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

For the given quadratic equation with equal roots, x = 1 is an obvious root because the sum of the coefficients (p-q) + (q-r) + (r-p) equals zero. Since the roots are equal, both roots must be 1, making the product of the roots 1 * 1 = 1. Using the formula for the product of the roots, which equals the constant term divided by the leading coefficient, we set (r-p)/(p-q) = 1. Solving r - p = p - q yields q + r = 2p.

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