Difference between revisions of "1970 AHSME Problems/Problem 18"

Latest revision as of 15:23, 2 July 2016

$\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}$ is equal to

$\text{(A) } 2\quad \text{(B) } 2\sqrt{3}\quad \text{(C) } 4\sqrt{2}\quad \text{(D) } \sqrt{6}\quad \text{(E) } 2\sqrt{2}$

Square the expression:

$(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}})^2=3+\sqrt{2}-2\sqrt{(3+\sqrt{2})(3-\sqrt{2})}+3-2\sqrt{2}=6-2\sqrt{9-8}=6-2\sqrt{1}=4$

$\Rightarrow\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}=\sqrt{4}=2\Rightarrow\boxed{A}$

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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

@@ Line 9: / Line 9: @@
 \text{(E) } 2\sqrt{2}</math>
-== Solution ==
+== Solution==
-<math>\fbox{C}</math>
+Square the expression:
+<math>(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}})^2=3+\sqrt{2}-2\sqrt{(3+\sqrt{2})(3-\sqrt{2})}+3-2\sqrt{2}=6-2\sqrt{9-8}=6-2\sqrt{1}=4</math>
+<math>\Rightarrow\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}=\sqrt{4}=2\Rightarrow\boxed{A}</math>
 == See also ==