Quick Reciprocal Problem

**Samson** · July 2nd 2010, 03:57 AM

Hello All,

I wanted to ask if someone could enlighten me with the reciprocal of 2+5*Sqrt(3).

I've just inverted all of it and I multiplied it by the conjugate, but i'm not sure if thats the right way to head with this thing.

Any help is appreciated!

**Sudharaka** · July 2nd 2010, 04:43 AM

Originally Posted by Samson

Hello All,

I wanted to ask if someone could enlighten me with the reciprocal of 2+5*Sqrt(3).

I've just inverted all of it and I multiplied it by the conjugate, but i'm not sure if thats the right way to head with this thing.

Any help is appreciated!

Dear Samson,

The reciprocal of $2+5\sqrt{3}$ is $\frac{1}{2+5\sqrt{3}}$ . You could multiply the numerator and denominator by the conjugate if you want to,

$\frac{1}{2+5\sqrt{3}}=\frac{1}{2+5\sqrt{3}}\times\ frac{2-5\sqrt{3}}{2-5\sqrt{3}}=-\frac{2-5\sqrt{3}}{71}$

Hope this will help you.

**Samson** · July 2nd 2010, 04:46 AM

Originally Posted by Sudharaka

Dear Samson,

The reciprocal of $2+5\sqrt{3}$ is $\frac{1}{2+5\sqrt{3}}$ . You could multiply the numerator and denominator by the conjugate if you want to,

$\frac{1}{2+5\sqrt{3}}=\frac{1}{2+5\sqrt{3}}\times\ frac{2-5\sqrt{3}}{2-5\sqrt{3}}=-\frac{2-5\sqrt{3}}{71}$

Hope this will help you.

That is what I was saying I needed to do, however I'm not sure if you still have that -71 in the denominator if that counts as still being in the form of a+b*Sqrt(3) . (For what is worth I got the same thing but i'm not sure if its right when I'm supposed to be able to put it into a form like that).

**Sudharaka** · July 2nd 2010, 05:29 AM

Originally Posted by Samson

That is what I was saying I needed to do, however I'm not sure if you still have that -71 in the denominator if that counts as still being in the form of a+b*Sqrt(3) . (For what is worth I got the same thing but i'm not sure if its right when I'm supposed to be able to put it into a form like that).

Dear Samson,

Yes. This is in the $a+b\sqrt{3}$ form, since $-\frac{2-5\sqrt{3}}{71}=-\frac{2}{71}+\frac{5}{71}\sqrt{3}$

**Samson** · July 2nd 2010, 05:34 AM

Originally Posted by Sudharaka

Dear Samson,

Yes. This is in the $a+b\sqrt{3}$ form, since $-\frac{2-5\sqrt{3}}{71}=-\frac{2}{71}+\frac{5}{71}\sqrt{3}$

Okay, so when the book mentions that form its okay that the numbers a and b are not integers?

**chiph588@** · July 2nd 2010, 07:15 AM

It depends on what ring you're in.

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