How do you simplify #sqrt(144x^6)#?

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Answer 1 · 2015-10-10T09:49:36

This can be written as #12x^3#

#sqrt(144x^6)=sqrt(144)*sqrt(x^6)=sqrt(12*12)*sqrt(x^6)=12*sqrt(x^6)=12x^3#

The final step: #sqrt(x^6)=x^3# comes from the properties of exponents, which say, that:

Using the properties above we can write, that:

#sqrt(x^6)=(x^6)^(1/2)=x^(6/2)=x^3#

Answer 2 · 2015-10-10T17:23:01

#sqrt(144x^6) = 12 abs(x^3)#

If #a, b >= 0# then #sqrt(ab) = sqrt(a)sqrt(b)#

If #a >= 0# then #sqrt(a^2) = a#

#sqrt(144x^6) = sqrt(144)sqrt(x^6) = sqrt(12^2)sqrt(abs(x^3)^2) = 12 abs(x^3)#

The modulus operation is necessary to deal with the case #x < 0#.

Going deeper:

Note that if #a < 0# then #sqrt(a^2) = -a#, so in general we can write #sqrt(x^2) = abs(x)#.

Any number has two square roots. If #a >= 0# then it has a positive square root: #sqrt(a)# and a negative square root: #-sqrt(a)#.

If #a < 0# then it has square roots #i sqrt(-a)# and #-i sqrt(-a)#. We call #i sqrt(-a)# the principal square root, defining #sqrt(a) = i sqrt(-a)#.

When it comes to square roots of negative numbers, the identity #sqrt(ab) = sqrt(a)sqrt(b)# breaks down. For example:

#1 = sqrt(1) = sqrt(-1 xx -1) != sqrt(-1) xx sqrt(-1) = -1#