How do you multiply #sqrt(2x^5y^2)/sqrt(14x^3y^8)#?

Question

1392 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-11T00:53:03

#(xsqrt7)/(7y^3)#

First, remember that:
#sqrt (a^2b)=asqrtb#

Using our rule, we simplify the radicals.
#(sqrt (2x^5y^2))/(sqrt (14x^3y^8))# becomes #(yx^2sqrt (2x))/(xy^4sqrt (14x))#

We now try to rationalize the denominator.

#(yx^2sqrt (2x))/(xy^4sqrt (14x))xx(sqrt (14x))/(sqrt (14x))=> (yx^2sqrt (2x)*sqrt(14x))/(xy^4(14x))#

Now, remember that:
#sqrt a *sqrtb=sqrt(ab)#

#(yx^2sqrt (2x)*sqrt(14x))/(xy^4(14x))# becomes
#(yx^2sqrt (28x^2))/(xy^4(14x))# Using our first rule, we can simplify this further.

#(yx^2sqrt (28x^2))/(xy^4(14x))=>(yx^2(2xsqrt7))/(xy^4(14x))#

We multiply this out to get:
#(2x^3ysqrt7)/(14x^2y^4)#

Lastly, remember that:
#(a^n)/(a^m)=a^(n-m)#

Using this rule, we now have:
#(xsqrt7)/(7y^3)#
This is the answer!