Complicated Quadratic Roots

The equation implies $x^n=6x^{n-1}+2x^{n-2}$ from where we get $\begin{aligned} a_n=\alpha^n-\beta^n &=&\left(6\alpha^{n-1}+2\alpha^{n-2}\right)-\left(6\beta^{n-1}+2\beta^{n-2}\right) \\ &=& 6\left(\alpha^{n-1}-\beta^{n-1}\right)+2\left(\alpha^{n-2}-\beta^{n-2}\right) \\ &=& 6a_{n-1}+2a_{n-2} \end{aligned}$ and hence $a_{10}-2a_8=\left(6a_9+2a_8\right)-2a_8=6a_9$ . Thus $X=6a_9/2a_9=3$ . So our quadratic is $x^2-\left(m-3\right)x+m=0$ . Now if both the roots lie in $(1,2)$ then their sum $m-3$ lies in $(2,4)$ . Hence $m$ lies in $(5,7)$ . Since $m\in\mathbb{Z}$ we must have $m=6$ which leads to non-real roots. So there's no such integral value of $m$ . $\square$

$\alpha =3+\sqrt { 11 } \\ \beta =3-\sqrt { 11 } \\ using\quad values\quad to\quad determine\quad X\quad :\\ \frac { { (3+\sqrt { 11 } ) }^{ 10 }-{ (3-\sqrt { 11 } ) }^{ 10 }-2{ (3+\sqrt { 11 } ) }^{ 8 }+2{ (3-\sqrt { 11 } ) }^{ 8 } }{ 2[{ (3+\sqrt { 11 } ) }^{ 9 }-{ (3-\sqrt { 11 } ) }^{ 9 }] } \\ after\quad manipulation\quad :\\ \frac { 6[{ (3+\sqrt { 11 } ) }^{ 9 }-({ 3-\sqrt { 11 } ) }^{ 9 }] }{ 2[({ 3+\sqrt { 11 } ) }^{ 9 }-{ (3-\sqrt { 11 } ) }^{ 9 }] } =\quad 3\quad =\quad X\\ equation:\quad \\ { x }^{ 2 }-(m-3)x+m=0\\ by\quad applying\quad two\quad conditions:\\ 1)\quad D\ge 0\\ 2)\quad 1<\frac { -b }{ 2a } <2\\ (1)\cap (2)\in \phi \\ \\$

${ x }^{ 2 }-6x-2=0\quad \Rightarrow { x }^{ 10 }-6{ x }^{ 9 }-2{ x }^{ 8 }=0\\ \\ \Rightarrow { \alpha }^{ 10 }-6{ { \alpha }^{ 9 } }-2{ \alpha }^{ 8 }={ \beta }^{ 10 }-6{ \beta }^{ 9 }-2{ \beta }^{ 8 }=0\\ \\ \Rightarrow ({ \alpha }^{ 10 }-{ \beta }^{ 10 })-6({ \alpha }^{ 9 }-{ \beta }^{ 9 })-2({ \alpha }^{ 8 }-{ \beta }^{ 8 })=0\\ \\ \Rightarrow X=3\quad now,\\ \\ for\quad root\quad of\quad equation\quad f\left( x \right) ={ x }^{ 2 }-(m-3)x-m=0\\ \\ to\quad lie\quad in\quad (1,2)\quad following\quad 3\quad condition\\ \\ should\quad be\quad fulfilled.\\ \\ (i)\quad f\left( 1 \right) f\left( 2 \right) >0\quad \Rightarrow m<10\\ \\ (ii){ b }^{ 2 }-4ac\ge 0\quad \Rightarrow m\notin \left( 1,9 \right) \\ \\ (iii)1<\cfrac { m-3 }{ 2 } <2\quad \Rightarrow 5<m<7\\ \\ so,it\quad is\quad clear\quad that\quad there\quad is\quad no\\ \\ possible\quad value\quad of\quad m.$

Exactly, Without calculating alpha and beta also the value of expression can be found out.