+0

# square root

0
479
3

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

#2
+15

### Better with 3.Binom :

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

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

Guest May 16, 2015
#1
+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
+15

### Better with 3.Binom :

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

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

Guest May 16, 2015
#3
+93654
+10

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