prove

• Jan 3rd 2008, 11:04 PM
perash
prove
without a calculator

prove
$\displaystyle \sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}}=\sqrt{5+\sqrt{21}}$
• Jan 3rd 2008, 11:09 PM
janvdl
Quote:

Originally Posted by perash
without a calculator

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

• Jan 4th 2008, 01:35 AM
wingless
Quote:

Originally Posted by janvdl

?

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

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

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

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

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

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

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

If you need help with the calculations I didn't show, just tell me :)
• Jan 4th 2008, 04:50 AM
janvdl
Quote:

Originally Posted by wingless
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He edited the post afterwards.
• Jan 4th 2008, 07:43 AM
Krizalid
Quote:

Originally Posted by perash
without a calculator

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

$\displaystyle \sqrt {2 + \sqrt 3 } = \frac{{\sqrt {4 + 2\sqrt 3 } }} {{\sqrt 2 }} = \frac{{\sqrt {\left( {1 + \sqrt 3 } \right)^2 } }} {{\sqrt 2 }} = \frac{{1 + \sqrt 3 }} {{\sqrt 2 }}.$

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

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