+0  
 
0
6
764
1
avatar

a=sqrt(6+sqrt(11))-sqrt(6-sqrt(11))

how much is a^2

Guest Dec 14, 2014

Best Answer 

 #1
avatar+92805 
+5

 

$${\mathtt{a}} = {\sqrt{{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}}}$$     

 

 find      $${{\mathtt{a}}}^{{\mathtt{2}}}$$

 

$${{\mathtt{a}}}^{{\mathtt{2}}} = {{\sqrt{{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}}}}^{\,{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\sqrt{{\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}}}}^{\,{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}}}$$

 

$${{\mathtt{a}}}^{{\mathtt{2}}} = \left({\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}\right){\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{\left(\left({\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}\right){\mathtt{\,\times\,}}\left({\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}\right)\right)}}$$

 

$${{\mathtt{a}}}^{{\mathtt{2}}} = {\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{\left({\mathtt{36}}{\mathtt{\,-\,}}{\mathtt{11}}\right)}}$$

 

$${{\mathtt{a}}}^{{\mathtt{2}}} = {\mathtt{12}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{25}}}}$$

 

$${{\mathtt{a}}}^{{\mathtt{2}}} = {\mathtt{12}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{5}}$$

 

$${{\mathtt{a}}}^{{\mathtt{2}}} = {\mathtt{2}}$$

 

and that means that   $$a=\pm \sqrt{ 2}$$          COOL !!!

Melody  Dec 14, 2014
 #1
avatar+92805 
+5
Best Answer

 

$${\mathtt{a}} = {\sqrt{{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}}}$$     

 

 find      $${{\mathtt{a}}}^{{\mathtt{2}}}$$

 

$${{\mathtt{a}}}^{{\mathtt{2}}} = {{\sqrt{{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}}}}^{\,{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\sqrt{{\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}}}}^{\,{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}}}$$

 

$${{\mathtt{a}}}^{{\mathtt{2}}} = \left({\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}\right){\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{\left(\left({\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}\right){\mathtt{\,\times\,}}\left({\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}\right)\right)}}$$

 

$${{\mathtt{a}}}^{{\mathtt{2}}} = {\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{11}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{\left({\mathtt{36}}{\mathtt{\,-\,}}{\mathtt{11}}\right)}}$$

 

$${{\mathtt{a}}}^{{\mathtt{2}}} = {\mathtt{12}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{25}}}}$$

 

$${{\mathtt{a}}}^{{\mathtt{2}}} = {\mathtt{12}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{5}}$$

 

$${{\mathtt{a}}}^{{\mathtt{2}}} = {\mathtt{2}}$$

 

and that means that   $$a=\pm \sqrt{ 2}$$          COOL !!!

Melody  Dec 14, 2014

15 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.