What is #sqrt(2x^3)*sqrt(6x^2)*sqrt(10x)# in simplified form?

1 Answer
Jan 28, 2017

Answer:

#sqrt(2x^3)*sqrt(6x^2)*sqrt(10x) = 2x^3sqrt(30)#

Explanation:

Note that if #a, b >= 0# then

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

By extension, we find if #a, b, c >= 0# then:

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

Note also that if #a>=0# then:

#sqrt(a^2) = a#

In our example, we will assume #x >= 0# in order that all of the original square roots are well defined Real numbers.

So in our example:

#sqrt(2x^3)*sqrt(6x^2)*sqrt(10x) = sqrt(2x^3*6x^2*10x)#

#color(white)(sqrt(2x^3)*sqrt(6x^2)*sqrt(10x)) = sqrt(2x^3*2x^3*30)#

#color(white)(sqrt(2x^3)*sqrt(6x^2)*sqrt(10x)) = sqrt((2x^3)^2)sqrt(30)#

#color(white)(sqrt(2x^3)*sqrt(6x^2)*sqrt(10x)) = 2x^3 sqrt(30)#

Footnote

It seems to be a common error to assume that:

#sqrt(x^2) = x#

This does hold, but only if #x >= 0#.

If we want to cover the case #x < 0# too then we could write:

#sqrt(x^2) = abs(x)#

In the given example, we can deduce that the case #x >= 0# is intended, since otherwise #sqrt(2x^3)# and #sqrt(10x)# could take imaginary values.