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sqrt(3)-4/(1+sqrt(3)

 Jun 6, 2014

Best Answer 

 #2
avatar+89 
+11

$${\sqrt{{\mathtt{3}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{4}}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{3}}}}\right)}} = {\mathtt{0.267\: \!949\: \!192\: \!431\: \!122\: \!7}}$$

or

$${\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,-\,}}{\mathtt{4}}\right)}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{3}}}}\right)}} = -{\mathtt{0.830\: \!127\: \!018\: \!922\: \!193\: \!2}}$$

or the formula from the topic

$${\mathtt{\,-\,}}{\frac{{\mathtt{4}}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{3}}}}\right)}} = -{\mathtt{1.464\: \!101\: \!615\: \!137\: \!754\: \!6}}$$

which one is you asking for?

 Jun 6, 2014
 #1
avatar+3451 
0

Nevermind, thought I could solve this but I can't.

 Jun 6, 2014
 #2
avatar+89 
+11
Best Answer

$${\sqrt{{\mathtt{3}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{4}}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{3}}}}\right)}} = {\mathtt{0.267\: \!949\: \!192\: \!431\: \!122\: \!7}}$$

or

$${\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,-\,}}{\mathtt{4}}\right)}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{3}}}}\right)}} = -{\mathtt{0.830\: \!127\: \!018\: \!922\: \!193\: \!2}}$$

or the formula from the topic

$${\mathtt{\,-\,}}{\frac{{\mathtt{4}}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{3}}}}\right)}} = -{\mathtt{1.464\: \!101\: \!615\: \!137\: \!754\: \!6}}$$

which one is you asking for?

problem Jun 6, 2014
 #3
avatar+114442 
+8

Since many people do not like to have any sort of radical in a denominator, we can get rid of this by multiplying the numerator and denominator by 1-√3....so we have

-4 / (1+√3) *  (1-√3) / (1-√3) 

The denominator of the resulting fraction becomes (1 -3) = -2

So we have ...   (-4)*(1-√3) /( -2) =  2 (1-√3)

Note how this procedure leaves us with no fraction at all......however, we still have a "nasty" radical hanging around....!!!

 Jun 6, 2014
 #4
avatar+112062 
+8

Thank you CPhill and Problem,

First the queston in the heading is different from the question in the body of the post so CPHill and Problem have answered different questions.

ANONYMOUS - Please be much more careful next time you make a post.

 

CPhill's answer is much better because Problem's answer is an estimate.  It is not an exact value for the question.

ALSO,

The second solution that Problem has given is wrong. 

 

$$\sqrt{3}-\frac{4}{1+\sqrt{3}}\\\\
=\frac{\sqrt{3}(1+\sqrt{3})}{1+\sqrt{3}}-\frac{4}{1+\sqrt{3}}\\\\
=\frac{\sqrt{3}+3-4}{1+\sqrt{3}}\\\\
=\frac{\sqrt{3}-1}{1+\sqrt{3}}\\\\
=\frac{\sqrt{3}-1}{1+\sqrt{3}}\times\frac{1-\sqrt{3}}{1-\sqrt{3}}\\\\
=\frac{-(\sqrt{3}-1)^2}{1-3}\\\\
=\frac{-(3-2\sqrt{3}+1)}{-2}\\\\
=\frac{-(4-2\sqrt{3})}{-2}\\\\
=2-\sqrt{3}\\\\$$

 Jun 7, 2014

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