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why does this: $${\sqrt{{\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}}}$$ equals 1?

Guest May 16, 2015

Best Answer 

 #2
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+15

Better with 3.Binom :

$$\left({\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}\left({\mathtt{7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}\right) = {\mathtt{49}}{\mathtt{\,-\,}}{\mathtt{16}}{\mathtt{\,\times\,}}{\mathtt{3}} = {\mathtt{1}}$$

 

$${\sqrt{{\mathtt{1}}}} = {\mathtt{1}}$$

radix !

Guest May 16, 2015
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3+0 Answers

 #1
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+10

$${\sqrt{{\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}}} = {\mathtt{0.267\: \!949\: \!192\: \!431\: \!122\: \!7}}$$

$${\sqrt{{\mathtt{7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}}} = {\mathtt{3.732\: \!050\: \!807\: \!568\: \!877\: \!3}}$$

 

$${\mathtt{0.267\: \!949\: \!192\: \!431\: \!122\: \!7}}{\mathtt{\,\times\,}}{\mathtt{3.732\: \!050\: \!807\: \!568\: \!877\: \!3}} = {\mathtt{1}}$$

 

You are right !!

Guest May 16, 2015
 #2
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+15
Best Answer

Better with 3.Binom :

$$\left({\mathtt{7}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}\left({\mathtt{7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}\right) = {\mathtt{49}}{\mathtt{\,-\,}}{\mathtt{16}}{\mathtt{\,\times\,}}{\mathtt{3}} = {\mathtt{1}}$$

 

$${\sqrt{{\mathtt{1}}}} = {\mathtt{1}}$$

radix !

Guest May 16, 2015
 #3
avatar+90968 
+10

Thanks Radix and anon,

 

I like this question, it is a difference of 2 squares question.   (Just like Radix said )

 

$$\\\sqrt{7-4\sqrt3}\times \sqrt{7+4\sqrt3}\\\\
=\sqrt{(7-4\sqrt3)(7+4\sqrt3)}\\\\
=\sqrt{(7)^2-(4\sqrt3)^2}\\\\
=\sqrt{49-16\times 3}\\\\
=\sqrt{49-48}\\\\
=\sqrt{1}\\\\
=1$$

Melody  May 17, 2015

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