The roots of the equation (p - q)x^2 + (q - r | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given the quadratic equation $(p - q)x^2 + (q - r)x + (r - p) = 0$.Observe that the sum of the coefficients is $(p - q) + (q - r) + (r - p) = 0$.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 $x_1$ and $x_2$. We know that the product of the roots is given by $\frac{c}{a}$.Therefore,$1 	imes x_2 = \frac{r - p}{p - q}$.Thus,the roots are $1$ and $\frac{r - p}{p - q}$.

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

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