Question

1/root(8-root(32)) is equal to- Also explain your answer

Answer 1

To Find:

$\frac{1}{ \sqrt{8 - \sqrt{32} } }$

By factoring out the perfect square integer from the denominator

$\frac{1}{ 2\sqrt{2 - \sqrt{2} } }$

By completing the perfect square in the denominator

$\frac{1}{2} \frac{ \sqrt{2 - \sqrt{2} } }{( \sqrt{2 - \sqrt{2} })( \sqrt{2 - \sqrt{2} } )}$

$\frac{1}{2} \frac{ \sqrt{2 - \sqrt{2} } }{2 - \sqrt{2} }$

By rationalising the denominator

$\frac{1}{2} \frac{ \sqrt{2 - \sqrt{2} } (2 + \sqrt{2} )}{(2 - \sqrt{2} )(2 + \sqrt{2}) }$

$\frac{1}{2} \frac{2 \sqrt{2 - \sqrt{2} } + \sqrt{4 - 2\sqrt{2} } }{4 - 2}$

$\frac{2 \sqrt{2 - \sqrt{2} } + 4 \sqrt{4 - 2 \sqrt{2} } }{4}$