Mathematics 64 Online
OpenStudy (anonymous):

3r^2+4r+10=0 solve by completing the square r= ??

jimthompson5910 (jim_thompson5910):

3r^2+4r+10=0 3(r^2+4/3r+10/3)=0 3(r^2+4/3r+4/9-4/9+10/3)=0 3((r+2/3)^2-4/9+10/3)=0 3((r+2/3)^2+26/9)=0 3(r+2/3)^2+3(26/9)=0 3(r+2/3)^2+26/3=0 So after completing the square, you should get $\large 3(r+\frac{2}{3})^2+\frac{26}{3}=0$

jimthompson5910 (jim_thompson5910):

$\large 3(r+\frac{2}{3})^2+\frac{26}{3}=0$ $\large 3(r+\frac{2}{3})^2=-\frac{26}{3}$ $\large (r+\frac{2}{3})^2=-\frac{26}{9}$ $\large r+\frac{2}{3}=\pm\sqrt{-\frac{26}{9}}$ $\large r+\frac{2}{3}=\pm\frac{i\sqrt{26}}{3}$ $\large r=-\frac{2}{3}\pm\frac{i\sqrt{26}}{3}$ $\large r=\frac{-2\pm i\sqrt{26}}{3}$ So the solutions are $\large r=\frac{-2+ i\sqrt{26}}{3} \textrm{ or }r=\frac{-2- i\sqrt{26}}{3}$

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