(B) Since $(x-2)$ is a factor of $x^2+ax+b$,we have $x=2$ as a root,so $(2)^2+2a+b=0$,which gives $2a+b=-4$ ... $(i)$.Similarly,since $(x-2)$ is a fac

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(B) Since $(x-2)$ is a factor of $x^2+ax+b$,we have $x=2$ as a root,so $(2)^2+2a+b=0$,which gives $2a+b=-4$ ... $(i)$.Similarly,since $(x-2)$ is a factor of $x^2+cx+d$,we have $(2)^2+2c+d=0$,which gives $2c+d=-4$ ... $(ii)$.Subtracting equation $(ii)$ from equation $(i)$:$(2a+b) - (2c+d) = -4 - (-4)$$2(a-c) + (b-d) = 0$$b-d = -2(a-c)$$b-d = 2(c-a)$Therefore,$\frac{b-d}{c-a} = 2$.

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Condition for common roots

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