If the equations 2ax^2 - 3bx + 4c = 0 and 3x^ | vedclass

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(C) Given the equations $2ax^2 - 3bx + 4c = 0$ and $3x^2 - 4x + 5 = 0$.Since the second equation $3x^2 - 4x + 5 = 0$ has a discriminant $D = (-4)^2 -

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$\frac{34}{15}$

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(C) Given the equations $2ax^2 - 3bx + 4c = 0$ and $3x^2 - 4x + 5 = 0$.Since the second equation $3x^2 - 4x + 5 = 0$ has a discriminant $D = (-4)^2 - 4(3)(5) = 16 - 60 = -44 If two quadratic equations have a common root and their coefficients are real,then both roots must be common.Thus,the ratios of the coefficients must be equal:$\frac{2a}{3} = \frac{-3b}{-4} = \frac{4c}{5} = k$$\frac{2a}{3} = k \Rightarrow a = \frac{3k}{2}$$\frac{3b}{4} = k \Rightarrow b = \frac{4k}{3}$$\frac{4c}{5} = k \Rightarrow c = \frac{5k}{4}$Now,calculate $\frac{a + b}{c}$:$\frac{a + b}{c} = \frac{\frac{3k}{2} + \frac{4k}{3}}{\frac{5k}{4}} = \frac{\frac{9k + 8k}{6}}{\frac{5k}{4}} = \frac{17k}{6} 	imes \frac{4}{5k} = \frac{17 	imes 2}{3 	imes 5} = \frac{34}{15}$.

If the equations $2ax^2 - 3bx + 4c = 0$ and $3x^2 - 4x + 5 = 0$ have a common root,then $\left( \frac{a + b}{c} \right)$ is equal to,where $a, b, c \in \mathbb{R}$.

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