Let $a,b,c$  be non zero real numbers such that ${\displaystyle \underset{0}{\overset{1}{\int }}\left(1+{\cos }^{8}x\right)\left(a{x}^2+bx+c\right)dx={\displaystyle \underset{0}{\overset{2}{\int }}\left(1+{\cos }^{8}x\right)\left(a{x}^2+bx+c\right)dx}}$  then the quadratic equation $a{x}^2+bx+c=0$  has

$\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\underset{0}{\overset{2}{\int }}f\left(x\right)dx =\underset{0}{\overset{1}{\int }}f\left(x\right)dx+\underset{1}{\overset{2}{\int }}f\left(x\right)dx\Rightarrow \underset{1}{\overset{2}{\int }}f\left(x\right)dx=0$ 

$\therefore {\displaystyle \underset{1}{\overset{2}{\int }}\left(1+{\cos }^{8}x\right)\left(a{x}^2+bx+c\right)dx=0}$

If $a{x}^2+bx+c=0$  has no root in (1,2) then the integrand is either positive or negative and the integral cannot vanish.

Hence $a{x}^2+bx+c=0$  has atleast one root in (1, 2)