Find the values of m for which both roots of equation x² - mx + 1 = 0 are less than unity

Let fix)= x² - mx + 1 Apply

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{D}\ge 0 \\ \begin{array}{l}\Rightarrow {\mathrm{m}}^2-4\ge 0\\ \Rightarrow (\mathrm{m}+2)(\mathrm{m}-2)\ge 0\\ \Rightarrow \mathrm{m}\le -2\text{ and }\mathrm{m}\ge 2\end{array} \end{gathered}$$

$\begin{array}{l}\left(\mathrm{ii}\right) \mathrm{af}\left(1\right)>\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(1-\mathrm{m}+1)<0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{m}<2\end{array}$

$\text{ (iii) }\alpha +\beta <2\Rightarrow m<2$

From (i), (ii) and (iii), we get

$m\le -2$