MHB Division with square roots at the base

AI Thread Summary
The discussion focuses on solving a division problem involving square roots: 3√3 divided by (6 - 2√3). The original poster expresses uncertainty about their method and shares their attempted solution. Another participant suggests rationalizing the denominator by multiplying by its conjugate, leading to a simplified result of 3(√3 + 1)/4. The original poster acknowledges this correction and expresses gratitude, indicating a willingness to apply the new technique in future exercises.
Anotherstudent
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Hi, I am new here and I don't know if anyone is going to answer to this post, but if you do so thank you very much. I have been frowning on these kind of problems!

I have been trying to solve some exercices from my homeworks. However, I don't know if I am doing them correctly. Here is one problem and how I solved it :

3√3
------- IS WHAT I HAD TO SOLVE
6 - 2√3

HOW I SOLVED IT :

3√3 √3 3√9
------- X ------ = ------ = 9
6 - 2√3 √3 6-2√9 Thanks for letting me know if I'm on the right track :D

Ps: sorry i don't know how people do the square roots
 
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Hello, Anotherstudent!

\text{Rationalize: }\:\frac{3\sqrt{3}}{6-2\sqrt{3}}
Multiply numerator and denominator
. . by the conjugate of the denominator.

\frac{3\sqrt{3}}{6-2\sqrt{3}}\cdot\frac{6+2\sqrt{3}}{6+2\sqrt{3}} \;=\;\frac{3\sqrt{3}(6+2\sqrt{3})}{(6-2\sqrt{3})(6+2\sqrt{3})}

. . =\;\frac{18\sqrt{3} + 18}{36-12} \;=\;\frac{18(\sqrt{3}+1)}{24} \;=\;\frac{3(\sqrt{3}+1)}{4}
 
Ahhhh this is it ! the conjugate! I knew something I was doing was wrong. Thank you so much for enlightening me, I will try to solve more exercice using the conjugate and I'll let you know how it did for me. Thanks a lot :) (heart)
 
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