Question 463055
Note that the roots of the quadratic are in the form


*[tex \LARGE r_1, r_2 = \frac{-(5p+1) \pm \sqrt{(5p+1)^2 - 4(4)(5p)}}{8}  = \frac{-5p - 1}{8} \pm \frac{\sqrt{25p^2 - 70p + 1}}{8}] (upon simplification). If we want these roots to differ by 1, we need to set the equality


*[tex \LARGE \frac{\sqrt{25p^2 - 70p + 1}}{8} = \frac{1}{2}]


This is so that the roots will be in the form k + (1/2) and k - (1/2), differing by 1. Solving this equation for p,


*[tex \LARGE \sqrt{25p^2 - 70p + 1} = 4]


*[tex \LARGE 25p^2 - 70p + 1 = 16]


*[tex \LARGE 25p^2 - 70p - 15 = 0] Luckily, this factors:


*[tex \LARGE 5(5p + 1)(p - 3) = 0]


p = -1/5 or p = 3.