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Verify that roots of the quadratic equation $(p-q)x^2 + (q - r)x + (r - p) = 0$ are equal when $q + r = 2p$.
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Sol. Discriminant $(D) = (q - r)^2 - 4 (p - q) (r - p)$ (I) ($\frac{1}{2}$ Mark)
$= (q + r - 2p)^2$ (II) ($\frac{1}{2}$ Mark)
Substituting, $q + r = 2p$
$\Rightarrow D = (2p - 2p)^2 = 0$ (III) ($\frac{1}{2}$ Mark)
$\therefore$ Roots of the given equation are equal. (IV) ($\frac{1}{2}$ Mark)
$= (q + r - 2p)^2$ (II) ($\frac{1}{2}$ Mark)
Substituting, $q + r = 2p$
$\Rightarrow D = (2p - 2p)^2 = 0$ (III) ($\frac{1}{2}$ Mark)
$\therefore$ Roots of the given equation are equal. (IV) ($\frac{1}{2}$ Mark)