Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 5, 2011"

Latest revision as of 15:48, 5 July 2011

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: $\boxed {3; \frac{-3 \pm i\sqrt{47}}{2}}$ .

Revision as of 13:54, 5 July 2011 (view source) Ahaanomegas (talk \| contribs) m (→‎Solution) ← Older edit		Latest revision as of 15:48, 5 July 2011 (view source) Parukia911 (talk \| contribs) (→‎Solution)
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	Use the quadratic formula to find the remaining roots to be <math>\frac{-3 \pm i\sqrt{47}}{2}</math>		Use the quadratic formula to find the remaining roots to be <math>\frac{-3 \pm i\sqrt{47}}{2}</math>

−	So, the answers are: <math> \boxed {-3; \frac{-3 \pm i\sqrt{47}}{2}} </math>.	+	So, the answers are: <math> \boxed {3; \frac{-3 \pm i\sqrt{47}}{2}} </math>.