How do you rationalize the denominator and simplify #5/(sqrt14-2)#?

1 Answer
May 9, 2018

Answer:

#\frac{5\sqrt{14}+10}{10}#

Explanation:

Use the identity #(a-b)(a+b)=a^2-b^2# in the following way:

#\frac{5}{\sqrt{14}-2} = \frac{5}{\sqrt{14}-2} \cdot 1 = \frac{5}{(\sqrt{14}-2)} \cdot \frac{(\sqrt{14}+2)}{(\sqrt{14}+2)}#

Now, the numerator is simply #5(\sqrt{14}+2) = 5\sqrt{14}+10#, while at the denominator we have the forementioned identity:

#(\sqrt{14}-2)(\sqrt{14}+2) = (\sqrt{14})^2 - 2^2 = 14-4 = 10#