If x^2+3x-2k=0 and x^2-2x-7k=0 have a non-zer | vedclass

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(D) Let $\alpha$ be the non-zero common root of the equations $x^2+3x-2k=0$ and $x^2-2x-7k=0$. Subtracting the two equations: $(x^2+3x-2k) - (x^2-2x-7

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$\frac{3}{5}$

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(D) Let $\alpha$ be the non-zero common root of the equations $x^2+3x-2k=0$ and $x^2-2x-7k=0$. Subtracting the two equations: $(x^2+3x-2k) - (x^2-2x-7k) = 0$ $\Rightarrow 5x+5k=0$ $\Rightarrow x=-k$. Since $x=-k$ is a root,substitute it into the first equation: $(-k)^2+3(-k)-2k=0$ $\Rightarrow k^2-5k=0$ $\Rightarrow k(k-5)=0$. Since the root is non-zero,$k 
eq 0$,so $k=5$. Substituting $k=5$ into the third equation: $5x^2+(5+2)x-(5+1)=0 \Rightarrow 5x^2+7x-6=0$. Factoring the quadratic: $5x^2+10x-3x-6=0$ $\Rightarrow 5x(x+2)-3(x+2)=0$ $\Rightarrow (5x-3)(x+2)=0$. The roots are $x=\frac{3}{5}$ and $x=-2$. The positive root is $x=\frac{3}{5}$.

If $x^2+3x-2k=0$ and $x^2-2x-7k=0$ have a non-zero common root,then the positive root of the equation $kx^2+(k+2)x-(k+1)=0$ is

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