Difference between revisions of "1990 AIME Problems/Problem 2"

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Suppose that <math>52+6\sqrt{43}</math> is in the form of <math>(a + b\sqrt{43})^2</math>. [[FOIL]]ing yields that <math>52 + 6\sqrt{43} = a^2 + 43b^2 + 2ab\sqrt{43}</math>. This implies that <math>a</math> and <math>b</math> equal one of <math>\pm1, \pm3</math>. The possible [[set]]s are <math>(3,1)</math> and <math>(-3,-1)</math>; the latter can be discarded since the [[square root]] must be positive. This means that <math>52 + 6\sqrt{43} = (\sqrt{43} + 3)^2</math>. Repeating this for <math>52-6\sqrt{43}</math>, the only feasible possibility is <math>(\sqrt{43} - 3)^2</math>.
 
Suppose that <math>52+6\sqrt{43}</math> is in the form of <math>(a + b\sqrt{43})^2</math>. [[FOIL]]ing yields that <math>52 + 6\sqrt{43} = a^2 + 43b^2 + 2ab\sqrt{43}</math>. This implies that <math>a</math> and <math>b</math> equal one of <math>\pm1, \pm3</math>. The possible [[set]]s are <math>(3,1)</math> and <math>(-3,-1)</math>; the latter can be discarded since the [[square root]] must be positive. This means that <math>52 + 6\sqrt{43} = (\sqrt{43} + 3)^2</math>. Repeating this for <math>52-6\sqrt{43}</math>, the only feasible possibility is <math>(\sqrt{43} - 3)^2</math>.
  
Rewriting, we get <math>(\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3</math>. Using the difference of [[cube]]s, we get that <math>[\sqrt{43} + 3\ - \sqrt{43} + 3]\ [(43 + 6\sqrt{43} + 9) + (43 - 9) + (43 - 6\sqrt{43} + 9)]</math> <math> = (6)(3 \cdot 43 + 9) = \boxed{828</math>.
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Rewriting, we get <math>(\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3</math>. Using the difference of [[cube]]s, we get that <math>[\sqrt{43} + 3\ - \sqrt{43} + 3]\ [(43 + 6\sqrt{43} + 9) + (43 - 9) + (43 - 6\sqrt{43} + 9)]</math> <math> = (6)(3 \cdot 43 + 9) = \boxed{828}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 23:58, 7 September 2011

Problem

Find the value of $(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}$.

Solution

Suppose that $52+6\sqrt{43}$ is in the form of $(a + b\sqrt{43})^2$. FOILing yields that $52 + 6\sqrt{43} = a^2 + 43b^2 + 2ab\sqrt{43}$. This implies that $a$ and $b$ equal one of $\pm1, \pm3$. The possible sets are $(3,1)$ and $(-3,-1)$; the latter can be discarded since the square root must be positive. This means that $52 + 6\sqrt{43} = (\sqrt{43} + 3)^2$. Repeating this for $52-6\sqrt{43}$, the only feasible possibility is $(\sqrt{43} - 3)^2$.

Rewriting, we get $(\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3$. Using the difference of cubes, we get that $[\sqrt{43} + 3\ - \sqrt{43} + 3]\ [(43 + 6\sqrt{43} + 9) + (43 - 9) + (43 - 6\sqrt{43} + 9)]$ $= (6)(3 \cdot 43 + 9) = \boxed{828}$.

See also

1990 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AIME Problems and Solutions