Finding roots using $\displaystyle\int_0^1(1+\cos^8x)(ax^2+bx+c)dx=\int_0^2(1+\cos^8x)(ax^2+bx+c)dx $ I have the following relation-

$$\displaystyle\int_0^1(1+\cos^8x)(ax^2+bx+c)dx=\int_0^2(1+\cos^8x)(ax^2+bx+c)dx $$ 
  $$\text{then what is the interval in which the root of the equation}\space ax^2+bx+c=0\space\text{lie?}$$

Where $a,b,c$ are non-zero numbers.
I think one should apply mean value theorem here. But I can't proceed. What to do?

$$\int_0^1(1+\cos^8x)(ax^2+bx+c)dx=\int_0^2(1+\cos^8x)(ax^2+bx+c)dx$$
$$\Leftrightarrow$$
$$\int_0^1(1+\cos^8x)(ax^2+bx+c)dx - \bigg(\int_0^1 (1+\cos^8x)(ax^2+bx+c)dx $$
$$+ \int_1^2 (1+\cos^8x)(ax^2+bx+c)dx \bigg)= 0$$
$$=$$
$$\int_1^2(1+\cos^8x)(ax^2+bx+c)dx=0$$
Observe that :
$$1+\cos^8(x) >0 \; \forall  \; x\in \mathbb R \; \text{since} \; -1 \leq \cos x \leq1 $$
But, since $1+\cos^8(x) > 0$ and the integral equals to zero, what can you conclude about $ax^2 + bx + c$ in the interval $[1,2]$ which the integral is defined to be zero ?
