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If one root of the equation \((l-\mathrm{m}) \mathrm{x}^2+l \mathrm{x}+1=0\) is double of the other and if \(l\) is real then the greatest value of \(\mathrm{m}\) is \((l \neq \mathrm{m})\) :
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The correct answer is:
\(\frac{9}{8}\)
Let the roots of the equation be \(\alpha\) and \(2 \alpha\), then
\(\alpha+2 \alpha=\frac{l}{\mathrm{~m}-l} \text { and } \alpha \cdot 2 \alpha=\frac{1}{l-\mathrm{m}}\)
\(\Rightarrow \alpha=\frac{l}{3(\mathrm{~m}-l)} \text { and } \alpha^2=\frac{1}{2(l-\mathrm{m})} \text {. }\)
Eliminating \(\alpha\), we get
\(\frac{l^2}{9(\mathrm{~m}-l)^2}=\frac{1}{2(l-\mathrm{m})} \Rightarrow 2 l^2-9 l+9 \mathrm{~m}=0\)
Since \(l\) is real, so \((-9)^2-4 \cdot 2 \cdot 9 \mathrm{~m} \geq 0\)
\(\Rightarrow \mathrm{m} \leq \frac{9}{8}\)
\(\alpha+2 \alpha=\frac{l}{\mathrm{~m}-l} \text { and } \alpha \cdot 2 \alpha=\frac{1}{l-\mathrm{m}}\)
\(\Rightarrow \alpha=\frac{l}{3(\mathrm{~m}-l)} \text { and } \alpha^2=\frac{1}{2(l-\mathrm{m})} \text {. }\)
Eliminating \(\alpha\), we get
\(\frac{l^2}{9(\mathrm{~m}-l)^2}=\frac{1}{2(l-\mathrm{m})} \Rightarrow 2 l^2-9 l+9 \mathrm{~m}=0\)
Since \(l\) is real, so \((-9)^2-4 \cdot 2 \cdot 9 \mathrm{~m} \geq 0\)
\(\Rightarrow \mathrm{m} \leq \frac{9}{8}\)
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