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

how much is a^2

Guest Dec 14, 2014

Best Answer 

 #1
avatar+91001 
+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
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1+0 Answers

 #1
avatar+91001 
+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

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