Ответ:

[tex]3 {x}^{2} - 2px - p + x = 0 \\ 3 {x}^{2} - x(2p - 1) - p = 0 \\ d = {(2p - 1)}^{2} + 12p = 4 {p }^{2} - 4p + 1 + 12p = 4 {p}^{2} + 8p + 1 \\ 4 {p}^{2} + 8p + 1 > 0 \\ d = 64 - 16 = 48 \\ p1 = \frac{ - 8 + 4 \sqrt{3} }{8} = \frac{ - 2 + \sqrt{3} }{2} \\ p2 = \frac{ - 8 - 4 \sqrt{3} }{8} = \frac{ - 2 - \sqrt{3} }{2} \\ 4(p - ( \frac{ - 2 + \sqrt{3} }{2} ))(p - ( \frac{ - 2 - \sqrt{3} }{2} )) > 0 \\ (p - \frac{ \sqrt{3} - 2}{2} )(p + \frac{ \sqrt{3} + 2 }{2} ) > 0 \\ ( - \infty \: \: \: \: .... \: \: \: \frac{ \sqrt{3} - 2 }{2} )u( \frac{ \sqrt{3} + 2}{2} \: \: \: \: .... \: \: \: \: + \infty )[/tex]