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

Explanation:

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)#

Related questions
See all questions in Square Root
Impact of this question
15400 views around the world
You can reuse this answer
Creative Commons License