if the difference between the roots of x^2+ax+1=0 is less than root 5 then the set of possible values of a is

Hi Ishitta,
Please find below the solution to the asked query:

$$\begin{gathered} {\mathrm{x}}^2+\mathrm{ax}+1=0 \\ \mathrm{Let} \mathrm{roots} \mathrm{of} \mathrm{this} \mathrm{equation} \mathrm{be} \mathrm{\alpha },\mathrm{\beta }. \\ \Rightarrow \mathrm{Sum} \mathrm{of} \mathrm{roots}=\mathrm{\alpha }+\mathrm{\beta }=-\mathrm{a} \\ \mathrm{and} \mathrm{product} \mathrm{of} \mathrm{roots}=\mathrm{\alpha \beta }=1 \\ \mathrm{Given} \mathrm{that}, \\ \left|\mathrm{\alpha }-\mathrm{\beta }\right|<\sqrt{5} \\ \mathrm{Squaring} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get}, \\ \Rightarrow {\left(\mathrm{\alpha }-\mathrm{\beta }\right)}^2<5 \\ \mathrm{Now} \mathrm{we} \mathrm{can} \mathrm{write} {\left(\mathrm{a}-\mathrm{b}\right)}^2={\left(\mathrm{a}+\mathrm{b}\right)}^2-4\mathrm{ab} \\ \Rightarrow {\left(\mathrm{\alpha }+\mathrm{\beta }\right)}^2-4\mathrm{\alpha \beta }<5 \\ \Rightarrow {\left(-\mathrm{a}\right)}^2-4<5 \\ \Rightarrow {\mathrm{a}}^2-4-5<0 \\ \Rightarrow {\mathrm{a}}^2-9<0 \\ \Rightarrow {\mathrm{a}}^2-{\left(3\right)}^2<0 \\ \mathrm{Using} \mathrm{identity} {\mathrm{a}}^2-{\mathrm{b}}^2=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right), \mathrm{we} \mathrm{get}, \\ \left(\mathrm{a}-3\right)\left(\mathrm{a}+3\right)<0 \\ \mathbf{\Rightarrow }\mathbf{a}\mathbf{\in }\left(\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{ }\mathbf{3}\right)\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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