Question

3squareroot2-2squareroot3 over 3squareroot2+2squareroot3
Rationalize the denominator and simplify.

Answer 1

\[3\sqrt{2}-2\sqrt{3}\over3\sqrt{2}+2\sqrt{3}\]

Answer 2

\[\frac{ a-b }{ a+b }*\frac{ a-b }{ a-b }\]
change the sign of denominator and multiply both in numerator and denominator.

Answer 3

have you followed..?
show me your work.

Answer 4

Sorry, just got back on.. And im lost bc im not timesing it

Answer 5

\[\frac{ 3\sqrt{2}-2\sqrt{3} }{3\sqrt{2}+2\sqrt{3} }*\frac{ 3\sqrt{2}-2\sqrt{3} }{ 3\sqrt{2}-2\sqrt{3} }\]
\[for denominator \left( a+b \right)\left( a-b \right)=a ^{2}-b ^{2}\]
\[for numerator \left( a-b \right)*\left( a-b \right)=a ^{2}-2*a*b+b ^{2}\]

Answer 6

But what do you do with the square roots??

Answer 7

\[example \left( 5\sqrt{6} \right)^{2}=5^{2}*6=25*6=150\]

Answer 8

\[5\sqrt{6}*3\sqrt{7}=5*3 \sqrt{6*7}=15\sqrt{42}\]

Answer 9

So I would take \[3\sqrt{2} (5\sqrt{6})?\]

Answer 10

these are examples only.
you stick to your question.

Answer 11

Yeah now I'm all confused.

Answer 12

1 have told how to square a term with square root
In the second example i have told how to multiply two terms with squareroots.

Answer 13

Well I'm sorry, I'm a dumbass and can't understand this worth pellet.

Answer 14

\[you have tofind thesquare of 3\sqrt{2} and 2\sqrt{3}\]

Answer 15

Okay, now how do I do that?

Answer 16

\[\left( 3\sqrt{2} \right)^{2}=3^{2}*2=9*2=18\]
\[similarly \left( 2\sqrt{3} \right)^{2}=2^{2}*3=4*3=12\]
\[3 \sqrt{2}*2\sqrt{3}=3*2\sqrt{2*3}=6\sqrt{6}\]
i think now you can complete.

Answer 17

Wouldn't that be \[3\sqrt{2}?\]

Answer 18

no, terms under square root are multiplied.
and terms out square root are multiplied separately.