How do you simplify #sqrt(4a^6)#?

3 Answers
May 25, 2017

Answer:

#sqrt4a^6 = 2a^3#

Explanation:

#sqrt(4a^6) = sqrt4*sqrta^6 = 2a^3#

The square root of #4# is #2#, and to obtain the square root of #a^6# we

need to divide the given exponent by the value of the root required:

So: #sqrta^6=a^(1/2)*a^6=a^(6/2)=a^3#

May 28, 2017

Answer:

#2##a^3#

Explanation:

The expression which we have is:-

#rArrsqrt(4a^6)#

It can also be written as

#rArrsqrt(4a^6)#

#rArrsqrt(2^2xxa^6)#

#rArr sqrt(2^2)xx sqrt(a^6)#

#rArr2xxa^3#

#rArr 2a^3#

The final answer is: #(2a^3)#

May 28, 2017

Answer:

#sqrt(4a^6) = abs(2a^3)#

Explanation:

Note that:

#(2a^3)^2 = 2^2(a^3)^2 = 4a^6#

So #2a^3# is a square root of #4a^6#.

Note that:

#(-2a^3)^2 = 4a^6#

So #-2a^3# is a square root of #4a^6# too.

What do we mean when we write #sqrt(4a^6)# ?

So long as the radicand is non-negative, then these symbols refer to the principal non-negative square root.

For any real value of #a#, we have #4a^6 >= 0#, so #sqrt(4a^6)# refers to its non-negative square root.

Note however that #2a^3# is positive, zero or negative according to whether #a# is positive, zero or negative.

So in order to cover all real values of #a# we can write:

#sqrt(4a^6) = abs(2a^3)#

If, in addition we are told that #a >= 0# then this simplifies to:

#sqrt(4a^6) = 2a^3#