Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 6"

(typo?? w/e, I'll just pretend that the 7 = 8)
(Standardized answer choices; minor edits)
 
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The value of the expression <math>K=\sqrt{19+8\sqrt{3}}-\sqrt{7+4\sqrt{3}}</math> is
 
The value of the expression <math>K=\sqrt{19+8\sqrt{3}}-\sqrt{7+4\sqrt{3}}</math> is
  
A. <math>4</math>
+
<math>\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 4\sqrt{3}\qquad\mathrm{(C)}\ 12+4\sqrt{3}\qquad\mathrm{(D)}\ -2\qquad\mathrm{(E)}\ 2</math>
  
B. <math>4\sqrt{3}</math>
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==Solution==
 +
Suppose that <math>19 + 8\sqrt{3}</math> can be written in the form of <math>(a+b\sqrt{3})^2</math>, in order to eliminate the [[square root]].
  
C. <math>12+4\sqrt{3}</math>
+
Then <math>19 = a^2 + 3b^2</math> and <math>2ab\sqrt{3} = 8\sqrt{3} \Longrightarrow ab = 4</math>, and we quickly find that <math>19 + 8\sqrt{3} = (4+\sqrt{3})^2</math>.
  
D. <math>-2</math>
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Doing the same on the second radical gets us <math>(2 + \sqrt{3})^2</math>.
  
E. <math>2</math>
+
Thus the expression evaluates to <math>\sqrt{(4+ \sqrt{3})^2} - \sqrt{(2 + \sqrt{3})^2} = 4 + \sqrt{3} - 2 - \sqrt{3} = 2\ \mathrm{(E)}</math>.
 
 
==Solution==
 
Suppose that <math>19 + 8\sqrt{3}</math> can be written in the form of <math>(a+b\sqrt{3})^2</math>, in order to eliminate the [[square root]]. Then <math>19 = a^2 + 3b^2</math> and <math>2ab\sqrt{3} = 8\sqrt{3} \Longrightarrow ab = 4</math>, and we quickly find that <math>19 + 8\sqrt{3} = (4+\sqrt{3})^2</math>. Doing the same on the second radical gets us <math>(2 + \sqrt{3})^2</math>. Thus the expression evaluates to <math>\sqrt{(4+ \sqrt{3})^2} - \sqrt{(2 + \sqrt{3})^2} = 4 + \sqrt{3} - 2 - \sqrt{3} = 2\ \mathrm{(E)}</math>.
 
  
 
==See also==
 
==See also==

Latest revision as of 09:43, 27 April 2008

Problem

The value of the expression $K=\sqrt{19+8\sqrt{3}}-\sqrt{7+4\sqrt{3}}$ is

$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 4\sqrt{3}\qquad\mathrm{(C)}\ 12+4\sqrt{3}\qquad\mathrm{(D)}\ -2\qquad\mathrm{(E)}\ 2$

Solution

Suppose that $19 + 8\sqrt{3}$ can be written in the form of $(a+b\sqrt{3})^2$, in order to eliminate the square root.

Then $19 = a^2 + 3b^2$ and $2ab\sqrt{3} = 8\sqrt{3} \Longrightarrow ab = 4$, and we quickly find that $19 + 8\sqrt{3} = (4+\sqrt{3})^2$.

Doing the same on the second radical gets us $(2 + \sqrt{3})^2$.

Thus the expression evaluates to $\sqrt{(4+ \sqrt{3})^2} - \sqrt{(2 + \sqrt{3})^2} = 4 + \sqrt{3} - 2 - \sqrt{3} = 2\ \mathrm{(E)}$.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 5
Followed by
Problem 7
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