Question

5x²-6x-2=0 find the root by complete square method​

Answer 1

$\huge\mathfrak\red{Solution:-}$

$5{x}^{2} - 6x - 2 = \\ \\ Dividing\: 5 \: on \: both \: sides\\ \\ \frac{5{x}^{2}}{5} - \frac{6x}{5} - \frac{2}{5} = 0 \\ \\ {x}^{2} - \frac{6x}{5} = \frac{2}{5} \\ \\ {x}^{2} - 2x × \frac{3}{5} = \frac{2}{5} \\ \\adding\: ({\frac{3}{5})}^{2} \: on \: both \: sides\\ \\ {x}^{2} - 2x × \frac{3}{5} + {(\frac{3}{5})}^{2} = \frac{2}{5} + {(\frac{3}{5})}^{2} \\ \\ {(x -\frac{3}{5})}^{2} = \frac{2}{5} + \frac{9}{25} \\ \\ {(x - \frac{3}{5})}^{2} = \frac{10+9}{25} \\ \\ {(x - \frac{3}{5})}^{2} = \frac{19}{25} \\ \\ (x - \frac{3}{5}) = \sqrt{\frac{19}{25}} \\ \\ ( x -\frac{3}{5}) = \frac{\sqrt{19}}{5}\\ \\ Taking \: + \\ \\ x - \frac{3}{5} = + \frac{\sqrt{19}}{5}\\ \\ x = \frac{\sqrt{19}}{5} + \frac{3}{5} \\ \\ x = \frac{\sqrt{19} + 3}{5} \\ \\ Taking\: - \\ \\ x - \frac{3}{5} = -\frac{\sqrt{19}}{5} \\ \\ x = \frac{\sqrt{19}}{5} + \frac{3}{5} \\ \\ x = \frac{-\sqrt{19} + 3 }{5}$