10.1.1. (12 points) The numbers $a$ and $b$ are such that the polynomial $x^{4}+x^{3}+2 x^{2}+a x+b$ is the square of some other polynomial. Find $b$.
Answer: $\frac{49}{64} \approx 0.77$.

Solution. Let $\left(x^{2}+A x+B\right)^{2}=x^{4}+x^{3}+2 x^{2}+a x+b$. Expanding the brackets, we get

$$
x^{4}+2 A x^{3}+\left(A^{2}+2 B\right) x^{2}+2 A B x+B^{2}=x^{4}+x^{3}+2 x^{2}+a x+b,
$$

therefore, by equating the coefficients of the powers, we find $2 A=1, A^{2}+2 B=2,2 A B=a$, $B^{2}=b$. Hence, $A=\frac{1}{2}, B=1-\frac{A^{2}}{2}=\frac{7}{8}, b=B^{2}=\frac{49}{64} \approx 0.77$.