Question 149990
As a reminder, let's review a basic rule of square roots:
{{{ sqrt( 4x ) }}}
Then, I can factor the 4 as:
{{{ sqrt( 2*2*x ) }}}
If you see a "pair" of terms, you can you can "pull it out" thus:
{{{ 2*sqrt( x ) }}}
.
For your problem then:
{{{ sqrt( -48y^3x^4 ) }}}
{{{ sqrt( -(4*4*3)yyyx^2x^2 ) }}}
Notice, the pair of "4's", "y's", and "x^2".
{{{ 4x^2y*sqrt( -3y ) }}}