Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 2, 2011"
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==Problem== | ==Problem== | ||
− | {{:AoPSWiki:Problem of the Day/July | + | {{:AoPSWiki:Problem of the Day/July 2, 2011}} |
==Solution== | ==Solution== | ||
− | {{ | + | 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 , so , or . Now, we square this to get , so . This factors as , so the solutions are and . is extraneous, though, so the only solution is .