Answer
$\approx 4.583$
Work Step by Step
We start with an approximation:
$a_0=4$
Compute $a_1,a_2,a_3,a_4,a_5$ using the formula:
$a_n=\dfrac{1}{2}\left(a_{n-1}+\dfrac{p}{a_{n-1}}\right)$, where $p=21$.
$a_1=\dfrac{1}{2}\left(a_0+\dfrac{5}{a_0}\right)=\dfrac{1}{2}\left(4+\dfrac{21}{4}\right)=4.625$
$a_2=\dfrac{1}{2}\left(a_1+\dfrac{21}{a_1}\right)=\dfrac{1}{2}\left(4.625+\dfrac{21}{4.625}\right)\approx 4.5827703$
$a_3=\dfrac{1}{2}\left(a_2+\dfrac{21}{a_2}\right)=\dfrac{1}{2}\left(4.5827703+\dfrac{21}{4.5827703}\right)\approx 4.5825757$
$a_4=\dfrac{1}{2}\left(a_3+\dfrac{21}{a_3}\right)=\dfrac{1}{2}\left(4.5825757+\dfrac{21}{4.5825757}\right)\approx 4.5825757$
$a_5=\dfrac{1}{2}\left(a_4+\dfrac{21}{a_4}\right)=\dfrac{1}{2}\left(4.5825757+\dfrac{21}{4.5825757}\right)\approx 4.5825757$
So $\sqrt {21}\approx 4.583$
