If the difference between the roots of the eq | vedclass

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(B) Let the roots of the quadratic equation be $\alpha$ and $\beta$. Given $|\alpha - \beta| = 3$.For the equation $(k - 2)x^2 - (k - 4)x - 2 = 0$, th

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$3, 3/2$

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(B) Let the roots of the quadratic equation be $\alpha$ and $\beta$. Given $|\alpha - \beta| = 3$.For the equation $(k - 2)x^2 - (k - 4)x - 2 = 0$, the sum of roots $\alpha + \beta = \frac{k - 4}{k - 2}$ and the product of roots $\alpha \beta = \frac{-2}{k - 2}$.We know that $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$.Substituting the values: $3^2 = \left(\frac{k - 4}{k - 2}\right)^2 - 4\left(\frac{-2}{k - 2}\right)$.$9 = \frac{(k - 4)^2 + 8(k - 2)}{(k - 2)^2}$.$9(k - 2)^2 = k^2 - 8k + 16 + 8k - 16$.$9(k^2 - 4k + 4) = k^2$.$9k^2 - 36k + 36 = k^2$.$8k^2 - 36k + 36 = 0$.Dividing by $4$: $2k^2 - 9k + 9 = 0$.$2k^2 - 6k - 3k + 9 = 0 \Rightarrow 2k(k - 3) - 3(k - 3) = 0$.$(2k - 3)(k - 3) = 0$.Thus, $k = 3$ or $k = 3/2$.

If the difference between the roots of the equation $(k - 2)x^2 - (k - 4)x - 2 = 0$ is $3$, find the value of $k$.

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