1. Prove it

Prove That
$\frac{1}{\sqrt7-\sqrt6}=\frac{3}{\sqrt6-\sqrt3}+\frac{4}{\sqrt7+\sqrt3}$

2. Originally Posted by nathan02079
Prove That
$\frac{1}{\sqrt7-\sqrt6}=\frac{3}{\sqrt6-\sqrt3}+\frac{4}{\sqrt7+\sqrt3}$
we can prove this by showing we can make the LHS and RHS look the same.

take the left hand side and multiply it by $\frac {(\sqrt {6} - \sqrt {3})(\sqrt {7} + \sqrt {3})}{(\sqrt {6} - \sqrt {3})(\sqrt {7} + \sqrt {3})}$ (which is 1, so you're not changing anything). simplify the expression

separately, take the right hand side, combine the fractions, then multiply by $\frac {\sqrt {7} - \sqrt {6}}{\sqrt {7} - \sqrt {6}}$ (which is 1, so you're not changing anything). simplify the expression and note that it is the same expression you got before when manipulating the left hand side. thus the equation is proved

now, how did i know what to multiply by?

3. lol i got the same that you got...but i guessed what to multiply by....

so is there really a way to know what to multiply by?

4. and i got this far
$\frac{(\sqrt6-\sqrt3)(\sqrt7+\sqrt3)}{(\sqrt7-\sqrt6)(\sqrt6-\sqrt3)(\sqrt7+\sqrt3)}=\frac{3(\sqrt7+\sqrt3)+4(\ sqrt6-\sqrt3)}{(\sqrt6-\sqrt3)(\sqrt7+\sqrt3)}\bullet\frac{\sqrt7-\sqrt6}{\sqrt7-\sqrt6}$

5. Originally Posted by nathan02079
lol i got the same that you got...but i guessed what to multiply by....
good guess...weird if it's a guess, but good

so is there really a way to know what to multiply by?
yes, you multiply each side by the LCD of the other side. this makes the denominators of both sides the same, so if they are equal, the numerators should end up being the same as well.

6. Originally Posted by nathan02079
and i got this far
$\frac{(\sqrt6-\sqrt3)(\sqrt7+\sqrt3)}{(\sqrt7-\sqrt6)(\sqrt6-\sqrt3)(\sqrt7+\sqrt3)}=\frac{3(\sqrt7+\sqrt3)+4(\ sqrt6-\sqrt3)}{(\sqrt6-\sqrt3)(\sqrt7+\sqrt3)}\bullet\frac{\sqrt7-\sqrt6}{\sqrt7-\sqrt6}$
i told you to do both sides separately.

say, "Consider the left hand side" ...and then do your stuff

then say, "Consider the right hand side" ... and do your stuff

then say, "We see that the right hand side simplifies to the same expression as the left hand side does. thus, the equation is true"

anyway, what's the hold up? expand the brackets and simplify

7. So now do I just multiply the 4 and 3 in then simplify?

EDIT:wait a sec i g2g for dinner...i'll get back to this right after...thanks for the help so far though =)

8. Originally Posted by nathan02079
Prove That
$\frac{1}{\sqrt7-\sqrt6}=\frac{3}{\sqrt6-\sqrt3}+\frac{4}{\sqrt7+\sqrt3}$
$\frac{3}
{{\sqrt 6 - \sqrt 3 }} + \frac{4}
{{\sqrt 7 + \sqrt 3 }} = \frac{{\left( {\sqrt 6 - \sqrt 3 } \right)\left( {\sqrt 6 + \sqrt 3 } \right)}}
{{\sqrt 6 - \sqrt 3 }} + \frac{{\left( {\sqrt 7 - \sqrt 3 } \right)\left( {\sqrt 7 + \sqrt 3 } \right)}}
{{\sqrt 7 + \sqrt 3 }}.$

So

$\frac{3}
{{\sqrt 6 - \sqrt 3 }} + \frac{4}
{{\sqrt 7 + \sqrt 3 }} = \sqrt 7 + \sqrt 6 .$

Now multiply the last term by $\frac{{\sqrt 7 - \sqrt 6 }}
{{\sqrt 7 - \sqrt 6 }}$
and the rest follows.

9. Originally Posted by Krizalid
$\frac{3}
{{\sqrt 6 - \sqrt 3 }} + \frac{4}
{{\sqrt 7 + \sqrt 3 }} = \frac{{\left( {\sqrt 6 - \sqrt 3 } \right)\left( {\sqrt 6 + \sqrt 3 } \right)}}
{{\sqrt 6 - \sqrt 3 }} + \frac{{\left( {\sqrt 7 - \sqrt 3 } \right)\left( {\sqrt 7 + \sqrt 3 } \right)}}
{{\sqrt 7 + \sqrt 3 }}.$

So

$\frac{3}
{{\sqrt 6 - \sqrt 3 }} + \frac{4}
{{\sqrt 7 + \sqrt 3 }} = \sqrt 7 + \sqrt 6 .$

Now multiply the last term by $\frac{{\sqrt 7 - \sqrt 6 }}
{{\sqrt 7 - \sqrt 6 }}$
and the rest follows.
nice!

we could also get the same manipulation by conjugating each term

10. xD sorry to break it to you jhevon, but i actually understood kriz's explanation right away....no offense but thanks for the help though you guys

11. Originally Posted by nathan02079
xD sorry to break it to you jhevon, but i actually understood kriz's explanation right away....no offense but thanks for the help though you guys
none taken. i like his way better too

the question is, can you "see" his method? you may want to try it the conjugation way as well, it's just one line difference between Krizalid's method, but it easier to "see"