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 $(p - q) + (q - r) + (r - p) = 0$. If the su

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

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

List-$I$	List-$II$
$A$. All the roots are negative	$I$. $(b - 3)^2 = 36 + P^2$ for $P \in R$
$B$. Two roots are complex	$II$. $-3 < b < 9$
$C$. Two roots are positive	$III$. $b \in (-\infty, -3) \cup (9, \infty)$
$D$. All roots are real and distinct	$IV$. $b = 9$
	$V$. $b = -3$

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

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