Question

Find the values of k for which the equation has real and distinct roots.

Answer 1

$$\begin{gathered} \text{Given:} \\ k{x}^2+2x+1=0 \\ \text{Here,} \\ a=k,b=2\mathrm{and}c=1 \\ \mathrm{Discriminant}D\mathrm{is}\mathrm{given}\mathrm{by}: \\ D=(b^2-4ac) \\ \text{=}{\left(2\right)}^2-4\times k\times 1 \\ \text{= 4}-4k \\ \mathrm{If}D>0,\mathrm{t}\text{he roots of the equation will be real and distinct.} \\ \Rightarrow 4-4k>0 \\ \Rightarrow 4(1-k)>0 \\ \Rightarrow (1-k)>0 \\ \Rightarrow k<1 \\ \mathrm{Thus},\mathrm{the}\mathrm{equation}\mathrm{has}\mathrm{real}\mathrm{and}\mathrm{distinct}\mathrm{roots}\mathrm{for}\mathrm{all}\mathrm{real}\mathrm{values}\mathrm{of}\mathrm{k}<1. \end{gathered}$$