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Division with square roots at the base

Anotherstudent

New member
Nov 12, 2013
2
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
 

soroban

Well-known member
Feb 2, 2012
409
Hello, Anotherstudent!

[tex]\text{Rationalize: }\:\frac{3\sqrt{3}}{6-2\sqrt{3}}[/tex]

Multiply numerator and denominator
. . by the conjugate of the denominator.

[tex]\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})} [/tex]

. . [tex]=\;\frac{18\sqrt{3} + 18}{36-12} \;=\;\frac{18(\sqrt{3}+1)}{24} \;=\;\frac{3(\sqrt{3}+1)}{4}[/tex]
 

Anotherstudent

New member
Nov 12, 2013
2
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)