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

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Answer 1 · 2018-05-09T07:41:22

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

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:

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

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