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

(Solution to Problem 13 (credit to gaussintraining))
 
(See Also)
 
(One intermediate revision by one other user not shown)
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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>
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==See Also==
 
==See Also==
{{AHSME 35p box|year=1973|num-b=12|num-a=14}}
+
{{AHSME 30p box|year=1973|num-b=12|num-a=14}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 13:59, 20 February 2020

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 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AHSME Problems and Solutions