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 equations:${\alp

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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 equations:${\alpha^2} - \alpha + k = 0$  $(1)$$(2\alpha)^2 - (2\alpha) + 3k = 0 \Rightarrow 4{\alpha^2} - 2\alpha + 3k = 0$  $(2)$Multiply equation $(1)$ by $2$:$2{\alpha^2} - 2\alpha + 2k = 0$  $(3)$Subtract equation $(3)$ from equation $(2)$:$(4{\alpha^2} - 2\alpha + 3k) - (2{\alpha^2} - 2\alpha + 2k) = 0$$2{\alpha^2} + k = 0 \Rightarrow {\alpha^2} = -k/2$Substitute ${\alpha^2} = -k/2$ into equation $(1)$:$-k/2 - \alpha + k = 0 \Rightarrow \alpha = k/2$Now,substitute $\alpha = k/2$ into ${\alpha^2} = -k/2$:$(k/2)^2 = -k/2$$k^2/4 = -k/2$$k^2 + 2k = 0$$k(k + 2) = 0$Thus,$k = 0$ or $k = -2$. Since $k=0$ makes both equations ${x^2} - x = 0$ (roots $0, 1$),the condition that one root is double the other ($2	imes 0 = 0$ or $2	imes 1 = 2$) is satisfied by $k = -2$.

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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