Let $ \alpha$ and $ \beta$ be the roots of equation $ px^{2}+qx+r=0,p\neq 0$. If $ p,q,r$ are in A.P. and $ \displaystyle \frac{1}{\alpha }+\displaystyle \frac{1}{\beta }=4$, then the value of $ |\alpha -\beta |$ is

Roots alpha, beta satisfy alpha+beta = -q/p and alpha*beta = r/p. 1/alpha + 1/beta = (alpha+beta)/(alpha*beta) = (-q/p) / (r/p) = -q/r = 4. So q = -4r. Since p, q, r are in AP, 2q = p+r. Substitute q: -8r = p+r => p = -9r. Equation: -9rx^2 - 4rx + r = 0. Divide by -r: 9x^2 + 4x - 1 = 0. |alpha-beta| = sqrt((alpha+beta)^2 - 4*alpha*beta) = sqrt((-4/9)^2 - 4*(-1/9)) = sqrt(16/81 + 36/81) = sqrt(52/81) = sqrt(4*13)/9 = 2*sqrt(13)/9.