+0

# rationalizing denominators

-2
78
7
+204

Find $\displaystyle{ \frac{2}{1 + 2\sqrt{3}} + \frac{3}{2 - \sqrt{3}}}$, and write your answer in the form $\displaystyle \frac{A + B\sqrt{3}}{C}$, with the fraction in lowest terms and $A > 0$. What is $A+B+C$?

Feb 24, 2021

#1
+939
0

$$\frac{2}{1+2\sqrt{3}} \cdot \frac{1-2\sqrt{3}}{1-2\sqrt{3}}+ \frac{3}{2-\sqrt{3}} \cdot \frac{2+\sqrt{3}}{2+\sqrt{3}}$$

$$\frac{2 - 4\sqrt{3}}{13} + \frac{6 + 3\sqrt{3}}{1}$$

$$\frac{(2 - 4\sqrt{3} )+ 13(6 + 3\sqrt{3})}{13}$$

$$\frac{80 +35\sqrt{3}}{13}$$

.
Feb 24, 2021
#2
+266
0

Hello oliviapow06...!

$$\frac{2}{1+2\sqrt{3}}+\frac{3}{2-\sqrt{3}}$$

$$=\frac{2\left(2-\sqrt{3}\right)}{\left(1+2\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{3\left(1+2\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(1+2\sqrt{3}\right)}$$

$$=\frac{2\left(2-\sqrt{3}\right)+3\left(1+2\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(1+2\sqrt{3}\right)}$$

$$=\frac{2\left(2-\sqrt{3}\right)+3\left(1+2\sqrt{3}\right)}{3\sqrt{3}-4}$$

$$=\frac{7+4\sqrt{3}}{3\sqrt{3}-4}$$

$$=\frac{37\sqrt{3}+64}{11}$$

$$A=64$$

$$B\sqrt{3}=37\sqrt{3}$$

$$C=11$$

$$64+37\sqrt{3}+11$$

$$=75+37\sqrt{3}$$

Feb 24, 2021
#3
+204
-2

I don't know how but this is apparently wrong

Feb 24, 2021
#4
+266
+1

Oh, both of these answers or just mine?

DewdropDancer  Feb 24, 2021
#5
+312
-1

both of them; it was 112

ellapow  Feb 24, 2021
#7
+266
+1

Oh, I'm sorry, I must have messed up on a part.

DewdropDancer  Feb 25, 2021
#6
0

Bruh I know you are a alt for clairepow STOP cheating on AOPS. This will not be my last warning.

Feb 24, 2021