How do you simplify #sqrt(-50x^2y^2)#?

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Answer 1 · 2016-11-11T03:34:55

#sqrt(-50x^2y^2)=sqrt(-1)sqrt50sqrt(x^2)sqrt(y^2)=5xyisqrt2#

I'm going to rewrite this to make it easier to work with:

#sqrt(-50x^2y^2)=sqrt(-1)sqrt50sqrt(x^2)sqrt(y^2)#

We have 4 square roots and I'll take them in turn:

First is #sqrt(-1)#. This is the definition of the term #i#, so we have

#sqrt(-1)=i#

Next up is #sqrt50#. We can break this down as follows:

#sqrt50=sqrt(25xx2)=sqrt25sqrt2=5sqrt2#

And lastly we have two variables, both are squared, so we can handle them this way:

#sqrt(x^2)=x# and #sqrt(y^2)=y#

So we end up with:

#sqrt(-50x^2y^2)=sqrt(-1)sqrt50sqrt(x^2)sqrt(y^2)=5xyisqrt2#