The value of k so that the equation 2x^2 + 5x | vedclass

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(A) The general equation of the second degree $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if the determinant $\Delta

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$4$

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(A) The general equation of the second degree $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if the determinant $\Delta = \begin{vmatrix} a & h & g \ h & b & f \ g & f & c \end{vmatrix} = 0$.Comparing the given equation $2x^2 + 5xy + 3y^2 + 6x + 7y + k = 0$ with the general form,we have $a = 2, b = 3, c = k, h = 5/2, g = 3, f = 7/2$.Substituting these values into the determinant condition:$\begin{vmatrix} 2 & 5/2 & 3 \ 5/2 & 3 & 7/2 \ 3 & 7/2 & k \end{vmatrix} = 0$Multiplying rows/columns to simplify:$\frac{1}{4} \begin{vmatrix} 4 & 5 & 6 \ 5 & 6 & 7 \ 6 & 7 & 2k \end{vmatrix} = 0$Expanding the determinant:$4(6(2k) - 49) - 5(5(2k) - 42) + 6(35 - 36) = 0$$4(12k - 49) - 5(10k - 42) - 6 = 0$$48k - 196 - 50k + 210 - 6 = 0$$-2k + 8 = 0$$2k = 8 \implies k = 4$.

The value of $k$ so that the equation $2x^2 + 5xy + 3y^2 + 6x + 7y + k = 0$ represents a pair of straight lines is:

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