Solution
= Solution
For this <polynomial>, $f'(x)=2x$, so the specified <Newton root-finding iteration> simplifies to
$$
x_{n+1}=\frac12\left(x_n+\frac2{x_n}\right).
$$
Starting at one gives $x_1=(1+2)/2=3/2$, followed by $x_2=(3/2+4/3)/2=17/12$. Consequently
$$
\boxed{x_1=1.5,\qquad x_2\approx1.42\text{ to two decimal places}.}
$$