Question

HELP ME! I cant get past this!
3(sqrt)2 - 2(sqrt)3 over 3(sqrt)2 + 2(sqrt)3

Answer 1

\[3\sqrt{2}-2\sqrt{3}\div3\sqrt{2}+2\sqrt{3}\]

Answer 2

\[\frac{ 3\sqrt2-2\sqrt3 }{ 3\sqrt2+2\sqrt3 }\]
like this?

Answer 3

yes

Answer 4

you multiply top and bottom by the conjugate of the bottom...

\[\frac{ 3\sqrt2-2\sqrt3 }{ 3\sqrt2+2\sqrt3 }=\frac{ (3\sqrt2-2\sqrt3) }{ (3\sqrt2+2\sqrt3) }\cdot\frac{ (3\sqrt2-2\sqrt3) }{ (3\sqrt2-2\sqrt3) }\]
work that out and you'll get rid of the square roots in the denominator.

Answer 5

Thats the part that im stuck on, i dont really understand how to work them out.

Answer 6

did you learn how to FOIL or distribute?

Answer 7

you there?

Answer 8

Yes, but im still struggling with it. I dont understand what to do with the square roots and how to combine the different ones.

Answer 9

okay...
\[\frac{ (3\sqrt2-2\sqrt3)(3\sqrt2-2\sqrt3) }{ (3\sqrt2+2\sqrt3)(3\sqrt2-2\sqrt3) }=\]\[\frac{3\cdot3\cdot \sqrt2 \cdot \sqrt2-2(2\cdot3\cdot \sqrt2 \cdot \sqrt3)+2\cdot 2\cdot \sqrt3 \cdot \sqrt3}{3\cdot3\cdot \sqrt2 \cdot \sqrt2+(2\cdot3\cdot \sqrt2 \cdot \sqrt3)-(2\cdot3\cdot \sqrt2 \cdot \sqrt3)+2\cdot 2\cdot \sqrt3 \cdot \sqrt3}\]\[=\frac{ 9\cdot 2-2(6\cdot \sqrt6)+4\cdot 3 }{ 9 \cdot 2-4 \cdot 3 }=\frac{ 18-12\cdot \sqrt6+12 }{ 18-12 }=\frac{ 30-12\cdot \sqrt6 }{ 6 }\]
and you should be able to simplify from there.

Answer 10

Okay thank you so much!
I'll do that now

Answer 11

do you understand how to combine them?

Answer 12

yes. would the answer be \[5-2\sqrt{6}\]

Answer 13

?

Answer 14

yes, but do you understand how to multiply out the factors to get there, now?

Answer 15

More than i did before but not completely

Answer 16

rememeber, \(\sqrt a\cdot\sqrt b=\sqrt{a\cdot b}\) and \(a\sqrt b =a\cdot\sqrt b\)

Answer 17

take it slow and write everything out and as you do more you'll get quicker and see shortcuts.

Answer 18

Okay :) thank you so much. I wouldnt have just passed that online lesseon with out you