How do you simplify #sqrt(72/3)#?

Question

5718 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-22T17:30:44

#2sqrt6#

#"using the "color(blue)"law of radicals"#

#•color(white)(x)sqrtaxxsqrtbhArrsqrt(ab)#

#rArrsqrt(72/3)=sqrt24=sqrt(4xx6)=sqrt4xxsqrt6=2sqrt6#

Answer 2 · 2018-05-22T17:31:50

#2sqrt(6)#

The goal in simplifying a square root is to divide the terms into their common factors.

This can be done in the following way.

Firstly you divide the radicand to get the simplest term: 24 (#72 / 3#)

Now, you find the common factors of 24.

6 factors into #2*3# and 4 factors into #2^2# == #2^3 * 3#
3 is a factor of itself and 8 factors into #2^3# == #2^3 * 3#

As you can see, either way you will get to the same result.

Adding this into our radical:
#sqrt(2^3 * 3) = (2^3 * 3)^(1/2) =# exponent rule = # 2^(3+(1/2)) * 3^(1/2)#

Rewriting this equation we get:
# 2^(5/2) * 3^(1/2)# == #2^2 * (2 ^(1/2)*3^(1/2)) #

Applying the square root (or factoring out the exponents)we get:

#sqrt(2^2 * (2 *3)# == #2sqrt(2*3)# == #2sqrt(6)#

Answer 3 · 2018-05-22T17:33:34

#2sqrt6#

dividing under a radical is allowed:

#sqrt(72/3) = sqrt(24) = sqrt(4*6) = sqrt(2^2*6) = 2sqrt6#

Answer 4 · 2018-05-22T17:39:16

#2sqrt6#

#sqrt(72/3)=sqrt24#

#rArr sqrt24=sqrt 4 xx sqrt6#

#rArr2sqrt6#