Question

For what values of k, the equation $k{x}^2-6x-2=0$ has real roots?
(a) $k\le \frac{-9}{2}$
(b) $k\ge \frac{-9}{2}$
(c) $k\le -2$
(d) None of these

Answer 1

(b) $k\ge \frac{-9}{2}$

$$\begin{gathered} \text{It is given that the roots of the equation (}k{x}^2-6x-2=0)\text{are real.} \\ \therefore D\ge 0 \\ \Rightarrow (b^2-4ac)\ge 0 \\ \Rightarrow {(-6)}^2-4\times k\times (-2)\ge 0 \\ \Rightarrow 36+8k\ge 0 \\ \Rightarrow k\ge \frac{-36}{8} \\ \Rightarrow k\ge \frac{-9}{2} \end{gathered}$$