Difference between revisions of "1973 AHSME Problems/Problem 13"

(Solution to Problem 13 (credit to gaussintraining))
 
m (Solution)
Line 12: Line 12:
  
 
Squaring the expression and taking the positive square root (since numerator and denominator are positive) of the result yields
 
Squaring the expression and taking the positive square root (since numerator and denominator are positive) of the result yields
<cmath>\sqrt{(\frac{2(\sqrt2+\sqrt6)}{3\sqrt{2+\sqrt3}})^2}</cmath>
+
<cmath>\sqrt{\left(\frac{2(\sqrt2+\sqrt6)}{3\sqrt{2+\sqrt3}}\right)^2}</cmath>
<cmath>\sqrt{\frac{4(2+4\sqrt{3}+6)}{9(2+\sqrt{3}}}</cmath>
+
<cmath>\sqrt{\frac{4(2+4\sqrt{3}+6)}{9(2+\sqrt{3})}}</cmath>
<cmath>\sqrt{\frac{4(8+4\sqrt{3})}{9(2+\sqrt{3}}}</cmath>
+
<cmath>\sqrt{\frac{4(8+4\sqrt{3})}{9(2+\sqrt{3})}}</cmath>
 
<cmath>\sqrt{\frac{16}{9}}</cmath>
 
<cmath>\sqrt{\frac{16}{9}}</cmath>
 
<cmath>\frac43</cmath>
 
<cmath>\frac43</cmath>

Revision as of 17:32, 27 July 2018

Problem

The fraction $\frac{2(\sqrt2+\sqrt6)}{3\sqrt{2+\sqrt3}}$ is equal to

$\textbf{(A)}\ \frac{2\sqrt2}{3} \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ \frac{2\sqrt3}3 \qquad \textbf{(D)}\ \frac43 \qquad \textbf{(E)}\ \frac{16}{9}$

Solution

Squaring the expression and taking the positive square root (since numerator and denominator are positive) of the result yields \[\sqrt{\left(\frac{2(\sqrt2+\sqrt6)}{3\sqrt{2+\sqrt3}}\right)^2}\] \[\sqrt{\frac{4(2+4\sqrt{3}+6)}{9(2+\sqrt{3})}}\] \[\sqrt{\frac{4(8+4\sqrt{3})}{9(2+\sqrt{3})}}\] \[\sqrt{\frac{16}{9}}\] \[\frac43\]

The answer is $\boxed{\textbf{(D)}}$.

See Also

1973 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
All AHSME Problems and Solutions