AoPS Wiki talk:Problem of the Day/July 2, 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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AoPS Wiki talk:Problem of the Day/July 2, 2011

Problem

Solution