Question

of the equation
$5 {x}^{2} - 6 {x }^{2} - 2$
by the method of completing the square number​

Answer 1

Step-by-step explanation:

$5 {x}^{2} - 6x - 2 = 0 \\ \\ dividing \: 5 \: on \: both \: sides \\ \\ \frac{5 {x}^{2} }{5} - \frac{6x}{5} - \frac{2}{5} = 0 \\ \\ {x}^{2} - \frac{6x}{5} = \frac{2}{5} \\ \\ {x}^{2} - 2x \times \frac{3}{5} = \frac{2}{5} \\ \\ adding \: {( \frac{3}{5}) }^{2} on \: both \: sides \\ \\ {x}^{2} - 2x \times \frac{3}{5} + { (\frac{3}{5}) }^{2} = \frac{2}{5} + {( \frac{3}{5} )}^{2} \\ \\ {(x - \frac{3}{5}) }^{2} = \frac{2}{5} + \frac{9}{25} \\ \\ {(x - \frac{3}{5} )}^{2} = \frac{10 + 9}{25} \\ \\ {(x - \frac{3}{5} )}^{2} = \frac{19}{25} \\ \\ x - \frac{3}{5} = \sqrt{ \frac{19}{25} } \\ \\ x - \frac{3}{5} = \frac{ \sqrt{19} }{5} \\ \\ on \: taking \: + \\ \\ x - \frac{3}{5} = + \frac{ \sqrt{19} }{5} \\ \\ x = \frac{ \sqrt{19} }{5} + \frac{3}{5} \\ \\ x = \frac{ \sqrt{19} + 3}{5} \\ \\ on \: taking \: - \\ \\ x - \frac{3}{5} = - \frac{ \sqrt{19} }{5} \\ \\ x = - \frac{ \sqrt{19} }{5} + \frac{3}{5} \\ \\ x = \frac{ - \sqrt{19} + 3}{5} \\ \\ \\ \\ here \: is \: your \: solution.$