Question

Rationalize the denominator​

Answer 1

$\huge \mathtt{\;\;\;\; \frac{1}{\sqrt{8} - 2}}$

$\huge \mathtt{ = \frac{1}{\sqrt{8} - 2} × \frac{\sqrt{8} + 2}{\sqrt{8} + 2}}$

$\huge \mathtt{ = \frac{\sqrt{8} + 2}{(\sqrt{8})² - (2)²}}$

$\huge \mathtt{ = \frac{\sqrt{8} + 2}{8 - 4}}$

$\huge \mathtt{ = \frac{2(\sqrt{2} + 1)}{4}}$

$\huge \mathtt{ = \frac{\not 2(\sqrt{2} + 1)}{\not 4 \;_{2}}}$

$\huge \mathtt{ = \frac{\sqrt{2} + 1}{2}}$

Answer 2

Answer:

$\frac{1}{ \sqrt{8} - 2 } \\ = \frac{1}{ \sqrt{8} - 2} \times \frac{ \sqrt{8} + 2 }{ \sqrt{8} + 2 } \\ = \frac{1 (\sqrt{8} + 2)}{( \sqrt{8} - 2)( \sqrt{8} + 2) } \\ = \frac{ \sqrt{8} + 2}{ {( \sqrt{8}) }^{2} - {(2)}^{2} } \\ = \frac{ \sqrt{8} + 2}{8 - 4} \\ = \frac{ \sqrt{8} + 2}{4} \\ = \frac{2 \sqrt{2} + 2}{4} \\ = \frac{2( \sqrt{2} + 1) }{4} \\ = \frac{ \sqrt{2} + 1 }{2}$