The value of k for which one of the roots of | vedclass

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(B) Let $\alpha$ be a root of $x^2 - x + k = 0$. Then,$2\alpha$ is a root of $x^2 - x + 3k = 0$.Substituting these roots into the respective equations

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$-2$

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(B) Let $\alpha$ be a root of $x^2 - x + k = 0$. Then,$2\alpha$ is a root of $x^2 - x + 3k = 0$.Substituting these roots into the respective equations:$1) \alpha^2 - \alpha + k = 0$$2) (2\alpha)^2 - (2\alpha) + 3k = 0 \Rightarrow 4\alpha^2 - 2\alpha + 3k = 0$Multiply equation $(1)$ by $2$: $2\alpha^2 - 2\alpha + 2k = 0$.Subtract this from equation $(2)$:$(4\alpha^2 - 2\alpha + 3k) - (2\alpha^2 - 2\alpha + 2k) = 0 - 0$$2\alpha^2 + k = 0 \Rightarrow \alpha^2 = -k/2$.Substitute $\alpha^2 = -k/2$ into equation $(1)$:$-k/2 - \alpha + k = 0 \Rightarrow \alpha = k/2$.Since $\alpha^2 = (k/2)^2$,we have $-k/2 = k^2/4$.$k^2/4 + k/2 = 0 \Rightarrow k^2 + 2k = 0$.$k(k + 2) = 0$,so $k = 0$ or $k = -2$.Since $k=0$ leads to trivial roots $(0, 0)$,the non-zero value is $k = -2$.

The value of $k$ for which one of the roots of $x^2 - x + 3k = 0$ is double of one of the roots of $x^2 - x + k = 0$ is

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