if one root of (l-m)x2+lx+1=0 be double of the other and if l be real, show that m≤9/8
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1708$Let $\mathbf{\alpha,2\alpha}$ be the Roots of the equation $\mathbf{(l-m)x^2+lx+1=0}$.Then \\\\ $\mathbf{\alpha+2\alpha=-\frac{-l}{(l-m)}\Leftrightarrow \alpha=-\frac{l}{3.(l-m)}\Leftrightarrow \alpha^2=\frac{l^2}{9.(l-m)^2}.............................(1)}$\\\\ $\mathbf{\alpha.2\alpha=\frac{1}{(l-m)}\Leftrightarrow \alpha^2=\frac{1}{2.(l-m)}..........................(2)}$\\\\ Now equating value of $\mathbf{\alpha^2}$ from $\mathbf{(1)}$ and $\mathbf{(2)},$ We Get\\\\ $\mathbf{\frac{l^2}{9.(l-m)^2}=\frac{1}{2.(l-m)}\Leftrightarrow 2l^2=9(l-m)}$\\\\\\ OR $\mathbf{2l^2-9l+9m=0}$\\\\ Now Given equation has Real Roots. So its Discriminant i.e $D\geq 0$\\\\ So $\mathbf{81-4.2.9m\geq 0\Leftrightarrow 9-8m\geq 0\Leftrightarrow m\leq \frac{9}{8}}$\\\\ So $\boxed{\boxed{\mathbf{m\leq \frac{9}{8}}}}$