# Math Help - Surds

1. ## Surds

I have been given a question to expand and simplify this: (3+root6)(2-root3)

i managed to get it down to 6-3root3-3root2+2root6 but i cannot get it any further, help would be much appreciated.

PS sorry i could find a square root sign.

2. Originally Posted by gustifo
I have been given a question to expand and simplify this: (3+root6)(2-root3)

i managed to get it down to 6-3root3-3root2+2root6 but i cannot get it any further, help would be much appreciated.

PS sorry i could find a square root sign.
Hello gustifo,

$(3+\sqrt{6})(2-\sqrt{3})=6-3\sqrt{3}+2\sqrt{6}-3\sqrt{2}$

I think you simplified it as much as you can.

3. Hello, gustifo!

Ya done good!

Simplify: . $(3 + \sqrt{6})(2-\sqrt{3})$

$(3+\sqrt{6})(2 - \sqrt{3}) \;=\;(3)(2) + (3)(\text{-}\sqrt{3}) + (\sqrt{6})(2) + (\sqrt{6})(\text{-}\sqrt{3})$

. . $= \;6 - 3\sqrt{3} - 2\sqrt{6} - \sqrt{18}$

But $\sqrt{18} \:=\:\sqrt{(9)(2)} \:=\:\sqrt{9}\,\sqrt{2} \:=\:3\sqrt{2}$

. . Therefore: . $6 - 3\sqrt{3} - 2\sqrt{6} - 3\sqrt{2}$

And that's as far as we can go . . .

Edit: Too slow again . . . *sigh*