One problem leads to another!

First, we find the value of p. We get: -

\ln { (100\quad -\quad { p }^{ 2 } } )\quad =\quad \ln { \quad 64 } \ 100\quad -\quad { p }^{ 2 }\quad =\quad 64\ { p }^{ 2 }\quad =\quad 36\ p\quad =\quad \pm 6 But p > 0, so p = 6. Now we simplify the integral.

\int _{ -2 }^{ \alpha }{ { -x }^{ 2 } } +\quad x\quad +\quad 6dx\quad =\quad \frac { 125 }{ 6 } \ \begin{matrix} \alpha \ -2 \end{matrix}\left[ \frac { { -x }^{ 3 } }{ 3 } \quad +\quad \frac { { x }^{ 2 } }{ 2 } \quad +\quad 6x \right] \quad =\quad \frac { 125 }{ 6 } \ \left[ \frac { { -\alpha }^{ 3 } }{ 3 } \quad +\quad \frac { { \alpha }^{ 2 } }{ 2 } \quad +\quad 6\alpha \right] -\left[ \frac { 8 }{ 3 } +\quad 2\quad -\quad 12\quad \right] \quad =\quad \frac { 125 }{ 6 } \ \frac { { -2\alpha }^{ 3 }\quad +\quad { 3\alpha }^{ 2 }\quad +\quad 36\alpha }{ 6 } \quad =\quad \frac { 125 }{ 6 } -\quad \frac { 22 }{ 3 } \ \frac { { -2\alpha }^{ 3 }\quad +\quad { 3\alpha }^{ 2 }\quad +\quad 36\alpha }{ 6 } \quad =\quad \frac { 81 }{ 6 } \ { -2\alpha }^{ 3 }\quad +\quad { 3\alpha }^{ 2 }\quad +\quad 36\alpha \quad -\quad 81\quad =\quad 0\

After solving the cubic equation, we get α = 3 or 4.5. As mentioned in the question, y has to be found only for integral values of α, so we substitute α = 3, to get the value of y as 6. Then we find f(z).

f(z)\quad =\quad { z }^{ 3 }\quad -\quad 6{ z }^{ 2 }\quad +\quad 18z\quad +\quad 25\quad =\quad 0\ By\quad factor\quad theorem,\quad (z\quad +\quad 1)\quad is\quad a\quad factor\quad of\quad the\quad polynomial.\ Dividing\quad by\quad (z\quad +\quad 1),\quad we\quad get:\quad -\quad \ (z\quad +\quad 1)({ z }^{ 2 }\quad -\quad 7z\quad +\quad 25)\quad =\quad 0\ After\quad solving\quad for\quad z,\quad we\quad get\quad \beta \quad =\quad -1\quad as\quad the\quad only\quad integral\quad root.\quad \ Therefore,\quad the\quad answer\quad is\quad -(\beta )\quad =\quad \boxed { 1 } \quad \