If the roots of equation $(p - q) x^{2} + (q - r) x + (r - p) = 0$ are equal, find $(q + r)$

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-2p

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Solution

The correct option is B 2p
$b^{2} - 4 a c = 0$ (2 roots are equal)
${(q - r)}^{2} - 4 (p - q) (r - p) = 0$
$q^{2} + r^{2} - 2 p r - 4 p r + 4 p^{2} + 4 q r - 4 q p = 0$
${(q + r - 2 p)}^{2} = 0$
$q + r = 2 p$

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