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If one root of the equation $(1-\mathrm{m}) \mathrm{x}^{2}+1 \mathrm{x}+1=0$ is double the other and 1 is real, then what is the greatest value of $\mathrm{m}$ ?
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The correct answer is:
$\frac{9}{8}$
Given equation is
$(\ell-m) x^{2}+\ell x+1=0$
Roots are $\alpha, \beta$. One root is double the other. $\beta=2 \alpha$
Sum of roots $=\alpha+\beta$
$\begin{array}{ll}3 \alpha=\frac{-\ell}{\ell-m} & \alpha(2 \alpha)=\frac{1}{(\ell-m)} \\ \Rightarrow \alpha^{2}=\frac{\ell^{2}}{9(\ell-m)^{2}} \quad 2 \alpha^{2}=\frac{1}{\ell-m} \\ \Rightarrow 2 \frac{\ell^{2}}{9(\ell-m)^{2}}=\frac{1}{(\ell-m)} \\ \Rightarrow \frac{2 \ell^{2}}{9(\ell-m)}=1 \\ \Rightarrow 2 \ell^{2}=9(\ell-m) \Rightarrow 2 \ell^{2}-9 \ell+9 m=0\end{array}$
For $\ell$ to be real discriminant should be $b^{2}-4 a c \geq 0$ $81-4 \times 2 \times 9 m \geq 0$
$m \leq \frac{9}{8}$
$(\ell-m) x^{2}+\ell x+1=0$
Roots are $\alpha, \beta$. One root is double the other. $\beta=2 \alpha$
Sum of roots $=\alpha+\beta$
$\begin{array}{ll}3 \alpha=\frac{-\ell}{\ell-m} & \alpha(2 \alpha)=\frac{1}{(\ell-m)} \\ \Rightarrow \alpha^{2}=\frac{\ell^{2}}{9(\ell-m)^{2}} \quad 2 \alpha^{2}=\frac{1}{\ell-m} \\ \Rightarrow 2 \frac{\ell^{2}}{9(\ell-m)^{2}}=\frac{1}{(\ell-m)} \\ \Rightarrow \frac{2 \ell^{2}}{9(\ell-m)}=1 \\ \Rightarrow 2 \ell^{2}=9(\ell-m) \Rightarrow 2 \ell^{2}-9 \ell+9 m=0\end{array}$
For $\ell$ to be real discriminant should be $b^{2}-4 a c \geq 0$ $81-4 \times 2 \times 9 m \geq 0$
$m \leq \frac{9}{8}$
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