If alpha, beta (alpha beta) are two values | vedclass

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(D) For two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to be perpendicular,the product of their slopes must be $-1$,or $a_1a_2 + b_1b_2

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$10$

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(D) For two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to be perpendicular,the product of their slopes must be $-1$,or $a_1a_2 + b_1b_2 = 0$.Here,$a_1 = 2, b_1 = (3 - 2k)$ and $a_2 = k, b_2 = (k - 1)$.Applying the condition $a_1a_2 + b_1b_2 = 0$:$2(k) + (3 - 2k)(k - 1) = 0$$2k + (3k - 3 - 2k^2 + 2k) = 0$$-2k^2 + 7k - 3 = 0$$2k^2 - 7k + 3 = 0$$(2k - 1)(k - 3) = 0$Thus,$k = 3$ or $k = \frac{1}{2}$.Given $\alpha > \beta$,we have $\alpha = 3$ and $\beta = \frac{1}{2}$.Therefore,$\alpha^2 + 2\beta = (3)^2 + 2(\frac{1}{2}) = 9 + 1 = 10$.

If $\alpha, \beta$ $(\alpha > \beta)$ are two values of $k$ such that the equations $2x + (3 - 2k)y + (2k + 1) = 0$ and $kx + (k - 1)y - 4 = 0$ represent two perpendicular lines,then $\alpha^2 + 2\beta =$

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