How do you simplify #sqrt216#?

Question

Autumn Arnold · Accepted Answer

#sqrt(216) = 6sqrt(6)#

The shortened version is as follows:

#sqrt(216) = sqrt(6^2*6) = 6sqrt(6)#

How do we find out that #216 = 6^2*6# ?

Recombining prime factors that have been separated one at a time is one method.

Here's a factor tree for #216#:

#color(white)(0000)216# #color(white)(0000)"/"color(white)(0)"\"# #color(white)(000)2color(white)(00)108# #color(white)(000000)"/"color(white)(0)"\"# #color(white)(00000)2color(white)(00)54# #color(white)(0000000)"/"color(white)(00)"\"# #color(white)(000000)2color(white)(000)27# #color(white)(000000000)"/"color(white)(00)"\"# #color(white)(00000000)3color(white)(0000)9# #color(white)(000000000000)"/"color(white)(0)"\"# #color(white)(00000000000)3color(white)(000)3#

Thus, we discover:

#216 = 2*2*2*3*3*3=(2*3*2*3)*(2*3) = 6^2*6#

By definition:

#sqrt(6^2) = 6#

For any positive numbers #a# and #b# we have:

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

Hence:

#sqrt(216) = sqrt(6^2*6) = sqrt(6^2)sqrt(6) = 6sqrt(6)#

Luna Angel · Answer

undefined To simplify $ \sqrt{216} $, you can find the prime factorization of 216 and then simplify the square root.

1. Find the prime factorization of 216:
$$ 216 = 2^3 \times 3^3 $$

2. Rewrite $ \sqrt{216} $ using the prime factorization:
$$ \sqrt{216} = \sqrt{2^3 \times 3^3} $$

3. Break up the square root:
$$ \sqrt{216} = \sqrt{2^3} \times \sqrt{3^3} $$

4. Simplify each factor under the square root:
$$ \sqrt{216} = 2 \times 3 \sqrt{3} $$

5. Multiply the factors outside the square root:
$$ \sqrt{216} = 6 \sqrt{3} $$

Therefore, $ \sqrt{216} $ simplifies to $ 6\sqrt{3} $.