Question

Find the value of $k$ so that the following equation has equal roots.
$2x^2-(k-2)x+1=0$

Answer 1

The given equation has equal roots

$\Rightarrow$ Discriminant$=b^2-4ac=0$

where $a=2,b-(k-2),c=-1$

$\Rightarrow {(k-2)}^2-4\times 2\times 1=0$

$\Rightarrow {(k-2)}^2=8$

$\Rightarrow k-2=\pm \sqrt{8}=\pm 2\sqrt{2}$

$\therefore k=2\pm 2\sqrt{2}$