Question

Solve by the method of completing the square $5x^2-6x-2=0$

• A. $\frac{3+\sqrt{19}}{5}$ $and$ $\frac{3-\sqrt{19}}{5}$
• B. $\frac{-3+\sqrt{19}}{5}$ $and$ $\frac{3-\sqrt{19}}{5}$
  
• C. $\frac{3+\sqrt{19}}{3}$ $and$ $\frac{3-\sqrt{19}}{5}$
• D. $\frac{3+\sqrt{19}}{5}$ $and$ $\frac{-3-\sqrt{19}}{3}$

Answer 1

The correct option is A $\frac{3+\sqrt{19}}{5}$ $and$ $\frac{3-\sqrt{19}}{5}$

$5x^2-6x-2=0x^2-\left(\frac{6}{5}\right)x-\left(\frac{2}{5}\right)=0(x-(\frac{3}{5}){)}^2-(\frac{3}{5}{)}^2-\left(\frac{2}{5}\right)=0(x-(\frac{3}{5}){)}^2-(\frac{9}{25})-(\frac{2}{5})=0(x-\left(\frac{3}{5}\right){)}^2=\left(\frac{9+10}{25}\right)(x-(\frac{3}{5}\left)\right)=\pm \sqrt{\left(\frac{19}{25}\right)}x=\left(\frac{3}{5}\right)\pm \sqrt{\left(\frac{19}{25}\right)}\therefore x=\left(\frac{3+\sqrt{19}}{5}\right)or\left(\frac{3-\sqrt{19}}{5}\right)$