root of $b,e,a,r$ , poor bear!

As $b,e,a,r$ are root of $P(x)=x^4-2x^3-7x^2-2x+1=0$ we can write :

$P(x)=(x-b)(x-e)(x-a)(x-r)$ $P(x)=x^4-(b+e+a+r)x^3+(be+ba+br+ea+er+ar)x^2-(bea+ber+ear+bar)x+bear$

$b+e+a+r=2$

$be+ba+br+ea+er+ar=-7$

$bea+ber+ear+bar=2$

$bear=1$

$\frac{1}{b}+\frac{1}{e}+\frac{1}{a}+\frac{1}{r}=\frac{bea+ber+ear+bar}{bear}=\frac{2}{1}$

$\boxed{\frac{1}{b}+\frac{1}{e}+\frac{1}{a}+\frac{1}{r}=2}$

Put $x=\frac{1}{y}$ in the given equation we get, $y^4-2y^3-7y^2-2y+1=0$ whose roots are $\frac{1}{b}, \frac{1}{e}, \frac{1}{a}, \frac{1}{r}$ .

So, $\frac{1}{b}+\frac{1}{e}+\frac{1}{a}+\frac{1}{r}=\boxed{2}$ .