Question

Show that the equation $x^2+ax-1=0$ has real and distinct roots for all real values of a.

Answer 1

$$\begin{gathered} \text{Given:} \\ {x}^2+ax-1=0 \\ \text{Here,} \\ a=1,b=a\mathrm{and}c=-1 \\ \mathrm{Discriminant}D\mathrm{is}\mathrm{given}\mathrm{by}: \\ D=(b^2-4ac) \\ \Rightarrow D=a^2-4\times 1\times (-1) \\ \Rightarrow D=(a^2+4) \\ Forallrealvaluesofa,wehave{a}^2>0or,a^2+4>0 \\ \therefore D>0\text{for all real values of}a. \\ \text{Thus, the roots of the equation are real and distinct for all real values of}a. \end{gathered}$$