Question

Find the values of $k$ for which the given equation has real roots
$k{x}^2-6x-2=0$

Answer 1

We have $k{x}^2-6x-2=0$

Here, $a=k,b=-6,c=-2$

$\therefore D=b^2-4ac={(-6)}^2-4\times k\times -2=36+8k$

The given equation will have real roots, if

$D\ge 0\Rightarrow 36+8k\ge 0\Rightarrow k\ge \frac{-36}{8}\Rightarrow k\ge -\frac{9}{2}$