# Thread: Surds question

1. ## Surds question

I need help with this question.

2. ## Re: Surds question

Give it a go.
Work out 1+c by getting a common denominator.
Then do the same for 1-c.
Then do the division.

3. ## Re: Surds question

Originally Posted by Britt00
I need help with this question.
$\dfrac{\sqrt3+1}{\sqrt3-1}=\dfrac{3+2\sqrt3+1}{2}$

4. ## Re: Surds question

Here is the method step by step.

$\dfrac{1 + \sqrt{3}}{1 - \sqrt{3}} = \dfrac{1 + \sqrt{3}}{1 - \sqrt{3}} * 1 = \dfrac{1 + \sqrt{3}}{1 - \sqrt{3}} * \dfrac{1 + \sqrt{3}}{1 + \sqrt{3}} =$

$\dfrac{1^2 + \sqrt{3} + \sqrt{3} + (\sqrt{3})^2}{1^2 + \sqrt{3} - \sqrt{3} - (\sqrt{3})^2} = \dfrac{1 + 2\sqrt{3} + 3}{1 - 3} = -\ \dfrac{2(2 + \sqrt{3})}{2}.$

EDIT: I believe Plato read the question incorrectly so he and I disagree in terms of sign only. Of course it is always possible I misread the problem, which was hard to parse on a tablet.

5. ## Re: Surds question

Please post your future algebra questions in the algebra forum. Thank you for your cooperation.

6. ## Re: Surds question

Jeff, why are you not using 1/sqrt(3)?

I get same as Sir Plato.

7. ## Re: Surds question

Because I did manage to misread the question. I hate it when people put those tiny jpegs in instead of writing out the question.

Yes of course Plato is correct.

$\dfrac{1 + \dfrac{1}{\sqrt{3}}}{1 - \dfrac{1}{\sqrt{3}}} = \dfrac{\dfrac{\sqrt{3} + 1}{\cancel{\sqrt{3}}}}{\dfrac{\sqrt{3} - 1}{\cancel{\sqrt{3}}}} = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}$.