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 $

Question

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?

Answer 1

Hint:

If $p(x)\gt 0$ for all $x$ and $$\int_m^np(x)f(x)dx=0$$ what can you say about $f(x)$ in the interval $[m,n]$?

Answer 2

$$\int_0^1(1+\cos^8x)(ax^2+bx+c)dx=\int_0^2(1+\cos^8x)(ax^2+bx+c)dx$$ $$\Leftrightarrow$$ $$\int_0^1(1+\cos^8x)(ax^2+bx+c)dx - \bigg(\int_0^1 (1+\cos^8x)(ax^2+bx+c)dx $$ $$+ \int_1^2 (1+\cos^8x)(ax^2+bx+c)dx \bigg)= 0$$ $$=$$ $$\int_1^2(1+\cos^8x)(ax^2+bx+c)dx=0$$

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 ?