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:
$\frac{2(1-{\sqrt 6})}{(1+{\sqrt 6)(1-{\sqrt 6})}}$
= $\frac{2{-2\sqrt 6 }}{-5}$
When I checked the answers It said the answer was
$$\frac{2({\sqrt 6} -1)}{5}$$
What did I do wrong
Thank you and help is appreciated
$$\frac{2}{\sqrt6+1}$$
$$\frac{2}{\sqrt6+1}\times \frac{\sqrt6-1}{\sqrt6-1}$$
$$\frac{2(\sqrt6-1)}{6-1}$$
$$\frac{2(\sqrt6-1)}{5}$$
Comments
Try with $\sqrt6-1$ instead of $1- \sqrt6$
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
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 $-1$:
$\frac{2-2\sqrt{6}}{-5}\times\frac{-1}{-1}=\frac{-(2-2\sqrt{6})}{5}=\frac{-2+2\sqrt{6}}{5}=\frac{2(-1+\sqrt{6})}{5}=\frac{2(\sqrt{6}-1)}{5}$
