The roots of the equation 7 x^2-6 x+1=0 are α and β , where 2 α and...

Since $\tan \alpha, \tan \beta$ are roots of $7 x^2-6 x+1=0\therefore \text{ Sum of roots }=\tan \alpha+\tan \beta=\frac{6}{7}$...(i) Product of roots $=\tan \alpha \cdot \tan \beta=\frac{1}{7}$....(ii) $\because \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}=\frac{\left(\frac{6}{7}\right)}{1-\frac{1}{7}}=1\Rightarrow \alpha+\beta=45^{{}^{\circ}} \Rightarrow 2 \alpha+2 \beta=90^{{}^{\circ}}$ Which means $3^{\text{rd }}$ angle of triangle is $90^{{}^{\circ}}$. Hence, triangle is right-angled triangle. Now consider, $$\begin{aligned} & (\tan \alpha-\tan \beta)=\sqrt{(\tan \alpha+\tan \beta)^2-4 \tan \alpha \cdot \tan \beta} \\ & =\sqrt{\left(\frac{6}{7}\right)^2-4\left(\frac{1}{7}\right)}=\frac{2 \sqrt{2}}{7} \end{aligned}$$ Thus, $\tan (\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \cdot \tan \beta}=\frac{1}{2 \sqrt{2}}$ and $(\alpha-\beta)=\frac{2 \tan (\alpha-\beta)}{1-\tan ^2(\alpha-\beta)}=\left(\frac{4 \sqrt{2}}{7}\right)\Rightarrow 2 \alpha \neq 2 \beta \Rightarrow$ triangle is not an isosceles triangles.