Let a, b, c, d in mathbb{R}. If the equations | vedclass

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(A) Let the common root be $\alpha$. Then we have: $2b\alpha^2 + 3c\alpha - d = 0$ $2a\alpha^2 + 3b\alpha + 4c = 0$ Using the method of cross-multipli

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$\frac{9}{2}$

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(A) Let the common root be $\alpha$. Then we have: $2b\alpha^2 + 3c\alpha - d = 0$ $2a\alpha^2 + 3b\alpha + 4c = 0$ Using the method of cross-multiplication for the system: $\frac{\alpha^2}{12c^2 + 3bd} = \frac{-\alpha}{8bc + 2ad} = \frac{1}{6b^2 - 6ac}$ From the first and second ratios: $\alpha = -\frac{12c^2 + 3bd}{8bc + 2ad}$ From the second and third ratios: $\alpha = -\frac{8bc + 2ad}{6b^2 - 6ac}$ Equating the two expressions for $\alpha$: $\frac{12c^2 + 3bd}{8bc + 2ad} = \frac{8bc + 2ad}{6(b^2 - ac)}$ $3(4c^2 + bd) \cdot 6(b^2 - ac) = (2(4bc + ad))^2$ $18(4c^2 + bd)(b^2 - ac) = 4(4bc + ad)^2$ $\frac{4bc + ad}{\frac{18}{4}(b^2 - ac)} = \frac{4c^2 + bd}{4bc + ad}$ $\frac{4bc + ad}{\frac{9}{2}(b^2 - ac)} = \frac{4c^2 + bd}{4bc + ad}$ Comparing this with the given equation,we get $k = \frac{9}{2}$.

Let $a, b, c, d \in \mathbb{R}$. If the equations $2bx^2 + 3cx - d = 0$ and $2ax^2 + 3bx + 4c = 0$ have a common root and $\frac{4bc + ad}{k(b^2 - ac)} = \frac{bd + 4c^2}{4bc + ad}$,then $k =$

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