Question

Solve the following quadratic equation by completing square method
$5y^2-4y-1=0$

Answer 1

$5y^2-4y-1=0$
$\Rightarrow y^2-\frac{4}{5}y-\frac{1}{5}=0$
$\Rightarrow y^2-\frac{4}{5}y+\frac{4}{25}-\frac{1}{5}-\frac{4}{25}=0$

$\Rightarrow {(y-\frac{2}{5})}^2-\frac{9}{25}=0$

$\Rightarrow {(y-\frac{2}{5})}^2-{\left(\frac{3}{5}\right)}^2=0$

$\Rightarrow (y-\frac{2}{5}-\frac{3}{5})(y-\frac{2}{5}+\frac{3}{5})=0$

$\Rightarrow (y-1)(y+\frac{1}{5})=0$

$\therefore y-1=0\Rightarrow y=1$

and $y+\frac{1}{5}=0\Rightarrow y=-\frac{1}{5}$