The roots of the equation a(b-c)x^2 + b(c-a)x | vedclass

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(C) Given the quadratic equation $a(b-c)x^2 + b(c-a)x + c(a-b) = 0$. Sum of coefficients $= a(b-c) + b(c-a) + c(a-b) = ab - ac + bc - ba + ca - cb = 0

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$\frac{c(a-b)}{a(b-c)}, 1$

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(C) Given the quadratic equation $a(b-c)x^2 + b(c-a)x + c(a-b) = 0$. Sum of coefficients $= a(b-c) + b(c-a) + c(a-b) = ab - ac + bc - ba + ca - cb = 0$. Since the sum of coefficients is $0$,$x = 1$ is a root of the equation. Let the roots be $\alpha$ and $\beta$. We know $\beta = 1$. From the product of roots formula,$\alpha 	imes \beta = \frac{	ext{constant term}}{	ext{coefficient of } x^2} = \frac{c(a-b)}{a(b-c)}$. Therefore,$\alpha 	imes 1 = \frac{c(a-b)}{a(b-c)}$. The roots are $1$ and $\frac{c(a-b)}{a(b-c)}$.

The roots of the equation $a(b-c)x^2 + b(c-a)x + c(a-b) = 0$ are

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