If alpha neq 0 and 0 are the roots of the equ | vedclass

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(C) Given the quadratic equation $x^2 - 5kx + (6k^2 - 2k) = 0$. Since $0$ is a root of the equation,it must satisfy the equation: $(0)^2 - 5k(0) + (6k

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$\frac{5}{3}$

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(C) Given the quadratic equation $x^2 - 5kx + (6k^2 - 2k) = 0$. Since $0$ is a root of the equation,it must satisfy the equation: $(0)^2 - 5k(0) + (6k^2 - 2k) = 0$ $6k^2 - 2k = 0$ $2k(3k - 1) = 0$ This gives $k = 0$ or $k = \frac{1}{3}$. If $k = 0$,the equation becomes $x^2 = 0$,which has roots $0, 0$. Since $\alpha 
eq 0$,$k$ cannot be $0$. If $k = \frac{1}{3}$,the equation becomes $x^2 - 5(\frac{1}{3})x + (6(\frac{1}{3})^2 - 2(\frac{1}{3})) = 0$ $x^2 - \frac{5}{3}x + (6(\frac{1}{9}) - \frac{2}{3}) = 0$ $x^2 - \frac{5}{3}x + (\frac{2}{3} - \frac{2}{3}) = 0$ $x^2 - \frac{5}{3}x = 0$ $x(x - \frac{5}{3}) = 0$ The roots are $0$ and $\frac{5}{3}$. Since $\alpha$ is the non-zero root,$\alpha = \frac{5}{3}$.

If $\alpha \neq 0$ and $0$ are the roots of the equation $x^2 - 5kx + (6k^2 - 2k) = 0$,then $\alpha = $

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