How do you simplify #16/(sqrt [5] - sqrt [7])#?

1 Answer
Oct 16, 2015

#-8(sqrt(5) + sqrt(7))#

Explanation:

To simplify this expression, you have to rationalize the denominator by using its conjugate expression.

For a binomial, you get its conjugate by changing the sign of the second term.

In your case, that would imply having

#sqrt(5) - sqrt(7) -> underbrace(sqrt(5) color(red)(+) sqrt(7))_(color(blue)("conjugate"))#

So, multiply the fraction by #1 = (sqrt(5) + sqrt(7))/(sqrt(5) + sqrt(7))# to get

#16/(sqrt(5) - sqrt(7)) * (sqrt(5) + sqrt(7))/(sqrt(5) + sqrt(7)) = (16(sqrt(5) + sqrt(7)))/((sqrt(5) - sqrt(7))(sqrt(5)+sqrt(7))#

The denominator is now in the form

#(a-b)(a+b) = a^2 - b^2#

This means that you can write

#(16(sqrt(5) + sqrt(7)))/( (sqrt(5))^2 - (sqrt(7))^2) = (16(sqrt(5) + sqrt(7)))/(5-7) = color(green)(-8(sqrt(5) + sqrt(7)))#