1. ## Rationalising surd....

okay, i seem to be pretty cool with the basics of rationalizing surds, just started doing a level maths, is going okay.

three questions, though, i keep trying to do and can't see where i'm going wrong, here they are.

don't know how to do the symbols so will just write it

********

square root of 6,

over

square root of six, plus square root of 3

i get 6 minus square root of 18,

over,

3

however, the book says ....

2 minus square root of 2,....

i really can't see how they got this, hopefully if its explained, i'll be able to do the other two.

thanks!

2. Have you tried using factors of 18 under the square root sign?

3. oh man i can't believe i missed that. thanks !

4. had missed that on the other question as well, fixed that proper good.

3rd question though, i still can't get;

root 7 + 5

over

root 7 + root 5

i get

7 minus root 35, plus 5 times root 7, minus 5 times root 5

over

2

where the hell am i going wrong!!

5. But its not root 7 + root 5, its (root7) + 5.

multiplying that by the rationalising factor (root 7 - root 5) gives;

7, - root35, + 5(root7), - 5(root5)

this book is known to have a few mistakes, by the way!

6. Originally Posted by artem1s
But its not root 7 + root 5, its (root7) + 5.

multiplying that by the rationalising factor (root 7 - root 5) gives;

7, - root35, + 5(root7), - 5(root5)

this book is known to have a few mistakes, by the way!
Hello, artem1s. This should clear it up for you.

$\frac{\sqrt{7}+5}{\sqrt{7}+\sqrt{5}}\cdot\frac{\sq rt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}} =\frac{7-\sqrt{35}+5\sqrt{7}-5\sqrt{5}}{7-5} = \frac{7-\sqrt{35}+5\sqrt{7}-5\sqrt{5}}{2}$

7. Yeah, that's what i got as well! Rotten book was wrong!

Thanks guys, i was tearing my hair out trying to figure this one out.