It posted before. \[x^4-x^3-1=(x^2-Sx+P)(x^2-x+Sx-\frac 1P)\to P^2=\frac{S}{1-S}, \ (1-2S)^2=S^3(1-S)^3.\] Let $z=2S-1>0$, then $64z^2=(1-z^2)^3$. Therefore from $z>z_1$ we get \[z<\frac{(1-z_1^2)^{3/2}}{8}, P<-\sqrt{\frac{1+z_1}{1-z_1}}\] and from $z<z_2$ we get \[z>\frac{(1-z_2^2)^{3/2}}{8}, P>-\sqrt{\frac{1+z_2}{1-z_2}}.\] If take $z_1=0$ we get \[z_2=\frac 18, S>\frac{2^{12}+63^{3/2}}{2^{13}}>\frac{1151}{2048},P<-\sqrt{\frac{3199}{897}}.\]

Another simple method : Note that : $f(x)=x^4-x^3-1$ We have : $f'(x)=x^2(4x-3) $ Now we can easily prove that the equation has only 2 distincts solution and : $\left\{\begin{matrix} 1.38<x_1<1.39\\ -0.82<x_2<-0.81 \end{matrix}\right.$ The rest is just some easy calculatation ..

p=-1.130685445.... s=0.5611050557....

(a) For $p(x)=x^4-x^3-1$ it is easily checked that $p(-1)>0$, $p(0)<0$, $p(2)>0$ and that the function is strictly decreasing for $x<\frac 3 4$ (we don't care about $x=0$) and increasing for $x>\frac 3 4$, so the Intermediate Value theorem ensures us that $p(x)$ has exactly two real roots. (b) Simple calculations show that $p(-0.82)>0$ and $p(-0.81)<0$ so $-0.82<x_1<-0.81$, and similarly $p(1.38)<0$ and $p(1.39)>0$ so $1.38<x_2<1.39$; it follows that $S=x_1+x_2 >1.38-0.82=0.56>\frac 6 {11}$, $P=x_1 x_2<(-0.81) \cdot 1.38=-1.1178<-\frac {11} {10}$. Of course, this requires knowing the roots in advance, but having to do that by hand one can use rougher bounds than two decimal digits for the roots to shorten calculations.

Sorry for bumping, but I just want to note that the easiest way to solve part $(a)$ is to notice that $$x^4-x^3-1=0 \implies x^3(x-1)=1 \implies x^3=\frac{1}{x-1}$$. Just graph $y=x^3$ and $y=\frac{1}{x-1}$ and notice there are $2$ intersection points.