Math Help - prove

1. prove

without a calculator

prove
$\sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}}=\sqrt{5+\sqrt{21}}$

2. Originally Posted by perash
without a calculator

prove
$\sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}}=\sqrt{5+\sqrt{21}}$

3. Originally Posted by janvdl
?

-----

$\sqrt{5+\sqrt{21}}=\sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}}$

$5+\sqrt{21}=(\sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}})^2$

$5+\sqrt{21}=2+\sqrt{3} + 4-\sqrt{7} + 2\sqrt{8 - 2\sqrt{7} + 4\sqrt{3} -\sqrt{21}}$

$2\sqrt{8 - 2\sqrt{7} + 4\sqrt{3} -\sqrt{21}} = \sqrt{21} - \sqrt{3} + \sqrt{7} - 1$

$8 - 2\sqrt{7} + 4\sqrt{3} -\sqrt{21} = (\frac{\sqrt{21} - \sqrt{3} + \sqrt{7} - 1}{2})^2$

............ (a few long but easy calculations here..)

$8 - 2\sqrt{7} + 4\sqrt{3} -\sqrt{21} = 8 - 2\sqrt{7} + 4\sqrt{3} -\sqrt{21} \text{ } \blacksquare$

If you need help with the calculations I didn't show, just tell me

4. Originally Posted by wingless
?

-----
He edited the post afterwards.

5. Originally Posted by perash
without a calculator

prove
$\sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}}=\sqrt{5+\sqrt{21}}$
$\sqrt {2 + \sqrt 3 } = \frac{{\sqrt {4 + 2\sqrt 3 } }}
{{\sqrt 2 }} = \frac{{\sqrt {\left( {1 + \sqrt 3 } \right)^2 } }}
{{\sqrt 2 }} = \frac{{1 + \sqrt 3 }}
{{\sqrt 2 }}.$

In the same way $\sqrt {4 - \sqrt 7 } = \frac{{\sqrt 7 - 1}}
{{\sqrt 2 }}.$

Finally $\sqrt {2 + \sqrt 3 } + \sqrt {4 - \sqrt 7 } = \frac{{\sqrt 3 + \sqrt 7 }}
{{\sqrt 2 }} = \frac{{\sqrt {10 + 2\sqrt {21} } }}
{{\sqrt 2 }} = \sqrt {5 + \sqrt {21} } \quad\blacksquare$