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If $x^2+3 x-2 k=0$ and $x^2-2 x-7 k=0$ have a non-zero common root, then the positive root of the equation $k x^2+$ $(k+2) \mathrm{x}-(\mathrm{k}+1)=0$ is
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$\frac{3}{5}$
$x^2+3 x-2 K=0$
$\begin{aligned} & x^2-2 x-7 K=0 \\ & \Rightarrow \quad \frac{x^2}{-21 K-4 K}=\frac{x}{-2 K+7 K}=\frac{1}{-2-3} \\ & \Rightarrow \quad \frac{x^2}{-25 K}=\frac{x}{5 K}=\frac{1}{-5} \\ & \therefore \quad x^2=\frac{-25 K}{-5}=5 K \\ & \quad x=\frac{5 K}{-5}=-K \\ & \therefore \quad K^2=5 K \Rightarrow K=0,5 \\ & \therefore \quad K=5(\text { As common root is non zero }) \\ & \Rightarrow \quad K x^2+(K+2) x-(K+1)=0 \Rightarrow 5 x^2+7 x-6=0 \\ & \Rightarrow \quad(5 x-3)(x+2)=0 \Rightarrow x=\frac{3}{5},-2 \\ & \therefore \quad x=\frac{3}{5} .\end{aligned}$
$\begin{aligned} & x^2-2 x-7 K=0 \\ & \Rightarrow \quad \frac{x^2}{-21 K-4 K}=\frac{x}{-2 K+7 K}=\frac{1}{-2-3} \\ & \Rightarrow \quad \frac{x^2}{-25 K}=\frac{x}{5 K}=\frac{1}{-5} \\ & \therefore \quad x^2=\frac{-25 K}{-5}=5 K \\ & \quad x=\frac{5 K}{-5}=-K \\ & \therefore \quad K^2=5 K \Rightarrow K=0,5 \\ & \therefore \quad K=5(\text { As common root is non zero }) \\ & \Rightarrow \quad K x^2+(K+2) x-(K+1)=0 \Rightarrow 5 x^2+7 x-6=0 \\ & \Rightarrow \quad(5 x-3)(x+2)=0 \Rightarrow x=\frac{3}{5},-2 \\ & \therefore \quad x=\frac{3}{5} .\end{aligned}$
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