(x - p)(2x + 1) = 0 2x2 + x - 2px - p = 0 2x2 + x (1 - 2p) - p = 0 2x2 - (2p - 1)x - p = 0 divide through by 2 x2 - \(\frac{(2p - 1)}{2}\)x - \(\frac{p}{2}\) = 0 compare to x2 - (sum of roots)x + product of roots = 0 sum of roots = \(\frac{2p - 1}{2}\) But sum of roots = 1 Given; \(\frac{2p - 1}{2}\) = 1 2p - 1 = 2 x 1 2p - 1 = 2 2p = 2 + 1 = 3 p = \(\frac{3}{2}\) p = 1\(\frac{1}{2}\)