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

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(C) Let $\alpha$ be the common root of the equations $2ax^2 - 3bx + 4c = 0$ and $3x^2 - 4x + 5 = 0$.Since $\alpha$ is a root,we have $2a\alpha^2 - 3b\

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$\frac{34}{31}$

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(C) Let $\alpha$ be the common root of the equations $2ax^2 - 3bx + 4c = 0$ and $3x^2 - 4x + 5 = 0$.Since $\alpha$ is a root,we have $2a\alpha^2 - 3b\alpha + 4c = 0$ and $3\alpha^2 - 4\alpha + 5 = 0$.Comparing the ratios of the coefficients for the common root $\alpha$,we have $\frac{2a}{3} = \frac{-3b}{-4} = \frac{4c}{5} = k$.This gives $2a = 3k$,$3b = 4k$,and $4c = 5k$.Thus,$a = \frac{3k}{2}$,$b = \frac{4k}{3}$,and $c = \frac{5k}{4}$.Now,calculate $\frac{a+b}{b+c} = \frac{\frac{3k}{2} + \frac{4k}{3}}{\frac{4k}{3} + \frac{5k}{4}}$.Simplifying the numerator: $\frac{9k + 8k}{6} = \frac{17k}{6}$.Simplifying the denominator: $\frac{16k + 15k}{12} = \frac{31k}{12}$.Therefore,$\frac{a+b}{b+c} = \frac{17k}{6} 	imes \frac{12}{31k} = \frac{17 	imes 2}{31} = \frac{34}{31}$.

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

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