Question

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

Answer 1

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 \[\large 3(r+\frac{2}{3})^2+\frac{26}{3}=0\]

Answer 2

\[\large 3(r+\frac{2}{3})^2+\frac{26}{3}=0\]


\[\large 3(r+\frac{2}{3})^2=-\frac{26}{3}\]


\[\large (r+\frac{2}{3})^2=-\frac{26}{9}\]


\[\large r+\frac{2}{3}=\pm\sqrt{-\frac{26}{9}}\]


\[\large r+\frac{2}{3}=\pm\frac{i\sqrt{26}}{3}\]


\[\large r=-\frac{2}{3}\pm\frac{i\sqrt{26}}{3}\]


\[\large r=\frac{-2\pm i\sqrt{26}}{3}\]
So the solutions are \[\large r=\frac{-2+ i\sqrt{26}}{3} \textrm{ or }r=\frac{-2- i\sqrt{26}}{3}\]