How do you write the expression #sqrt(48x^6y^3)# in simplified radical form.

1 Answer
Feb 4, 2015

First of all, let's deal with the numeric part: by factoring 48 with prime numbers, #\sqrt{48}=\sqrt{2^4 \cdot 3}#. Now use the fact that the square root of a product is the product of the square roots, so we have #\sqrt{48}=\sqrt{16}\cdot \sqrt{3}=4\sqrt{3}#

As for the variables, we can again say that #\sqrt{x^6y^3}=\sqrt{x^6}\sqrt{y^3}#. You can write #x^6# as #x^2\cdot x^2\cdot x^2#, so #\sqrt{x^6}=\sqrt{x^2\cdot x^2\cdot x^2}=\sqrt{x^2}\cdot\sqrt{x^2}\cdot\sqrt{x^2}=x\cdot x \cdot x =x^3#

The same reason bring us to write #y^3# as #y^2 \cdot y#, so #\sqrt{y^3}=\sqrt{y^2\cdot y}=\sqrt{y^2}\cdot\sqrt{y}=y\sqrt{y}#.

Putting all the pieces together, we have that
#\sqrt{48x^6y^3}=4x^3y\sqrt{3y}#