The roots of the given equation (p - q){x^2} | vedclass

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(C) Given equation is $(p - q)x^2 + (q - r)x + (r - p) = 0$.Notice that the sum of the coefficients is $(p - q) + (q - r) + (r - p) = p - q + q - r +

Vedclass · Accepted Answer

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

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(C) Given equation is $(p - q)x^2 + (q - r)x + (r - p) = 0$.Notice that the sum of the coefficients is $(p - q) + (q - r) + (r - p) = p - q + q - r + r - p = 0$.If the sum of the coefficients of a quadratic equation $ax^2 + bx + c = 0$ is zero,then $x = 1$ is always one of the roots.Let the roots be $\alpha$ and $\beta$. We know that the product of the roots $\alpha \cdot \beta = \frac{c}{a}$.Here,$\alpha = 1$,so $1 \cdot \beta = \frac{r - p}{p - q}$.Therefore,the roots are $1$ and $\frac{r - p}{p - q}$.

The roots of the given equation $(p - q){x^2} + (q - r)x + (r - p) = 0$ are

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