AoPS Wiki talk:Problem of the Day/July 5, 2011

Revision as of 14:48, 5 July 2011 by Parukia911 (talk | contribs) (Solution)
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Problem

AoPSWiki:Problem of the Day/July 5, 2011

Solution

Try the only possible integer roots which are $\pm$ {1,2,3,6,7,14,21,42}

3 works so divide out (x-3) to get $x^2 + 3x + 14$

Use the quadratic formula to find the remaining roots to be $\frac{-3 \pm i\sqrt{47}}{2}$

So, the answers are: $\boxed {3; \frac{-3 \pm i\sqrt{47}}{2}}$.