# Rationalising the denominator surds

I have been struggling on this question. I don't understand how to change a negative surd fraction to a positive surd fraction.

Question: Rationalise and simply $$\frac{2}{1+{\sqrt 6}}$$

What I did:

$\frac{2(1-{\sqrt 6})}{(1+{\sqrt 6)(1-{\sqrt 6})}}$

= $\frac{2{-2\sqrt 6 }}{-5}$

$$\frac{2({\sqrt 6} -1)}{5}$$

What did I do wrong

Thank you and help is appreciated

• change sign in numerator and denominator – Riccardo.Alestra Jul 25 '17 at 7:12

Nothing went wrong

If you take your result $\dfrac{2-2\sqrt 6 }{-5}$

Multiply num and den by $-1$ and collect $2$ in the numerator, you get

$$\dfrac{2(\sqrt 6 -1)}{5}$$ which is the result of the book

• Thank you I understand – AspaeringStar7 Jul 25 '17 at 7:21

$$\frac{2}{\sqrt6+1}$$

$$\frac{2}{\sqrt6+1}\times \frac{\sqrt6-1}{\sqrt6-1}$$

$$\frac{2(\sqrt6-1)}{6-1}$$

$$\frac{2(\sqrt6-1)}{5}$$

Try with $\sqrt6-1$ instead of $1- \sqrt6$

• Thank you I understand – AspaeringStar7 Jul 25 '17 at 7:21

There's nothing wrong with what you did.

If you want to get rid of the minus sign from the denominator, multiply top and bottom by $-1$:

$\frac{2-2\sqrt{6}}{-5}\times\frac{-1}{-1}=\frac{-(2-2\sqrt{6})}{5}=\frac{-2+2\sqrt{6}}{5}=\frac{2(-1+\sqrt{6})}{5}=\frac{2(\sqrt{6}-1)}{5}$

• Thank you I understand – AspaeringStar7 Jul 25 '17 at 7:22

this is easy to deal . like $\sqrt 2 =-(-\sqrt2)$.