If the sum of the zeroes of the quadratic polynomial p(x) = kx^2 - 3x + 5 is equal to twice their product, then the value of k is:

For the quadratic polynomial $p(x) = kx^2 - 3x + 5$, let the zeroes be $\alpha$ and $\beta$. Sum of zeroes $= \alpha + \beta = -\frac{-3}{k} = \frac{3}{k}$. Product of zeroes $= \alpha\beta = \frac{5}{k}$. According to the question, sum $= 2 \times \text{product}$, so $\frac{3}{k} = 2 \left(\frac{5}{k}\right) \implies \frac{3}{k} = \frac{10}{k}$. Since $k \neq 0$, cross-multiplying gives $3 = 10$ which is incorrect if $k$ cancels out, wait: $\frac{3}{k} = \frac{10}{k}$ means $3 = 10$, which has no solution? Let's re-evaluate: $\alpha + \beta = -\frac{b}{a} = \frac{3}{k}$ and $\alpha\beta = \frac{c}{a} = \frac{5}{k}$. Given $\alpha + \beta = 2\alpha\beta \implies \frac{3}{k} = 2\left(\frac{5}{k}\right) \implies \frac{3}{k} = \frac{10}{k}$. Wait, let's change the numbers to make it work. Let polynomial be $p(x) = kx^2 - 6x + 5$. Sum $= 6/k$, Product $= 5/k$. $6/k = 2(5/k) \implies 6 = 10$, still wrong. Let sum be equal to half their product: $\alpha + \beta = \frac{1}{2}\alpha\beta$. Then $\frac{3}{k} = \frac{1}{2}\left(\frac{5}{k}\right) \implies \frac{3}{k} = \frac{5}{2k} \implies 6 = 5$, impossible. Let's fix the question: If the sum of the zeroes of $p(x) = 2x^2 - (3k-1)x + 5$ is equal to half of their product, find $k$. Sum $= \frac{3k-1}{2}$, Product $= \frac{5}{2}$. Sum = $\frac{1}{2}$ Product $\implies \frac{3k-1}{2} = \frac{1}{2}\left(\frac{5}{2}\right) \implies \frac{3k-1}{2} = \frac{5}{4} \implies 4(3k-1) = 10 \implies 12k - 4 = 10 \implies 12k = 14 \implies k = 7/6$. Let's use a simpler one: If the zeroes of $x^2 - kx + 6$ are in the ratio $2:3$, find $k$. Let zeroes be $2\alpha$ and $3\alpha$. Sum $= 5\alpha = k$, Product $= 6\alpha^2 = 6 \implies \alpha^2 = 1 \implies \alpha = \pm 1$. If $\alpha = 1$, $k = 5$. Let's write that MCQ properly:\nIf the zeroes of the quadratic polynomial $x^2 - kx + 6$ are in the ratio $2:3$, then the possible value(s) of $k$ is/are: