Rationalising surd....

• Nov 19th 2008, 03:06 PM
artem1s
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!
• Nov 19th 2008, 04:14 PM
Theoretically!
Have you tried using factors of 18 under the square root sign?
• Nov 20th 2008, 02:33 PM
artem1s
oh man i can't believe i missed that. thanks !
• Nov 20th 2008, 03:23 PM
artem1s
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!! :(
• Nov 22nd 2008, 09:40 AM
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!
• Nov 22nd 2008, 10:03 AM
masters
Quote:

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.

$\frac{\sqrt{7}+5}{\sqrt{7}+\sqrt{5}}\cdot\frac{\sq rt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}} =\frac{7-\sqrt{35}+5\sqrt{7}-5\sqrt{5}}{7-5} = \frac{7-\sqrt{35}+5\sqrt{7}-5\sqrt{5}}{2}$
• Nov 22nd 2008, 03:00 PM
artem1s
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. :)