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Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 2, 2011"

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==Solution==
 
==Solution==
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We simplify the LHS to be <math>\sqrt{x} + \tfrac{3x}{2} + 2 = 10</math>, so <math>\sqrt{x} = 8 - \tfrac{3x}{2}</math>, or <math>2\sqrt{x} = 16 - 3x</math>.  Now, we square this to get <math>4x = 9x^2 - 96x + 256</math>, so <math>9x^2 - 100x + 256 = 0</math>.  This factors as <math>(x - 4)(9x - 64)</math>, so the solutions are <math>4</math> and <math>\tfrac{64}{9}</math>.  <math>\tfrac{64}{9}</math> is extraneous, though, so the only solution is <math>\boxed{4}</math>.

Latest revision as of 10:23, 3 July 2011

Problem

AoPSWiki:Problem of the Day/July 2, 2011

Solution

We simplify the LHS to be $\sqrt{x} + \tfrac{3x}{2} + 2 = 10$, so $\sqrt{x} = 8 - \tfrac{3x}{2}$, or $2\sqrt{x} = 16 - 3x$. Now, we square this to get $4x = 9x^2 - 96x + 256$, so $9x^2 - 100x + 256 = 0$. This factors as $(x - 4)(9x - 64)$, so the solutions are $4$ and $\tfrac{64}{9}$. $\tfrac{64}{9}$ is extraneous, though, so the only solution is $\boxed{4}$.

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