What are the roots of the given equation (p - | vedclass

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(C) Given the quadratic equation $(p - q)x^2 + (q - r)x + (r - p) = 0$.Let $f(x) = (p - q)x^2 + (q - r)x + (r - p)$.Observe the sum of the coefficient

Vedclass · Accepted Answer

$\frac{r - p}{p - q}, 1$

Vedclass · Answer

(C) Given the quadratic equation $(p - q)x^2 + (q - r)x + (r - p) = 0$.Let $f(x) = (p - q)x^2 + (q - r)x + (r - p)$.Observe the sum of the coefficients: $(p - q) + (q - r) + (r - p) = p - q + q - r + r - p = 0$.Since the sum of the coefficients is $0$,$x = 1$ is always a root of the equation.Let the roots be $\alpha$ and $\beta$. We know that the product of the roots $\alpha \beta = \frac{c}{a} = \frac{r - p}{p - q}$.Since $\alpha = 1$,we have $1 	imes \beta = \frac{r - p}{p - q}$,so $\beta = \frac{r - p}{p - q}$.Thus,the roots are $1$ and $\frac{r - p}{p - q}$.

What are the roots of the given equation $(p - q)x^2 + (q - r)x + (r - p) = 0$?

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