How do you write #y^(-1/2)/x^(1/2)# in radical form?
1 Answer
Explanation:
Your starting expression looks like this
#y^(-1/2)/x^(1/2)#
The first thing to do is rewrite the negative exponent as a positive exponent. You know that
#color(blue)(n^(-a) = 1/n^a)#
In your case, you have
#y^(-1/2) = 1/y^(1/2)#
The expression becomes
#y^(-1/2)/x^(1/2) = 1/x^(1/2) * 1/y^(1/2)#
Take a look at the denominator. You have
#x^(1/2) * y^(1/2) = (x * y)^(1/2)#
The expression is now equaivalent to
#1/x^(1/2) * 1/y^(1/2) = 1^(1/2)/(x * y)^(1/2) = (1/(x * y))^(1/2)#
You know that
#color(blue)( n^(a/b) = root(b)(n^a))#
In your case, you will have
#(1/(xy))^(1/2) = sqrt(1/(xy))#
Extra step
You can rationalize the denominator and simplify this expression further
#sqrt(1/(xy)) = sqrt(1)/sqrt(xy) = 1/sqrt(xy) * sqrt(xy)/sqrt(xy) = sqrt(xy)/(sqrt(xy) * sqrt(xy)) = sqrt(xy)/(xy)#
