How do you write y^(-1/2)/x^(1/2)y−12x12 in radical form?
1 Answer
Explanation:
Your starting expression looks like this
y^(-1/2)/x^(1/2)y−12x12
The first thing to do is rewrite the negative exponent as a positive exponent. You know that
color(blue)(n^(-a) = 1/n^a)n−a=1na
In your case, you have
y^(-1/2) = 1/y^(1/2)y−12=1y12
The expression becomes
y^(-1/2)/x^(1/2) = 1/x^(1/2) * 1/y^(1/2)y−12x12=1x12⋅1y12
Take a look at the denominator. You have
x^(1/2) * y^(1/2) = (x * y)^(1/2)x12⋅y12=(x⋅y)12
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)1x12⋅1y12=112(x⋅y)12=(1x⋅y)12
You know that
color(blue)( n^(a/b) = root(b)(n^a))nab=b√na
In your case, you will have
(1/(xy))^(1/2) = sqrt(1/(xy))(1xy)12=√1xy
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)√1xy=√1√xy=1√xy⋅√xy√xy=√xy√xy⋅√xy=√xyxy
